Rotational and circular motion

Class 11 Physics, unit 4, book pages 53–76

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This unit describes turning and circular motion. It defines angular position, displacement, velocity and acceleration, links them to the linear quantities through $S = r θ$ , $v = r ω$ and $a_{T} = r α$ , and shows how the rotational equations mirror the linear ones. It then explains centripetal acceleration and force with examples such as banked roads, strings, orbits and centrifuges, introduces moment of inertia, angular momentum and its conservation, and relates torque to angular acceleration. It ends with weightlessness in satellites and artificial gravity.

Learning outcomes (12)

Express angles in radians.
Define and calculate angular displacement, angular velocity and angular acceleration [This involves use of $S = r θ$ , $v = r ω$ , $ω = 2 π / T$ , $a = r ω^{2}$ and $a = v^{2} / r$ to solve problems].
Use equations of angular motion to solve problems involving rotational motions.
Analyse qualitatively motion in a curved path due to a perpendicular force.
Define and calculate centripetal force [Use $F = m r ω^{2}$ , $F = m v^{2} / r$ ].
Analyse situations involving circular motion in terms of centripetal force [e.g. situations in which centripetal acceleration is caused by a tension force, a frictional force, a gravitational force, or a normal force].
Define and calculate moment of inertia of a body and angular momentum.
State and apply the law of conservation of angular momentum. Illustrate the applications of conservation of angular momentum in real life. [Such as by flywheels to store rotational energy, by gyroscopes in navigation systems, by ice skaters to adjust their angular velocity].
Justify how a centrifuge is used to separate materials using centripetal force.
Derive and apply the relation between torque, moment of inertia and angular acceleration.
Explain why the objects in orbiting satellites appear to be weightless.
Describe how artificial gravity is created to counter weightlessness.

Rotational motion

Rotational motion is the turning or spinning motion of an object about an axis that passes through it. We consider rigid objects: these are non-deformable, so the particles forming the object stay at fixed positions relative to one another as the object rotates.

To describe the rotation we use an axis of rotation. It is a line about which rotation takes place. This line remains fixed during the rotational motion, while the other points of the body move in circles about it. The axis may be a pivot, a hinge or any other support.

Figure 4.1: axis of rotation. Every point of the rigid object (P1, P2, P3) moves in a circle, shown dashed, whose centre lies on the axis.

Every point of a rotating rigid object moves in a circle (shown dashed in Figure 4.1 for points $P_{1}$ , $P_{2}$ and $P_{3}$ ) whose centre is on the axis of rotation. A straight line drawn from the axis to any point of the object sweeps out the same angle in the same time interval.

4.1.1Angular position (θ)

Rotational kinematics deals with the motion of objects along a circular path without any reference to forces or torques.

Angular position: The angle through which the position vector of a moving object is displaced with respect to some chosen reference direction is called angular position $θ$ .

Let an object A be rotated through an arc length $S$ from a certain reference axis, along a circle of radius $r$ , as shown in Figure 4.2. The angular position of the object is the angle $θ$ between its radial line (the position vector $r$ ) and the fixed reference line in space, often chosen as the $+ x$ axis.

Figure 4.2: angular position. The angle

θ

between the position vector

r

of the object A and the fixed reference line (the

+ x

axis).

θ = \frac{S}{r}

(4.1)

This resembles the way we identify the position of an object in translational motion as the distance $x$ between the object and the reference position, which is the origin ( $x = 0$ ). The angle plays the part of $x$ in rotation.

4.1.2Angular displacement (Δθ)

Angular displacement: The change in angular position with respect to a chosen reference direction is termed angular displacement $Δ θ$ .

As a particle on a rigid object travels from position A to position B in a time interval $t$ , as in Figure 4.3, the reference line fixed to the object sweeps out an angle:

Figure 4.3: angular displacement. As the object moves from A to B the reference line fixed to it sweeps out the angle

Δ θ = θ_{2} - θ_{1}

Δ θ = θ_{2} - θ_{1}

(4.2)

Conventionally, positive angular displacements represent anti-clockwise motion and negative angular displacements represent clockwise motion.

Units of angular displacement

The SI unit of angular displacement is the radian (rad). Other units are degrees and revolutions.

Relation between radian and degree

In one complete rotation there are $36 0^{\circ}$ .

number of degrees in one revolution = 36 0^{\circ}

To find the number of radians in one revolution, put $S$ equal to the circumference of the circle, $2 π r$ , in Eq. (4.1):

θ = \frac{2 π r}{r} = 2 π rad so number of radians in one revolution = 2 π rad

For one complete revolution the number of radians must equal the number of degrees, therefore:

2 π rad = 36 0^{\circ} or 1 rad = \frac{36 0 ^{\circ}}{2 π} = \frac{36 0 ^{\circ}}{2 \times 3.14} = 57. 3^{\circ}

Figure 4.4: radian measurement. One radian is the angle subtended at the centre of a circle by an arc whose length is equal to the radius of the circle.

Radian: One radian (1 rad) is the angle subtended at the centre of a circle by an arc whose length is equal to the radius of the circle.

Direction of angular displacement

Angular displacement is a vector quantity: it has both magnitude and direction. The right-hand rule is used to specify its direction.

Right-hand rule: Holding the axis of rotation in the right hand with the fingers curling in the direction of rotation, the thumb gives the direction of the angular displacement.

Right-hand rule: with the fingers curling in the direction of rotation (anti-clockwise as seen from above), the thumb points along the axis, the direction of the angular displacement.

4.1.3Angular velocity (ω)Often examined

Angular velocity: The time rate of change of angular displacement of a body is called angular velocity $ω$ .

If $Δ θ$ is the small angular displacement in time $Δ t$ , then the angular velocity $ω$ is:

ω = \frac{Δ θ}{Δ t}

(4.3)

Units of angular velocity. The SI unit of angular velocity is radian per second ( $rad s^{- 1}$ ). Other units are $deg s^{- 1}$ , $rev s^{- 1}$ and $rev mi n^{- 1}$ (rpm). The direction of angular velocity is the same as that of the angular displacement (right-hand rule).

Figure 4.5: angular velocity.

ω

points in the positive

z

direction for anti-clockwise rotation (as seen from above) and in the negative

z

direction for clockwise rotation.

Average angular velocity ( $ω_{a v}$ )

The total angular displacement $θ$ of a body during time $t$ , divided by that time, is called average angular velocity.

ω_{a v} = \frac{θ}{t}

(4.4)

Instantaneous angular velocity ( $ω_{in s t}$ )

The limiting value of the ratio between a small angular displacement $Δ θ$ and a small time interval $Δ t$ , as the time interval approaches zero, is called instantaneous angular velocity. (The book heads this 'instantaneous average velocity'.)

ω_{in s t} = Δ t \to 0 lim \frac{Δ θ}{Δ t}

(4.5)

4.1.4Angular acceleration (α)Often examined

Angular acceleration: The time rate of change of angular velocity is called angular acceleration $α$ .

If $Δ ω$ is the change of angular velocity in time $Δ t$ , then the angular acceleration $α$ is:

α = \frac{Δ ω}{Δ t}

(4.6)

Units of angular acceleration. The SI unit is $rad s^{- 2}$ . Other units are $deg s^{- 2}$ and $rev s^{- 2}$ . The direction of angular acceleration is determined by the right-hand rule.

It is taken as positive when the angular velocity of a body increases. In this case angular velocity and angular acceleration have the same direction.
It is taken as negative when the angular velocity of a body decreases. In this case angular velocity and angular acceleration are anti-parallel.

Figure 4.6: angular acceleration. When the rotation speeds up,

α

and

ω

point the same way; when it slows down, they point opposite ways.

Average angular acceleration ( $α_{a v}$ )

The change of angular velocity of a body during time $t$ , divided by $t$ , is called average angular acceleration. The book writes the numerator as the 'total angular velocity' $ω$ , which means the change in $ω$ over the time $t$ (for a body starting from rest it is the final $ω$ ).

α_{a v g} = \frac{ω}{t}

(4.7)

Instantaneous angular acceleration ( $α_{in s t}$ )

The limiting value of the ratio between a small change in angular velocity and a small time interval, as the time interval approaches zero, is called instantaneous angular acceleration.

α_{in s t} = Δ t \to 0 lim \frac{Δ ω}{Δ t}

(4.8)

4.1.5Relationship between linear and angular kinematic quantitiesOften examined

Linear kinematic quantities such as displacement, velocity and acceleration can be related to their rotational analogues.

Derivation Relation between linear and angular displacement, $S = r θ$

Aim. Connect the arc length $S$ travelled by a particle on a circle of radius $r$ with the angle $θ$ it turns through.

Figure 4.7: linear and angular displacement. The arc AC equals the radius

r

and subtends 1 radian; the arc AB is the distance

S

and subtends

θ

. (The book's caption reads 'linear and angular acceleration'; the figure shows displacement.)

A particle moves in a circle of radius $r$ with centre O from point A to point B. Choose another point C so that $∠ A O C = 1$ radian. By the definition of the radian, arc AC is equal to $r$ .
Arcs on the same circle are proportional to the angles they subtend at the centre.
$\frac{Arc AB}{Arc AC} = \frac{∠ A O B}{∠ A O C}$
Arc AB is the linear displacement $S$ and $∠ A O B$ is the angular displacement $θ$ . Put arc AC $= r$ and $∠ A O C = 1$ rad.
$\frac{S}{r} = \frac{θ}{1 rad} or θ = \frac{S}{r}$
For $θ$ in radians this is written as

Result

S = r θ (4.9)

The distance moved along the circle is the radius times the angle in radians. The relation is valid only when $θ$ is in radians.

Derivation Relation between linear and angular velocity, $v = r ω$

Aim. Connect the speed of the particle along the circle with its angular velocity.

Take the change per unit time of both sides of $S = r θ$ : multiply by $Δ/Δ t$ and let $Δ t \to 0$ .
$Δ t \to 0 lim \frac{Δ}{Δ t} S = Δ t \to 0 lim \frac{Δ}{Δ t} (r θ)$
The radius does not change with time, so $r$ comes out of the limit.
$Δ t \to 0 lim \frac{Δ S}{Δ t} = r \times Δ t \to 0 lim \frac{Δ θ}{Δ t} (4.10)$
By the definitions of linear and angular velocity, the two limits are
$v = Δ t \to 0 lim \frac{Δ S}{Δ t} (4.11) ω = Δ t \to 0 lim \frac{Δ θ}{Δ t} (4.12)$
Substitute Eqs. (4.11) and (4.12) into Eq. (4.10).

Result

v = r ω (4.13)

As $Δ t \to 0$ the points A and B move closer together and the direction of the linear velocity becomes the tangent to the circle. So this velocity is also called the tangential velocity. Points farther from the axis (larger $r$ ) move faster, though all have the same $ω$ .

Derivation Relation between linear and angular acceleration, $a_{T} = r α$

Aim. Connect the tangential acceleration of the particle with its angular acceleration.

Figure 4.8: acceleration components. The total acceleration

a

has a tangential part

a_{T}

(along the velocity) and a radial part

a_{R}

(towards the centre).

In angular motion the linear acceleration has two components, a tangential component and a radial component (Figure 4.8). In vector form and in magnitude:
$a = a_{T} + a_{R} a = a_{T}^{2} + a_{R}^{2}$
Apply $\frac{Δ}{Δ t}$ and the limit $Δ t \to 0$ to both sides of $v = r ω$ . Since $r$ is constant it comes out of the limit.
$Δ t \to 0 lim \frac{Δ v}{Δ t} = r \times Δ t \to 0 lim \frac{Δ ω}{Δ t} (4.14)$
By the definitions of the tangential and the angular acceleration:
$a = Δ t \to 0 lim \frac{Δ v}{Δ t} = a_{T} (4.15) α = Δ t \to 0 lim \frac{Δ ω}{Δ t} (4.16)$
Put Eqs. (4.15) and (4.16) in Eq. (4.14).

Result

a_{T} = r α (4.17)

This allows all the kinematic equations to be written in rotational form (Table 4.1).

Tangential and radial components of acceleration

Tangential component. The component of the acceleration that is parallel to the linear instantaneous velocity is the tangential component. Tangential acceleration occurs due to a change in the magnitude of the linear velocity.

Radial component. The component of the acceleration along the radius of the circular path is the radial component. It arises due to the change in the direction of the linear instantaneous velocity. For an object moving in a circular path with constant speed there is only the radial acceleration, also called the centripetal acceleration.

Kinematic equations of rotational motion

Kinematics for rotational motion is similar to translational kinematics. Replacing $S \to θ$ , $v \to ω$ and $a \to α$ in the linear equations gives the angular equations.

Table 4.1: kinematic equations for rotational motion
Equations for linear motion	Equations for angular motion
$S = v t$	$θ = ω t$
$v_{f} = v_{i} + a t$	$ω_{f} = ω_{i} + α t$
$2 a S = v_{f}^{2} - v_{i}^{2}$	$2 α θ = ω_{f}^{2} - ω_{i}^{2}$
$S = v_{i} t + \frac{1}{2} a t^{2}$	$θ = ω_{i} t + \frac{1}{2} α t^{2}$

Example 4.1: a rolling bicycle tyre

In a workshop, a bicycle tyre of radius

33.1 cm

is rolled across the level floor with an initial velocity of

6.80 m s^{- 1}

. Assuming constant angular acceleration, the tyre comes to rest at a distance of

74.8 m

. Determine (a) the initial angular velocity of the tyre; (b) the total number of revolutions it made before coming to rest; (c) the angular acceleration of the tyre; and (d) the time it took before coming to rest.

Given

$v_{i} = 6.80 m s^{- 1}$ initial velocity
$r = 33.1 cm = 0.331 m$ radius
$ω_{f} = 0 rad s^{- 1}$ comes to rest
$S = 74.8 m$ distance

To find

(a) $ω_{i}$
(b) number of revolutions $N$
(c) angular acceleration $α$
(d) time $t$

(a) The linear and angular velocity are related by $v = r ω$ , so $ω = v / r$ .
$ω_{i} = \frac{6.80}{0.331} = 20.54 rad s^{- 1}$
(b) With no slipping, in one revolution the tyre moves a distance equal to its circumference $2 π r$ . The number of revolutions is the total distance divided by $2 π r$ .
$N = \frac{S}{2 π r} = \frac{74.8}{2 \times 3.14 \times 0.331} = 35.9 rev$
(c) Each revolution is $2 π$ radians, so the total angular displacement is $θ = 35.9 \times 2 π = 225.6$ rad. Use the equation that has no time (the third one of Table 4.1).
$2 α θ = ω_{f}^{2} - ω_{i}^{2} \Rightarrow α = \frac{ω _{f}^{2} - ω _{i}^{2}}{2 θ} = \frac{0 - ( 20.54 ) ^{2}}{2 \times 225.6} = - 0.94 rad s^{- 2}$
(d) Any equation with time will do; the simplest is $ω_{f} = ω_{i} + α t$ . Rearrange it for $t$ .
$t = \frac{ω _{f} - ω _{i}}{α} = \frac{0 - 20.54}{- 0.94} = 21.9 s$

So the tyre takes about $22$ seconds to come to rest. The negative sign of $α$ shows that it slows down (anti-parallel to $ω$ ).

ω_{i} = 20.54 rad s^{- 1}, N = 35.9 rev, α = - 0.94 rad s^{- 2}, t = 21.9 s

Assignment 4.1: tractor wheels

The front wheel of a tractor travels

700

revolutions while the rear wheel travels

280

in a time interval of

40

seconds. Find their angular velocities.

Given

$N_{f} = 700 rev$
$N_{r} = 280 rev$
$t = 40 s$

To find

$ω_{f}$ (front)
$ω_{r}$ (rear)

Each revolution is $2 π$ rad, so the angular displacement is $θ = 2 π N$ . Use $ω = θ / t$ (Eq. 4.4).
$ω_{f} = \frac{2 π \times 700}{40} = 110 rad s^{- 1}$
Same for the rear wheel.
$ω_{r} = \frac{2 π \times 280}{40} = 44 rad s^{- 1}$

In revolutions per second these are $17.5 rev s^{- 1}$ and $7 rev s^{- 1}$ (the front wheel is $2.5$ times faster).

ω_{f} = 110 rad s^{- 1}, ω_{r} = 44 rad s^{- 1}

Worked example: total acceleration of a point on a wheel

A wheel of radius

0.50 m

is speeding up. At one instant its angular velocity is

2.0 rad s^{- 1}

and its angular acceleration is

3.0 rad s^{- 2}

. For a point P on the rim find the speed, the tangential and radial accelerations, and the total acceleration. What would they be for a point

0.25 m

from the axis?

Given

$r = 0.50 m$
$ω = 2.0 rad s^{- 1}$
$α = 3.0 rad s^{- 2}$

To find

$v$ , $a_{T}$ , $a_{R}$ , $a$

Tangential speed from the book's relation $v = r ω$ .
$v = r ω = 0.50 (2.0) = 1.0 m s^{- 1}$
The tangential acceleration comes from the change in speed, so it uses $α$ .
$a_{T} = r α = 0.50 (3.0) = 1.5 m s^{- 2}$
The radial (centripetal) acceleration comes from the change in direction, so it uses $ω$ and acts towards the axis.
$a_{R} = r ω^{2} = 0.50 (2.0)^{2} = 2.0 m s^{- 2}$
The two components are at right angles, so combine them with Pythagoras.
$a = a_{T}^{2} + a_{R}^{2} = 1. 5^{2} + 2. 0^{2} = 6.25 = 2.5 m s^{- 2}$
It points between the inward radius and the direction of motion, at $tan^{- 1} (1.5/2.0) = 36. 9^{\circ}$ from the radius.
For $r = 0.25 m$ the same $ω$ and $α$ apply to every point of a rigid body, only $r$ halves, so every acceleration halves.
$a_{T} = 0.75, a_{R} = 1.0, a = 1.25 m s^{- 2}$

Accelerations of the point P on the rim, drawn to scale. The tangential part

a_{T}

and the radial part

a_{R}

add at right angles to give the total acceleration

a

v = 1.0 m s^{- 1}, a_{T} = 1.5, a_{R} = 2.0, a = 2.5 m s^{- 2}

4.2Centripetal acceleration and centripetal forceOften examined

Consider a particle moving in a circular path of radius r with constant speed. The direction of its velocity is changing. This change of velocity produces an acceleration which is directed towards the centre of the circle. This type of acceleration is called centripetal acceleration.

Derivation Centripetal acceleration a_c=\dfrac{v^{2}}{r}

Aim. Find the acceleration of a particle moving with constant speed v on a circle of radius r.

Assumptions.

The speed v is constant (uniform circular motion).
A and B are so close that \Delta t\to0.

Figure 4.9: centripetal acceleration. (a) From A to B the velocity changes from \vec v_i to \vec v_f while the position vector turns through \Delta\theta. (b) The velocity vectors redrawn tail to tail: \Delta\vec v=\vec v_f-\vec v_i is the change of velocity.

In Figure 4.9(a) the particle is at A at time t_i with velocity \vec v_i and at B at a later time t_f with velocity \vec v_f. For uniform circular motion \vec v_i and \vec v_f differ only in direction, so their magnitudes are equal.
|\vec v_i|=|\vec v_f|=|\vec v|\qquad |\vec r_i|=|\vec r_f|=|\vec r|
Redraw the velocity vectors tail to tail (Figure 4.9(b)). The vector \Delta\vec v joining their heads is the vector change of velocity.
\vec v_f=\vec v_i+\Delta\vec v
The velocity is perpendicular to the position vector at each point. So the angle \Delta\theta between \vec v_i and \vec v_f equals the angle between \vec r_i and \vec r_f. The two triangles (position triangle with side \Delta r, velocity triangle with side \Delta v) are similar, so their sides are in proportion.
\dfrac{\Delta v}{v}=\dfrac{\Delta r}{r}\qquad\text{or}\qquad\Delta v=v\,\dfrac{\Delta r}{r}
Divide both sides by \Delta t to get the average acceleration.
a_{av}=\dfrac{\Delta v}{\Delta t}=\dfrac{v}{r}\,\dfrac{\Delta r}{\Delta t}
Now let A and B come extremely close together, so \Delta t\to0 and the acceleration becomes instantaneous.
a=\dfrac{v}{r}\lim_{\Delta t\to0}\dfrac{\Delta r}{\Delta t}
As \Delta t\to0 the chord \Delta r becomes the arc, so \lim\dfrac{\Delta r}{\Delta t}=v.
a=\dfrac{v}{r}\times v=\dfrac{v^{2}}{r}
This is the centripetal acceleration a_c. It points towards the centre while the radial unit vector \hat r points outwards, so in vector form a negative sign is included. Using v=r\omega we get the second form.

Resulta_c=-\dfrac{v^{2}}{r}\,\hat r\ (4.18)\qquad a_c=-r\omega^{2}\,\hat r\ (4.19)

Magnitude a_c=v^{2}/r=r\omega^{2}, directed towards the centre of the circle at every instant. Because its direction keeps changing, it is not a constant vector even when the speed is constant.

According to Newton's second law, an object that is accelerating must have a net force acting on it. For example, to open a door, a force must be applied to produce tangential acceleration, thereby creating torque. Similarly, for an object to move in a circle, a force must be applied to it to keep it moving in that circle, thereby giving it the necessary radial (centripetal) acceleration.

4.2.1Centripetal forceOften examined

Centripetal force: The net force that causes a particle to undergo centripetal acceleration is called centripetal force F_c.

When Newton's second law is applied to a particle moving in uniform circular motion we can write:

F_c=m\,a_c

(4.20)

Putting Eq. (4.18) or Eq. (4.19) in Eq. (4.20) gives the centripetal force:

F_c=-\dfrac{mv^{2}}{r}\,\hat r\qquad\text{and}\qquad F_c=-mr\omega^{2}\,\hat r

(4.21)

The direction of the centripetal force is always towards the centre of the circle. Its magnitude is mv^{2}/r=mr\omega^{2}.

To swing a ball in a circle at the end of a string, the tension in the string acts as the centripetal force.
For a moon revolving around the Earth, or planets around the Sun, gravity acts as the centripetal force.
In other situations it can be a normal force, or even an electric force (as in CD players and computer hard disks).

Frictional force and normal force as centripetal force

When a car travels without skidding around an un-banked curve, the static frictional force between the tyres and the road provides the necessary centripetal force. The reliance on friction can be eliminated completely for a given speed if the roads are banked at an angle relative to the horizontal while making a turn (Figure 4.10).

Because the roadbed makes an angle with the horizontal, the normal force has a component F_N\sin\theta that points toward the centre C of the circle and provides the centripetal force.

Derivation Banked curve: \tan\theta=\dfrac{v^{2}}{rg}

Aim. Find the banking angle \theta at which a car can turn a curve of radius r at speed v without relying on friction.

Assumptions.

The road is frictionless.
The car moves with constant speed v in a horizontal circle of radius r.

Figure 4.10(b): a car on a friction-free banked curve. The normal force F_N is perpendicular to the road; its horizontal component F_N\sin\theta points towards the centre C and is the centripetal force, and F_N\cos\theta balances the weight mg.

Figure 4.10(a) shows a car going around a friction-free banked curve. The radius r of the curve is measured parallel to the horizontal. The only forces on the car are its weight mg and the normal force F_N of the road, which is perpendicular to the road surface.
The roadbed makes an angle \theta with the horizontal, so F_N has a horizontal component F_N\sin\theta pointing to the centre. It provides the centripetal force.
F_c=F_N\sin\theta=\dfrac{mv^{2}}{r}\qquad(4.22)
The car does not accelerate vertically. The vertical component F_N\cos\theta only balances the weight.
F_N\cos\theta=mg
Divide Eq. (4.22) by this equation. F_N cancels.
\dfrac{F_N\sin\theta}{F_N\cos\theta}=\dfrac{mv^{2}/r}{mg}

Result\tan\theta=\dfrac{v^{2}}{rg}\qquad(4.23)

For a given speed v the centripetal force needed for a turn of radius r at angle \theta is independent of the mass of the vehicle. Higher speeds and smaller radii need more steeply banked curves (larger \theta). At a speed too low for a given \theta a car would slide down a frictionless banked curve; at a speed too high it would slide off the curve.

Tension force as centripetal force

When objects are connected by a string or rope and move in a circle, the tension in the string acts as the centripetal force. It pulls the object towards the centre of the circle, preventing it from moving in a straight line and causing it to follow a curved path instead. For example, to swing a ball in a circle on the end of a string, the tension in the string acts as centripetal force, as shown in Figure 4.10(c). In this case the centripetal force F_c is provided by the tension T in the string such that:

Figure 4.10(c): top view of a ball swung in a horizontal circle. The tension T in the string pulls the ball towards the centre and acts as the centripetal force F_c.

T=\dfrac{mv^{2}}{r}

The magnitude of the tension is directly related to the speed of the object and the radius of the circle. As the speed of the object or the radius of the circle increases, the tension in the string must also increase to provide the necessary centripetal force to keep the object moving in a circular path.

Gravitational force as centripetal force

The force of gravity keeps planets in orbit around the Sun and satellites in orbit around the Earth, serving as the centripetal force. Without the force of gravity acting as the centripetal force, planets and satellites would not be able to maintain their orbits and would instead drift off into space. This force is essential for maintaining the balance between the inward force of gravity and the outward tendency of the object's inertia. In this case the gravitational force F_g is responsible for providing the centripetal acceleration required for the circular motion. Mathematically:

F_g=\dfrac{mv^{2}}{r}

Centrifuge

A centrifuge is a device that separates substances suspended in a liquid mixture by spinning a sample of the liquid mixture very quickly around an axle. Any small denser particles found in the liquid travel in a straight line inside the test tube, obeying Newton's first law. The liquid in the test tube applies a centripetal force on these particles to keep them moving in a circle. After running the centrifuge at high speed for a period of time, the particles become clumped together at the bottom of the test tube, which can be collected and the sample is analysed (Figure 4.11).

Figure 4.11: centrifuge. At high speed the tubes swing outwards and the denser particles collect at the bottom (outer) end of each tube.

The same centrifugation principle is applied in the following commonly used devices:

Cream separator: a centrifugal device that separates milk into cream and skimmed milk.
Washing machine dryer: it consists of a long cylinder with small holes on its walls. Wet clothes are placed in this cylinder, which is then rotated rapidly to dry them.

Example 4.2: radius of a curve

The centripetal force on a car of mass 856\,\mathrm{kg} moving along a curve is 7250\,\mathrm{N}. If its speed is 12.0\,\mathrm{m\,s^{-1}}, what is the radius of the curve?

Given

m = 856\,\mathrm{kg}
F_c = 7250\,\mathrm{N}
v = 12.0\,\mathrm{m\,s^{-1}}

To find

radius r

The centripetal force is F_c=mv^{2}/r in magnitude. Rearrange for r. (The book keeps the minus sign, F_c=-7250\,\mathrm{N}, which shows the force points to the centre; the radius comes out positive.)
r=\dfrac{mv^{2}}{F_c}
Put in the values.
r=\dfrac{856\times(12.0)^{2}}{7250}=\dfrac{123264}{7250}=17.0\,\mathrm{m}

r=17\,\mathrm{m}

Assignment 4.2: radius of a curve

A car, with its centre of gravity 0.4\,\mathrm{m} above the Earth's surface, is passing through a curve whose speed limit is 15\,\mathrm{m\,s^{-1}}. Find the radius of the curve.

Given

v = 15\,\mathrm{m\,s^{-1}} speed limit
h = 0.4\,\mathrm{m} height of the centre of gravity
g = 9.8\,\mathrm{m\,s^{-2}}

To find

radius of the curve

The book does not say what the curve is. We read it as in numerical problem 4.2 of the exercise: the speed limit is the speed at which the car just stays in contact with the road, so gravity alone provides the centripetal force, mg=mv^{2}/R, where R is the radius of the path of the centre of gravity.

Gravity provides the centripetal force at the speed limit.
mg=\dfrac{mv^{2}}{R}\ \Rightarrow\ R=\dfrac{v^{2}}{g}=\dfrac{(15)^{2}}{9.8}=23.0\,\mathrm{m}
The centre of gravity is 0.4\,\mathrm{m} above the road, so the radius of the road itself is smaller by that amount.
r=R-h=23.0-0.4=22.6\,\mathrm{m}

r\approx22.6\,\mathrm{m}\ (R=23\,\mathrm{m}\text{ for the centre of gravity})

Worked example: whirling a stone, and what if the string breaks

A 0.20\,\mathrm{kg} stone is whirled in a horizontal circle of radius 0.80\,\mathrm{m} at 2.0 revolutions per second. (a) Find the tension in the string. (b) What is the tension if the speed is doubled? (c) What path does the stone follow if the string breaks, and why?

Given

m = 0.20\,\mathrm{kg}
r = 0.80\,\mathrm{m}
f = 2.0\,\mathrm{rev\,s^{-1}}

To find

(a) T
(b) T' for 2v
(c) path after breaking

(a) Convert revolutions per second to angular velocity: one revolution is 2\pi rad.
\omega=2\pi f=2\pi(2.0)=12.57\,\mathrm{rad\,s^{-1}}
The tension supplies the centripetal force, T=mr\omega^{2}.
T=0.20(0.80)(12.57)^{2}=25.3\,\mathrm{N}
Check with the speed form: v=r\omega=10.05\,\mathrm{m\,s^{-1}}.
T=\dfrac{mv^{2}}{r}=\dfrac{0.20(10.05)^{2}}{0.80}=25.3\,\mathrm{N}
(b) T\propto v^{2} at fixed radius, so doubling the speed multiplies the tension by 2^{2}=4.
T'=4(25.3)=101\,\mathrm{N}
(c) If the string breaks, the tension (the centripetal force) disappears. No net force acts on the stone, so by Newton's first law it keeps moving in a straight line in the direction of its velocity at that moment, along the tangent to the circle. It does not fly outwards along the radius.

Top view. While the string holds, the tension T turns the stone along the circle. When it breaks the stone leaves along the tangent, the direction of v at that moment.

T=25.3\,\mathrm{N},\quad T'=101\,\mathrm{N},\quad \text{straight tangent}

4.3Moment of inertia

Moment of inertia (rotational inertia): The property of a body by which it maintains its state of rest or uniform rotational motion about a fixed axis is called moment of inertia.

The moment of inertia is the rotational equivalent of mass. Objects with a larger mass have a larger inertia, meaning that they are harder to accelerate linearly. Similarly, an object with a larger moment of inertia is harder to accelerate angularly. For a single particle of mass m at distance r from the axis, the moment of inertia is:

I=mr^{2}

(4.24)

Derivation Moment of inertia of a rigid body, I=\sum m_ir_i^{2}

Aim. Extend I=mr^{2} from a single particle to a whole rigid body.

Figure 4.12: moment of inertia. A single mass m at distance r from the axis has I=mr^{2}; a rigid object is divided into small masses m_1,m_2,\dots,m_n at distances r_1,r_2,\dots,r_n and I=\sum m_ir_i^{2}.

A rigid body is divided into a large number of small portions of masses m_1,m_2,m_3,\dots,m_n at distances r_1,r_2,r_3,\dots,r_n from the axis of rotation (Figure 4.12).
Each portion is like a point mass going round the axis, so by Eq. (4.24) its moment of inertia is m_ir_i^{2}.
Since the portions are fixed relative to one another (rigid body), the moments of inertia about the same axis simply add.
I=m_1r_1^{2}+m_2r_2^{2}+\dots+m_nr_n^{2}

ResultI=\sum_{i=1}^{n}m_ir_i^{2}\qquad(4.25)

The moment of inertia depends on the mass of the body and on how that mass is distributed about the axis: mass farther from the axis counts more, because of the factor r^{2}. Its SI unit is \mathrm{kg\,m^{2}}.

Moments of inertia of uniform objects

Figure 4.13 of the book shows the calculated moments of inertia for various objects of uniform composition, each of mass M. They are collected below.

Figure 4.13: moment of inertia for uniform objects (given in the book as a figure)
Object	Rotation axis (as drawn in the book)	Moment of inertia
Cylinder rod (length L)	through the centre, perpendicular to the rod	I=\dfrac{1}{12}ML^{2}
Cylinder rod (length L)	through one end, perpendicular to the rod	I=\dfrac{1}{3}ML^{2}
Rectangular plate (sides a, b)	through the centre, perpendicular to the plate	I=\dfrac{1}{12}M(a^{2}+b^{2})
Rectangular plate (sides a, b)	through a corner (end of the plate), perpendicular to the plate	I=\dfrac{1}{3}M(a^{2}+b^{2})
Solid cylinder or disc (radius R)	through the centre, along the symmetry axis	I=\dfrac{1}{2}MR^{2}
Ring or hoop (radius R)	through the centre, along the symmetry axis	I=MR^{2}
Hollow cylinder (radii R_1, R_2)	along the symmetry axis	I=\dfrac{1}{2}M(R_1^{2}+R_2^{2})
Solid sphere (radius R)	through the centre (a diameter)	I=\dfrac{2}{5}MR^{2}
Hollow sphere (radius R)	through the centre (a diameter)	I=\dfrac{2}{3}MR^{2}

Worked example: same dumbbell, two different axes

Two 2.0\,\mathrm{kg} masses are fixed 1.0\,\mathrm{m} apart on a light rod (treat the rod as massless). Find the moment of inertia (a) about an axis through the middle of the rod and perpendicular to it, (b) about a parallel axis through one of the masses. Explain why the values differ.

Given

m_1=m_2 = 2.0\,\mathrm{kg}
d = 1.0\,\mathrm{m}

To find

I for each axis

Use I=\sum m_ir_i^{2}, where r_i is the perpendicular distance of each mass from the axis. (a) Each mass is 0.5\,\mathrm{m} from the axis.
I=2.0(0.5)^{2}+2.0(0.5)^{2}=0.50+0.50=1.0\,\mathrm{kg\,m^{2}}
(b) The mass on the axis has r=0 and contributes nothing. The other is 1.0\,\mathrm{m} away.
I=2.0(0)^{2}+2.0(1.0)^{2}=2.0\,\mathrm{kg\,m^{2}}
Why do the values differ? Each mass contributes mr^{2}, and r is measured from the axis. Moving the axis to one mass puts that mass at r=0 and the other at r=1.0\,\mathrm{m} instead of 0.5\,\mathrm{m}. Since r is squared, that mass now contributes four times as much, so the total doubles. The moment of inertia depends on where the axis is.

The same two masses give different moments of inertia for different axes: I=1.0\,\mathrm{kg\,m^{2}} about the centre and 2.0\,\mathrm{kg\,m^{2}} about one end mass.

I_a=1.0\,\mathrm{kg\,m^{2}},\qquad I_b=2.0\,\mathrm{kg\,m^{2}}

4.4Angular momentumOften examined

The angular momentum \vec L of an object is defined as:

Angular momentum: The cross product of the position vector \vec r (with respect to the axis of rotation) and the linear momentum \vec p of an object: \vec L=\vec r\times\vec p.

The SI unit of angular momentum is \mathrm{kg\,m^{2}\,s^{-1}} (equivalent to \mathrm{J\,s}) and its dimensions are [ML^{2}T^{-1}].

For a point mass

Derivation Angular momentum of a point mass, L=I\omega

Aim. Express the angular momentum of a mass m moving in a circle of radius r in terms of its angular velocity.

Figure 4.14: mass m rotating at distance r from the axis of rotation. The momentum \vec p is tangent to the circle, so the angle between \vec r and \vec p is 90^\circ.

By the definition of angular momentum, for \theta the angle between \vec r and \vec p (the magnitude is rp\sin\theta, along \hat n perpendicular to both).
\vec L=\vec r\times\vec p\qquad\text{or}\qquad \vec L=rp\sin\theta\,\hat n
The momentum is tangent to the circle and \vec r is along the radius, so \theta=90^\circ and \sin90^\circ=1.
L=rp\qquad(4.26)
Linear momentum is p=mv (4.27) and the linear and angular velocities are related by v=r\omega (4.28). Combining them:
p=mr\omega\qquad(4.29)
Put Eq. (4.29) in Eq. (4.26).
L=r(mr\omega)=mr^{2}\omega
Since I=mr^{2} for a point mass, mr^{2}\omega=I\omega.

ResultL=mr^{2}\omega=I\omega\qquad(4.30)

The angular momentum of an object can also be defined as the product of its moment of inertia and its angular velocity, just like linear momentum is the product of mass and velocity.

For a rigid body

Derivation Angular momentum of a rigid body, L=I\omega

Aim. Find the total angular momentum of a rigid body rotating about a fixed axis.

Divide the rigid body into a large number of small masses m_1,m_2,m_3,\dots,m_n at distances r_1,r_2,r_3,\dots,r_n from the axis of rotation. The net angular momentum is the sum of the individual angular momenta.
L_{net}=L_1+L_2+L_3+\dots+L_n\qquad(4.31)
Each small mass is a point mass, so Eq. (4.30) applies to each one.
\begin{aligned}L_1&=m_1r_1^{2}\omega_1&&(4.32)\\ L_2&=m_2r_2^{2}\omega_2&&(4.33)\\ L_3&=m_3r_3^{2}\omega_3&&(4.34)\\ &\ \ \vdots\\ L_n&=m_nr_n^{2}\omega_n&&(4.35)\end{aligned}
Put Eqs. (4.32) to (4.35) in Eq. (4.31).
L_{net}=m_1r_1^{2}\omega_1+m_2r_2^{2}\omega_2+\dots+m_nr_n^{2}\omega_n\qquad(4.36)
All points of the same rigid body rotate with the same angular velocity, so \omega_1=\omega_2=\dots=\omega_n=\omega. Take \omega out as a common factor.
L_{net}=(m_1r_1^{2}+m_2r_2^{2}+\dots+m_nr_n^{2})\,\omega
The term in brackets is the moment of inertia of the rigid body, Eq. (4.25).

ResultL_{net}=\left(\sum_{i=1}^{n}m_ir_i^{2}\right)\omega=I\omega\qquad(4.37)

The same result L=I\omega holds for a point mass and for a rigid body about a fixed axis.

4.4.3Relation between torque and angular momentumOften examined

Derivation Torque equals the rate of change of angular momentum

Aim. Show that \tau=\Delta L/\Delta t.

The angular momentum of an object is the cross product of its position vector with respect to the axis and its linear momentum.
\vec L=\vec r\times\vec p
Take the change per unit time of both sides.
\dfrac{\Delta\vec L}{\Delta t}=\dfrac{\Delta}{\Delta t}(\vec r\times\vec p)
The position vector \vec r (the distance from the axis) is not changing for the particle moving in its circle, so it is taken outside the change.
\dfrac{\Delta\vec L}{\Delta t}=\vec r\times\dfrac{\Delta\vec p}{\Delta t}\qquad(4.38)
Newton's second law in terms of momentum says that force is the rate of change of momentum.
\vec F=\dfrac{\Delta\vec p}{\Delta t}\qquad(4.39)
Put Eq. (4.39) in Eq. (4.38).
\dfrac{\Delta\vec L}{\Delta t}=\vec r\times\vec F\qquad(4.40)
By the definition of torque, \vec\tau=\vec r\times\vec F (4.41). So the right side of Eq. (4.40) is the torque.

Result\dfrac{\Delta\vec L}{\Delta t}=\vec\tau\qquad(4.42)

Torque is the time rate of change of angular momentum, the rotational form of Newton's second law F=\Delta p/\Delta t. If no torque acts, \Delta L/\Delta t=0: the angular momentum does not change.

4.4.4Conservation of angular momentumOften examined

Law of conservation of angular momentum: In the absence of any external torque, the angular momentum of a system remains constant.

From Eq. (4.42), if there is no external torque then:

\dfrac{\Delta L}{\Delta t}=0\ \Rightarrow\ \Delta L=0\ \Rightarrow\ L_f-L_i=0\ \Rightarrow\ L_f=L_i

I_f\omega_f=I_i\omega_i

(4.43)

Equation (4.43) implies that the final angular momentum should be equal to the initial angular momentum.

Spinning ice skater

A spinning ice skater is an interesting example of conservation of angular momentum. When the skater's arms are extended, the rotational inertia I is relatively large and the angular velocity \omega is relatively small (Figure 4.16). Often at the end of a spin, the skater pulls the arms close to the body, resulting in a much faster spin (larger angular velocity) because of a much smaller rotational inertia. When a rotating body contracts, its angular velocity increases; and when a rotating body expands, its angular velocity decreases. This phenomenon is the result of the conservation of angular momentum.

Figure 4.16: a spinning ice skater. With the arms extended I is large and \omega small; pulling the arms in makes I small and \omega large, so that I\omega stays the same.

As L_f=L_i, or I_f\omega_f=I_i\omega_i, and the moment of inertia is I=mr^{2}:

m_fr_f^{2}\,\omega_f=m_ir_i^{2}\,\omega_i

As r_f<r_i and m_f=m_i, the rotational speed increases to compensate for the decrease in rotational inertia.

Gymnasts and divers

Gymnasts and divers generate their spins (torque) from a solid base or a diving board, after which the angular momentum remains unchanged (Figure 4.17). The usual somersaults and twists result from making variations in their rotational inertia: curled up (small I) they spin fast, stretched out (large I) they spin slowly.

Gyroscope

A gyroscope is a device that utilises the principle of angular momentum to maintain its orientation relative to the Earth's axis or resist changes in its orientation. It usually consists of a wheel mounted on an axle which can rotate freely and is secured in a metal frame (Figure 4.18(a)). When the wheel is made to spin, the gyroscope can be balanced on a flat surface; as the wheel stops spinning, the gyroscope will fall. If the gyroscope is tilted it also can stay levitated without falling, but will start precession about the gravitational force axis. This is the gravity-defying part of a gyroscope (Figure 4.18(b)).

This unusual behaviour can be explained by the vector nature of angular momentum: a change in the direction of the gyroscope requires a torque. The torque is provided by the gravitational force, its weight towards the ground. The angular momentum starts to follow the torque, as shown in Figure 4.18(c); the change in angular momentum \Delta L is:

\Delta L=\tau\times\Delta t

where the torque has the same direction as \Delta L and \Delta t is the duration of time. The same effects can also be observed even if the gyroscope is lifted by a string looped around its lower end.

Figure 4.18(c): in a gyroscope the angular momentum \vec L turns while its size stays the same. The change \Delta\vec L=\vec\tau\Delta t is in the direction of the torque (horizontal), so the axis moves round the vertical, which is precession.

Flywheel

A flywheel (Figure 4.19) is a mechanical component that stores energy by spinning a heavy disc or wheel about an axle. When torque is applied, the rotational speed increases, storing kinetic energy that can be used for different tasks. The concept of a flywheel is based on the principle of conservation of energy, where the energy input is stored in the form of rotational motion. This stored energy can then be released when needed, such as during power outages or to provide additional power for machinery.

Flywheels are commonly used in engines, industrial machinery and energy storage systems. In vehicles, flywheels can help smooth out the power delivery and improve fuel efficiency. In industrial settings, they can provide backup power and help regulate the speed of machinery.

The design of a flywheel involves careful consideration of the material, size and shape of the disc or wheel, as well as the bearings and axle, to minimise friction and maximise energy storage. The speed and torque at which the flywheel operates must also be carefully controlled to ensure safe and efficient operation.

In recent years there has been growing interest in using flywheels as a form of energy storage for renewable energy sources, such as wind and solar power. By storing excess energy generated during peak production times, flywheels can help balance the supply and demand of electricity on the grid.

Example 4.3: grinding wheel

What is the angular momentum of a 3.6\,\mathrm{kg} uniform cylindrical grinding wheel of radius 31\,\mathrm{cm} rotating at 1150\,\mathrm{rpm}? (b) How much torque is required to stop it in 7.8\,\mathrm{s}?

Given

m = 3.6\,\mathrm{kg}
R = 31\,\mathrm{cm}=0.31\,\mathrm{m}
\omega_i = 1150\,\mathrm{rpm}=120.4\,\mathrm{rad\,s^{-1}}
\Delta t = 7.8\,\mathrm{s}

To find

(a) angular momentum L
(b) torque \tau

Convert the rotation rate to \mathrm{rad\,s^{-1}}.
\omega_i=1150\times\dfrac{2\pi}{60}=120.4\,\mathrm{rad\,s^{-1}}
(a) The angular momentum is L=I\omega. For a solid cylinder (disc) I=\tfrac12mR^{2} (Figure 4.13).
L=\dfrac12mR^{2}\omega=\dfrac12\times3.6\times(0.31)^{2}\times120.4=20.83\,\mathrm{kg\,m^{2}\,s^{-1}}
(b) Torque is the rate of change of angular momentum, \tau=\dfrac{L_f-L_i}{\Delta t}. The wheel is brought to rest, so L_f=0.
\tau=\dfrac{0-20.83}{7.8}=-2.67\,\mathrm{kg\,m^{2}\,s^{-2}}=-2.67\,\mathrm{N\,m}

The negative sign shows that the torque is opposite to the angular momentum, as it must be to slow the wheel down. (The book writes the unit of L as J s, which is the same as \mathrm{kg\,m^{2}\,s^{-1}}.)

L=20.83\,\mathrm{kg\,m^{2}\,s^{-1}},\quad\tau=-2.67\,\mathrm{N\,m}

Assignment 4.3: angular momentum of the Earth

The Earth rotates about its own axis. What will be its angular momentum when its average angular speed around its axis is 7.29\times10^{-5}\,\mathrm{rad\,s^{-1}}?

The book gives no data except \omega. We treat the Earth as a uniform solid sphere and use I=\tfrac25MR^{2} from Figure 4.13, with M=5.98\times10^{24}\,\mathrm{kg} and R=6.38\times10^{6}\,\mathrm{m} (standard values).

Given

\omega = 7.29\times10^{-5}\,\mathrm{rad\,s^{-1}}
M = 5.98\times10^{24}\,\mathrm{kg}
R = 6.38\times10^{6}\,\mathrm{m}

To find

angular momentum L

Moment of inertia of a solid sphere.
I=\dfrac25MR^{2}=0.4\times5.98\times10^{24}\times(6.38\times10^{6})^{2}=9.74\times10^{37}\,\mathrm{kg\,m^{2}}
Angular momentum.
L=I\omega=9.74\times10^{37}\times7.29\times10^{-5}=7.1\times10^{33}\,\mathrm{kg\,m^{2}\,s^{-1}}

L\approx7.1\times10^{33}\,\mathrm{kg\,m^{2}\,s^{-1}}

Worked example: a boy on a turntable

A platform of moment of inertia 60\,\mathrm{kg\,m^{2}} turns freely at 2.0\,\mathrm{rad\,s^{-1}}. A 40\,\mathrm{kg} boy gently steps on at the rim, 1.5\,\mathrm{m} from the axis. (a) Find the new angular velocity. (b) He then walks to a point 0.50\,\mathrm{m} from the axis. Find the angular velocity now.

Given

I_p = 60\,\mathrm{kg\,m^{2}}
\omega_i = 2.0\,\mathrm{rad\,s^{-1}}
m = 40\,\mathrm{kg}

To find

(a) \omega_f at the rim
(b) \omega_f at 0.50\,\mathrm{m}

There is no external torque on the platform and boy (friction is neglected), so angular momentum is conserved: I_f\omega_f=I_i\omega_i. The boy is treated as a point mass, I=mr^{2}.
(a) Initial angular momentum and final moment of inertia.
L_i=60(2.0)=120\,\mathrm{kg\,m^{2}\,s^{-1}},\qquad I_f=60+40(1.5)^{2}=150\,\mathrm{kg\,m^{2}}
Solve for the new angular velocity.
\omega_f=\dfrac{L_i}{I_f}=\dfrac{120}{150}=0.80\,\mathrm{rad\,s^{-1}}
(b) Walking inwards lowers I but L is still 120.
I_f=60+40(0.50)^{2}=70\,\mathrm{kg\,m^{2}},\qquad\omega_f=\dfrac{120}{70}=1.7\,\mathrm{rad\,s^{-1}}
Compare the two answers. In (a) the platform slows from 2.0 to 0.80\,\mathrm{rad\,s^{-1}} because I rises from 60 to 150\,\mathrm{kg\,m^{2}}. In (b) I falls to 70\,\mathrm{kg\,m^{2}} and the platform speeds up again, but not back to 2.0\,\mathrm{rad\,s^{-1}}, because the boy is still on the platform.

\omega=0.80\,\mathrm{rad\,s^{-1}}\ \text{(rim)},\qquad 1.7\,\mathrm{rad\,s^{-1}}\ \text{(at }0.50\,\mathrm{m})

4.5Torque and angular acceleration

A relationship exists between torque and angular acceleration, just like force and acceleration in Newton's second law of motion.

For a point mass

Derivation Torque and angular acceleration, \tau=I\alpha

Aim. Relate the torque on a point mass moving in a circle to its angular acceleration.

Figure 4.20: mass m rotating at distance r from the axis of rotation under a tangential force F (perpendicular to \vec r).

A mass m rotates at a distance r from the axis (Figure 4.20). The force F that rotates it is a tangential force. By the definition of torque:
\vec\tau=rF\sin\theta\,\hat n\qquad(4.44)
The force is perpendicular to \vec r, so \theta=90^\circ and \sin90^\circ=1. The magnitude becomes
\tau=rF\qquad(4.45)
By Newton's second law F=ma (4.46), where a is the tangential acceleration, related to the angular acceleration by
a=r\alpha\qquad(4.47)
Put Eq. (4.47) in Eq. (4.46).
F=mr\alpha\qquad(4.48)
Put Eq. (4.48) in Eq. (4.45).
\tau=r(mr\alpha)\quad\text{or}\quad\tau=mr^{2}\alpha\qquad(4.49)
The term mr^{2} is the moment of inertia.

Result\tau=I\alpha\qquad(4.50)

Torque is the moment of inertia times the angular acceleration. This is similar to Newton's second law F=ma, which gives force as inertia (mass) times acceleration. For a rigid body, with I=\sum m_ir_i^{2}, the same relation holds.

Linear and rotational analogues used in this unit
Linear motion	Rotational motion
mass m	moment of inertia I
force F	torque \tau
acceleration a	angular acceleration \alpha
F=ma	\tau=I\alpha
momentum p=mv	angular momentum L=I\omega

4.6Weightlessness in satellites

Weightlessness occurs when the feeling of weight is completely or almost completely absent, meaning there is zero apparent weight. This occurs during free fall, when the force of gravity is balanced out by the inertial force from orbital flight, like the centrifugal force.

In a weightless environment, objects and individuals float freely, as there is no force pulling them towards the ground. This phenomenon is commonly experienced by astronauts in space, where they appear to float inside their spacecraft with objects around them.

The term zero gravity is often used incorrectly to describe weightlessness. Astronauts in space stations are not in a gravity-free environment: at about 250 miles above the Earth, where most space stations orbit, the gravitational field is still quite strong, roughly 95\% of its value at the surface of the Earth.

Weightlessness can be achieved in two ways:

Travel millions of miles away from the gravitational force of a large object, where the gravitational force reduces to nearly zero.
Much more practical: create a weightless environment through the act of free fall. Space stations are in constant free fall, having the right speed and being at the right altitude. Inside the space station the astronaut is also falling freely, so he appears to float (Figure 4.21), and physicists call it weightlessness.

Weightlessness can be enjoyed in amusement parks momentarily. It can also be simulated on Earth through techniques such as parabolic flights or neutral buoyancy tanks, allowing researchers to study the effects of microgravity on the human body and various materials.

Effects on astronauts

Living in a space station is not easy. Besides the dangers of space travel and the time spent away from family in isolation, astronauts feel many health issues related to microgravity: their bones and muscles get weakened, the cardiovascular system is affected and the immune system is compromised. Some everyday activities become near-impossible: their basic necessities like eating, sleeping and showering habits are modified. They cannot even cry; they face difficulty in digesting food and even in urination and excretion. Rotational simulated gravity has been proposed as a solution in human spaceflight to the adverse health effects caused by prolonged weightlessness.

Even though things and people may feel weightless in a weightless setting, they still possess mass and inertia, causing them to maintain their straight-line motion unless an outside force intervenes.

4.7Artificial gravity

Artificial gravity: The gravity produced artificially in satellites to counteract the effect of weightlessness is called artificial gravity.

It can be generated by rotating a space station around its own axis (Figure 4.22). The surface of the rotating space station exerts a force on objects in contact with it and thereby provides the centripetal force that keeps the object moving on a circular path. In space stations the astronauts feel weightless and cannot work effectively. To overcome this difficulty, artificial gravity can be provided by rotating the station about its own axis.

Figure 4.22: artificial gravity. The rim of the rotating space station pushes on the astronaut with the centripetal force; the astronaut feels this contact force as weight.

Derivation Speed, angular velocity, period and frequency for artificial gravity

Aim. Find how fast a space station of radius R must rotate to give an acceleration equal to g at its rim.

Assumptions.

The station rotates uniformly about its own axis.
The required artificial gravity equals g at the outer rim.

Consider a circular tube-shaped part of the space station of radius R rotating with speed v. The centripetal acceleration experienced at any point on the outer rim is
a_c=\dfrac{v^{2}}{R}\qquad(4.52)
Rearranging Eq. (4.52) gives the linear velocity.
v^{2}=a_cR\ \Rightarrow\ v=\sqrt{a_cR}
To give the same effect as gravity, the centripetal acceleration (and so the centripetal force) must equal the acceleration due to gravity, a_c=g.
v=\sqrt{gR}\qquad(4.53)
The relation between linear and angular velocity is v=\omega R (4.54). Compare it with Eq. (4.53).
\omega R=\sqrt{gR}\ \Rightarrow\ \omega=\dfrac{\sqrt{gR}}{R}=\sqrt{\dfrac{gR}{R^{2}}}=\sqrt{\dfrac{g}{R}}\qquad(4.55)
The time period is the time for one rotation. Use v=\omega R.
T=\dfrac{2\pi R}{v}=\dfrac{2\pi R}{\omega R}=\dfrac{2\pi}{\omega}\qquad(4.56)
Put the value of \omega from Eq. (4.55) into Eq. (4.56).
T=2\pi\sqrt{\dfrac{R}{g}}\qquad(4.57)
Frequency is the reciprocal of the time period, f=1/T.
f=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{R}}

Resultv=\sqrt{gR},\quad\omega=\sqrt{\dfrac{g}{R}},\quad T=2\pi\sqrt{\dfrac{R}{g}},\quad f=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{R}}

A larger station can give the same artificial gravity with a slower rotation (smaller \omega, longer T). The astronauts stand on the inside of the rim and the rim's push on their feet is the centripetal force, which they feel as weight.

Example 4.4: ring-shaped space station

An 80.0\,\mathrm{kg} astronaut stands on the rim of a rotating ring-shaped space station providing him sufficient artificial gravity g=9.8\,\mathrm{m\,s^{-2}}. If the radius of the space station is 1.5\,\mathrm{km}, calculate his (a) angular velocity, (b) time period and (c) frequency of rotation.

Given

m = 80.0\,\mathrm{kg} mass of the astronaut (not needed)
R = 1.50\,\mathrm{km}=1500\,\mathrm{m}
g = 9.8\,\mathrm{m\,s^{-2}}

To find

(a) \omega
(b) T
(c) f

(a) Angular velocity of the artificial satellite.
\omega=\sqrt{\dfrac{g}{R}}=\sqrt{\dfrac{9.8}{1500}}=0.08\,\mathrm{rad\,s^{-1}}
(b) Time period.
T=2\pi\sqrt{\dfrac{R}{g}}=2\times3.14\times\sqrt{\dfrac{1500}{9.8}}=77.73\,\mathrm{s}
(c) Frequency is the reciprocal of the time period.
f=\dfrac1T=\dfrac{1}{77.73}=0.013\,\mathrm{Hz}

The mass 80.0\,\mathrm{kg} is not needed: \omega, T and f depend only on g and R.

\omega=0.08\,\mathrm{rad\,s^{-1}},\quad T=77.7\,\mathrm{s},\quad f=0.013\,\mathrm{Hz}

Assignment 4.4: rotating spaceship

A spaceship having cylindrical shape is rotated at a speed of 20\,\mathrm{rpm} about its axis in order to provide artificial gravity to its inhabitants. If the spaceship has a diameter of 8\,\mathrm{m}, find the artificial gravity it provides.

Given

f = 20\,\mathrm{rpm}
D = 8\,\mathrm{m}\Rightarrow R=4\,\mathrm{m}

To find

artificial gravity g' at the rim

Convert the rotation rate to rad s^{-1}.
\omega=20\times\dfrac{2\pi}{60}=2.09\,\mathrm{rad\,s^{-1}}
The artificial gravity is the centripetal acceleration at the rim, a_c=\omega^{2}R (from g=\omega^{2}R, Eq. 4.55).
g'=\omega^{2}R=(2.09)^{2}\times4=17.5\,\mathrm{m\,s^{-2}}

This is about 1.8 times the normal gravity, so 20\,\mathrm{rpm} is too fast for a comfortable 1\,g.

g'\approx17.5\,\mathrm{m\,s^{-2}}

Worked example: size of the station and the artificial gravity

A ring-shaped space station of radius 200\,\mathrm{m} must give g=9.8\,\mathrm{m\,s^{-2}} at its rim. Find the angular velocity, the time period, the number of turns per minute, and the artificial gravity felt at a point half-way between the axis and the rim.

Given

R = 200\,\mathrm{m}
g = 9.8\,\mathrm{m\,s^{-2}}

To find

\omega, T, rpm
g' at R/2

The rim must supply g=\omega^{2}R as the centripetal acceleration.
\omega=\sqrt{\dfrac{g}{R}}=\sqrt{\dfrac{9.8}{200}}=0.221\,\mathrm{rad\,s^{-1}}
Time for one turn.
T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{0.2214}=28.4\,\mathrm{s}
Turns per minute.
\dfrac{60}{28.4}=2.11\ \text{rev per minute}
Rim speed as a check: v=\omega R=0.2214(200)=44.3\,\mathrm{m\,s^{-1}}, and \sqrt{gR}=\sqrt{1960}=44.3\,\mathrm{m\,s^{-1}}.
Every point of the station has the same \omega, so at r=R/2 the centripetal acceleration is \omega^{2}r, half the rim value.
g'=\omega^{2}\dfrac{R}{2}=0.04900(100)=4.9\,\mathrm{m\,s^{-2}}

Artificial gravity points outwards (towards the floor at the rim) and is proportional to the distance from the axis, so it is smaller closer to the axis.

\omega=0.221\,\mathrm{rad\,s^{-1}},\ T=28.4\,\mathrm{s},\ 2.11\,\mathrm{rpm},\ g'=4.9\,\mathrm{m\,s^{-2}}

Formula sheet

Angular position

\theta=\dfrac{S}{r}

S: arc length; r: radius

Unit: \mathrm{rad} (dimensionless)

Angular displacement

\Delta\theta=\theta_2-\theta_1

\theta_1,\theta_2: initial and final angular positions

Unit: \mathrm{rad}

Radians and degrees

2\pi\,\mathrm{rad}=360^\circ,\quad 1\,\mathrm{rad}=57.3^\circ

one revolution

Angular velocity

\omega=\dfrac{\Delta\theta}{\Delta t}

\Delta\theta: angular displacement in time \Delta t

Unit: \mathrm{rad\,s^{-1}}

Angular acceleration

\alpha=\dfrac{\Delta\omega}{\Delta t}

\Delta\omega: change of angular velocity in time \Delta t

Unit: \mathrm{rad\,s^{-2}}

Arc length

S=r\theta

\theta in radians

Unit: \mathrm{m}

Tangential velocity

v=r\omega

r: radius

Unit: \mathrm{m\,s^{-1}}

Tangential acceleration

a_T=r\alpha

\alpha: angular acceleration

Unit: \mathrm{m\,s^{-2}}

Total acceleration

a=\sqrt{a_T^{2}+a_R^{2}}

a_T: tangential part; a_R: radial part

Unit: \mathrm{m\,s^{-2}}

Angular displacement at constant \omega

\theta=\omega t

Unit: \mathrm{rad}

Rotational kinematics: final velocity

\omega_f=\omega_i+\alpha t

constant \alpha

Unit: \mathrm{rad\,s^{-1}}

Rotational kinematics: no time

2\alpha\theta=\omega_f^{2}-\omega_i^{2}

constant \alpha

Rotational kinematics: displacement

\theta=\omega_it+\tfrac12\alpha t^{2}

constant \alpha

Unit: \mathrm{rad}

Centripetal acceleration

a_c=\dfrac{v^{2}}{r}=r\omega^{2}

directed towards the centre

Unit: \mathrm{m\,s^{-2}}

Centripetal force

F_c=\dfrac{mv^{2}}{r}=mr\omega^{2}

m: mass; v: speed; r: radius

Unit: \mathrm{N}

Banked curve

\tan\theta=\dfrac{v^{2}}{rg}

\theta: banking angle (no friction)

Tension as centripetal force

T=\dfrac{mv^{2}}{r}

string with a mass moving in a circle

Unit: \mathrm{N}

Moment of inertia of a point mass

I=mr^{2}

r: distance from the axis

Unit: \mathrm{kg\,m^{2}}

Moment of inertia of a rigid body

I=\sum_{i=1}^{n}m_ir_i^{2}

m_i: small masses at distances r_i from the axis

Unit: \mathrm{kg\,m^{2}}

Moment of inertia: rod

I=\dfrac{1}{12}ML^{2}\ \text{(centre)},\quad I=\dfrac13ML^{2}\ \text{(end)}

M: mass; L: length

Unit: \mathrm{kg\,m^{2}}

Moment of inertia: rectangular plate

I=\dfrac{1}{12}M(a^{2}+b^{2})\ \text{(centre)},\quad I=\dfrac13M(a^{2}+b^{2})\ \text{(corner)}

a,b: sides

Unit: \mathrm{kg\,m^{2}}

Moment of inertia: disc, ring, hollow cylinder

I=\dfrac12MR^{2},\quad I=MR^{2},\quad I=\dfrac12M(R_1^{2}+R_2^{2})

solid cylinder or disc; ring or hoop; hollow cylinder

Unit: \mathrm{kg\,m^{2}}

Moment of inertia: spheres

I=\dfrac25MR^{2}\ \text{(solid)},\quad I=\dfrac23MR^{2}\ \text{(hollow)}

R: radius

Unit: \mathrm{kg\,m^{2}}

Angular momentum

\vec L=\vec r\times\vec p

\vec r: position vector; \vec p: linear momentum

Unit: \mathrm{kg\,m^{2}\,s^{-1}}

Angular momentum of a rotating body

L=I\omega

I: moment of inertia; \omega: angular velocity

Unit: \mathrm{kg\,m^{2}\,s^{-1}}

Torque and angular momentum

\vec\tau=\vec r\times\vec F=\dfrac{\Delta\vec L}{\Delta t}

\vec F: force; \Delta\vec L/\Delta t: rate of change of angular momentum

Unit: \mathrm{N\,m}

Conservation of angular momentum

I_f\omega_f=I_i\omega_i

no external torque

Change in angular momentum (gyroscope)

\Delta L=\tau\,\Delta t

\tau: torque; \Delta t: time

Unit: \mathrm{kg\,m^{2}\,s^{-1}}

Torque and angular acceleration

\tau=I\alpha

I: moment of inertia; \alpha: angular acceleration

Unit: \mathrm{N\,m}

Artificial gravity: speed

v=\sqrt{gR}

R: radius of the station; g: artificial gravity wanted

Unit: \mathrm{m\,s^{-1}}

Artificial gravity: angular velocity

\omega=\sqrt{\dfrac{g}{R}}

Unit: \mathrm{rad\,s^{-1}}

Artificial gravity: period and frequency

T=\dfrac{2\pi}{\omega}=2\pi\sqrt{\dfrac{R}{g}},\quad f=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{R}}

Units: \mathrm{s}; \mathrm{Hz}

Key terms

Rotational motion: The turning or spinning motion of an object about an axis that passes through it.
Rigid object: A non-deformable object whose particles stay in fixed positions relative to one another as it rotates.
Axis of rotation: The line about which rotation takes place; it remains fixed during rotational motion while the other points move in circles about it.
Angular position: The angle through which the position vector of a moving object is displaced with respect to a chosen reference direction: \theta=S/r.
Angular displacement: The change in angular position with respect to a chosen reference direction: \Delta\theta=\theta_2-\theta_1.
Radian: The angle subtended at the centre of a circle by an arc whose length is equal to the radius; 2\pi rad =360^\circ, 1\,\text{rad}=57.3^\circ.
Right-hand rule: Holding the axis of rotation in the right hand with the fingers curling in the direction of rotation, the thumb gives the direction of the angular displacement.
Angular velocity: The time rate of change of angular displacement: \omega=\Delta\theta/\Delta t.
Angular acceleration: The time rate of change of angular velocity: \alpha=\Delta\omega/\Delta t.
Tangential acceleration: The acceleration in a direction tangent to the circle at the point of interest; it changes the magnitude of the velocity: a_T=r\alpha.
Centripetal acceleration: The acceleration of an object moving in a circle, directed toward the centre: a_c=v^{2}/r=r\omega^{2}.
Centripetal force: The net force that causes a particle to undergo centripetal acceleration, directed toward the centre: F_c=mv^{2}/r=mr\omega^{2}.
Centrifuge: A device that separates substances suspended in a liquid by spinning the sample very quickly around an axle, so that denser particles collect at the bottom of the tube.
Moment of inertia: The property of a body by which it maintains its state of rest or uniform rotational motion about a fixed axis (rotational inertia): I=mr^{2} for a point mass, I=\sum m_ir_i^{2} for a rigid body.
Angular momentum: The cross product of the position vector and the linear momentum, \vec L=\vec r\times\vec p; for rotation about an axis L=I\omega.
Torque: The turning effectiveness of a force, \vec\tau=\vec r\times\vec F; it equals the rate of change of angular momentum and the product of moment of inertia and angular acceleration, \tau=I\alpha.
Conservation of angular momentum: In the absence of any external torque the angular momentum of a system remains constant: I_f\omega_f=I_i\omega_i.
Gyroscope: A device with a spinning wheel on an axle in a frame that uses angular momentum to maintain its orientation or resist changes in it.
Flywheel: A heavy spinning disc or wheel on an axle that stores energy as rotational kinetic energy and releases it when needed.
Weightlessness: The state of zero apparent weight, occurring in free fall, as for objects inside an orbiting satellite.
Artificial gravity: The gravity produced artificially in satellites, by rotating them about their own axis, to counteract the effect of weightlessness.

Multiple choice questionsp. 74–75

Encircle the correct option.

The term "centrifugation" means separation:

Athrough spinningAnswer
Bof components at higher temperature
Cthrough evaporation
Dof components at lower temperature

Why

Centrifugation separates the parts of a mixture by spinning it very fast, so that the centripetal force of the liquid holds the lighter parts in the circle while the denser particles move outwards and collect at the bottom of the tube (section 4.2.1, Fig. 4.11).

B and D mention temperature, which the process does not use. C is evaporation, a different method of separation that does not need spinning.

A car turns around a curve at 30\,\mathrm{km\,h^{-1}}. If it turns at double the speed, the tendency to overturn is:

Adoubled
BquadrupledAnswer
Chalved
Dunchanged

Why

In a turn the car needs a centripetal force F_c=\dfrac{mv^{2}}{r} (equation 4.21 and section 4.2.1). The tendency to overturn grows with this required force, so it depends on v^{2}.

Doubling v multiplies v^{2} by 2^{2}=4, so the tendency becomes four times larger (B). A assumes the effect grows only in step with the speed, which it does not, and C and D ignore the change in speed.

Note. The book prints no key for this question. B is worked out from F_c=mv^{2}/r, where the squared speed gives the factor of four.

32025 Annual I, Set D, MCQ 4

The moment of inertia of a spinning body about a certain axis doesn't depend on:

Adistribution of mass around the axis
Borientation of the axis
Cmass of the body
Dangular velocity of the bodyAnswer

Why

For a rigid body I=\sum m_i r_i^{2} (equation 4.25). It depends on the mass of each part (C) and on how that mass is spread out from the axis (A). The axis itself fixes the distances r_i, so the choice of axis, including its orientation (B), changes I.

The angular velocity \omega does not appear in the formula for I. A spinning body and a body at rest have the same moment of inertia about the same axis, so D is the answer.

Note. The book prints no key for this question. D is worked out from I=\sum m_i r_i^{2}; the orientation in B matters only because it sets the distances r_i from the chosen axis.

The change in angular momentum of a rod, when a torque of 2.5\,\mathrm{N\,m} is acted upon it for 2\,\mathrm{s}, is:

A1.25\,\mathrm{J\,s}
B2.5\,\mathrm{J\,s}
C5\,\mathrm{J\,s}Answer
D0\,\mathrm{J\,s}

Why

Torque is the rate of change of angular momentum (equation 4.42): \tau=\dfrac{\Delta L}{\Delta t}, so \Delta L=\tau\,\Delta t.

\Delta L=2.5\,\mathrm{N\,m}\times2\,\mathrm{s}=5\,\mathrm{J\,s}

B ignores the time interval, and A divides the torque by the time instead of multiplying. D would be right only if no torque acted.

52025 Annual I, Set B, MCQ 5

If size (length) of the wings of a fan is increased, its rotational speed, for the same voltage and current, will:

Aincrease
BdecreaseAnswer
Cremain constant
Dmay increase or decrease

Why

Same voltage and current means the same electrical power, so the motor supplies the same power to the blades. Longer wings have a larger moment of inertia (I=\sum m_i r_i^{2}) and meet more air, so the air opposes the rotation with a larger torque.

With \alpha=\tau/I (equation 4.50), a larger I and a larger opposing torque both reduce the rotational speed the fan can reach, so the speed decreases (B).

Note. The book prints no key for this question. B is worked out from the power supplied and the larger moment of inertia and air torque of longer wings.

In a body, angular acceleration is produced by:

Anet force
Bpower
Cpressure
Dnet torqueAnswer

Why

Newton's second law for rotation is \tau=I\alpha (equation 4.50), so the angular acceleration \alpha=\tau/I is produced by the net torque (D).

A net force gives linear acceleration; it turns a body only through the torque it produces about the axis. Power and pressure do not turn a body.

Note. The book prints no key for this question. D is worked out from \tau=I\alpha.

An astronaut feels weightless inside the International Space Station. It is because the International Space Station is:

Aoutside the gravitational field of Earth
Bfreely fallingAnswer
Cat rest
Din motion

Why

The station and the astronaut are both falling freely towards the Earth while moving sideways fast enough to keep orbiting. Nothing pushes the astronaut onto the floor, so the apparent weight is zero (section 4.6). Free fall is the reason, so B is correct.

A is wrong: the gravitational field at the height of most space stations is still about 95\,\% of its surface value, so gravity is strong there. C and D are wrong because weightlessness comes from free fall, not from being at rest or from motion alone.

How many radians account for circumference of a circle?

A1\,\mathrm{rad}
B2\,\mathrm{rad}
C\pi\,\mathrm{rad}
D2\pi\,\mathrm{rad}Answer

Why

The arc length is s=r\theta (equation 4.9). For the whole circle s=2\pi r, so \theta=\dfrac{2\pi r}{r}=2\pi\,\mathrm{rad}, which is 360^{\circ} (one revolution, section 4.1.2).

\pi\,\mathrm{rad} is half a circle (180^{\circ}), and 1\,\mathrm{rad} or 2\,\mathrm{rad} are arbitrary angles that do not make a full circle.

Short questionsp. 75

Give short answers of the following questions.

4.1

What is the value of angular acceleration of the minute hand of your wrist watch?

Solution

The minute hand turns through 2\pi\,\mathrm{rad} in one hour, so its angular velocity is constant at \omega=\dfrac{2\pi}{3600\,\mathrm{s}}\approx1.7\times10^{-3}\,\mathrm{rad\,s^{-1}}. Angular acceleration is the rate of change of angular velocity (equation 4.6), and this does not change.

\alpha=\dfrac{\Delta\omega}{\Delta t}=\dfrac{0}{\Delta t}=0

\alpha=0

4.2

Is there a real force that removes water from wet clothes in a washing machine? Explain how the water is removed.

Solution

No real outward force removes the water. The drum wall exerts a real inward (centripetal) force on the clothes, and this force is what keeps them moving in a circle (section 4.2.1). The cloth can supply only a limited inward force, which is not enough to keep the water on the circular path.

The water therefore tends to keep moving in a straight line, as Newton's first law says, and it leaves the cloth through the small holes in the drum wall. The outward effect is only the water's inertia, not a force pushing it out.

4.3

Determine the relation between (a) linear and angular displacement. (b) linear and angular velocity. (c) linear and angular acceleration.

Solution

(a) Linear displacement s is the arc length of a circle of radius r swept through the angle \theta in radians (equation 4.9).
s=r\theta
(b) Linear velocity is the angular velocity times the radius (equation 4.13).
v=r\omega
(c) The tangential component of linear acceleration is the angular acceleration times the radius (equation 4.17). The radial component is a_c=r\omega^{2}.
a_T=r\alpha

s=r\theta,\qquad v=r\omega,\qquad a_T=r\alpha

4.4

Is centripetal force a fundamental force or a force provided by any of the fundamental forces? Can any combination of the fundamental forces provide centripetal force?

Solution

Centripetal force is not a fundamental force. It is the name given to the net force that makes a body move in a circle towards the centre (equation 4.21). Whatever force supplies it, such as tension, gravity or friction, must be one of the fundamental forces or a combination of them.

Yes, combinations can provide it. Gravity supplies it for planets and satellites (F_g=\dfrac{mv^{2}}{r}). The electromagnetic forces supply it through the tension in a string, the normal force and friction on a curve, and the electric force in CD players and hard disks. The nuclear forces act only over very short distances inside atoms and do not supply it in these examples.

4.5

There are generally double tyres in heavy vehicles on one side of an axle. Will its moment of inertia be different from that of a single tyre?

Solution

Yes. The moment of inertia is I=\sum m_i r_i^{2} (equation 4.25), which depends on the total mass and on how far that mass is from the axle. Two identical tyres on the same axle are each at the same distance r from the axle, so the pair has twice the mass at that distance.

I_{\text{double}}=2mr^{2}=2\,I_{\text{single}}

So the double tyre has a moment of inertia about twice that of a single tyre, which makes it harder to start or stop the wheel from rotating.

I_{\text{double}}=2\,I_{\text{single}}

4.6

Why is it best to have the blades rotate in opposite directions for a helicopter having two sets of lifting blades?

Solution

The angular momentum of the whole system is conserved (section 4.4.4). Before the rotors start, the total angular momentum is zero. If the two sets of blades turn in opposite directions with equal angular momenta, their total is zero, so the helicopter body is not given any angular momentum and does not spin.

If both sets turned the same way, their angular momentum would be transferred to the body, which would then rotate the other way to keep the total at zero. Opposite rotation cancels this turning effect.

L_{\text{blades 1}}+L_{\text{blades 2}}=0\;\Rightarrow\;L_{\text{body}}=0

4.7

If diameter of Earth becomes half and there is no change in its mass, what effect will there be on the rotational speed of Earth around its own axis?

Solution

No external torque acts on the Earth, so its angular momentum is conserved: I_i\omega_i=I_f\omega_f (equation 4.43).
I_i\omega_i=I_f\omega_f
Treat the Earth as a uniform solid sphere of mass M and radius R, so that I=\dfrac{2}{5}MR^{2} (Fig. 4.13).
I=\dfrac{2}{5}MR^{2}
Halving the diameter halves R, so the moment of inertia becomes one quarter of its original value, because I\propto R^{2}.
I_f=\dfrac{2}{5}M\left(\dfrac{R}{2}\right)^{2}=\dfrac{1}{4}I_i
Substitute into the conservation law to find the new angular velocity.
\omega_f=\dfrac{I_i}{I_f}\,\omega_i=4\,\omega_i

The Earth would spin four times faster, so one rotation (a day) would take about a quarter of its present length, about 6 hours. This follows from the uniform-sphere model used in Fig. 4.13.

\omega_f=4\,\omega_i

4.8

Why can a tangential acceleration in circular motion change the magnitude of the velocity but not its direction?

Solution

In circular motion the linear velocity is along the tangent to the circle at each point (section 4.1.5). The tangential acceleration a_T=r\alpha (equation 4.17) also acts along that tangent, so it is parallel or anti-parallel to the velocity.

A component of acceleration along the velocity changes only its magnitude, v_f=v_i+a_T\,t. The direction of the velocity changes only through the radial (centripetal) component a_c=r\omega^{2}, which is perpendicular to the velocity.

4.9

Why is the value of artificial gravity usually smaller than 9.8\,\mathrm{m\,s^{-2}}?

Solution

Artificial gravity is the centripetal acceleration at the rim of a rotating station, a_c=\dfrac{v^{2}}{R}=R\omega^{2} (equations 4.52 and 4.53). To reach 9.8\,\mathrm{m\,s^{-2}} the station would need a larger radius or a faster rotation than is usually practical.

The size of a station and the rate at which it can safely spin are limited in practice, so a_c is usually less than g. The artificial gravity is then only a partial substitute for the Earth's gravity.

4.10

Why is a gyroscope used in aeroplanes?

Solution

A gyroscope is a spinning wheel with a large angular momentum. Unless an external torque acts on it, its angular momentum stays constant, so its axis keeps pointing in the same direction (section 4.4.4).

This gives aeroplanes a reference direction that does not change with the motion of the aircraft, which the navigation systems use to keep track of the aeroplane's orientation.

4.11

How does the rotation of a flywheel help to even out the power delivery from the engine?

Solution

A flywheel is a heavy wheel with a large moment of inertia. When the engine delivers more energy than is needed, the flywheel absorbs it and speeds up, storing the energy as rotational energy. When the engine delivers less, the flywheel gives this energy back and slows down slightly (section 4.4.4, Fig. 4.19).

Because its moment of inertia is large, a given torque changes its angular velocity only slowly (\alpha=\tau/I). The output shaft therefore receives an almost steady power, and the power delivery is smoothed.

4.12

A wall clock's arms show time as 09:15. Express the angle between the arms in radians.

Solution

The clock face has 12 hours in 360^{\circ}, so each hour mark is 30^{\circ} apart.
30^{\circ}\text{ per hour}
The minute hand at 15 minutes is a quarter of the way round from 12.
\theta_{\text{min}}=\dfrac{15}{60}\times360^{\circ}=90^{\circ}
The hour hand at 9:15 has passed 9 hour marks, and it has moved a quarter of the way to the next mark.
\theta_{\text{hour}}=9\times30^{\circ}+\dfrac{15}{60}\times30^{\circ}=277.5^{\circ}
The difference is the angle between the arms: the smaller of the two angles is 360^{\circ}-187.5^{\circ}.
\Delta\theta=277.5^{\circ}-90^{\circ}=187.5^{\circ}\;\Rightarrow\;360^{\circ}-187.5^{\circ}=172.5^{\circ}
Convert to radians using 180^{\circ}=\pi\,\mathrm{rad}.
\theta=172.5^{\circ}\times\dfrac{\pi}{180^{\circ}}\approx3.01\,\mathrm{rad}

\theta\approx3.01\,\mathrm{rad}

Note. The arms enclose 172.5^{\circ} on the smaller side, which is 3.01\,\mathrm{rad}. The reflex angle on the other side is 187.5^{\circ}\approx3.27\,\mathrm{rad}. The book prints no answer.

Comprehensive questionsp. 75–76

Answer the following questions in detail.

4.1

What is centripetal force? Explain. Write down at least two applications where centripetal force plays its role.

Solution

Centripetal force

The net force that causes a particle to undergo centripetal acceleration is called the centripetal force F_c. It is always directed towards the centre of the circle. In uniform circular motion it changes the direction of the velocity but not its speed.

By Newton's second law, F_c=ma_c (equation 4.20). With a_c=\dfrac{v^{2}}{r} this gives:

F_c=\dfrac{mv^{2}}{r}=mr\omega^{2}

Why a centripetal acceleration exists

Derivation Centripetal acceleration

Aim. Show that a body moving with constant speed v in a circle of radius r has acceleration a_c=\dfrac{v^{2}}{r} directed to the centre.

The velocity is a tangent at P, and the centripetal acceleration points at the centre O.

The velocity is along the tangent, so it is perpendicular to the radius at every point. Its magnitude stays constant, so |\vec v_i|=|\vec v_f|=v.
\vec v\perp\vec r,\qquad |\vec v_i|=|\vec v_f|=v
The velocity triangle (tail to tail, Fig. 4.9) and the position triangle are similar, because the angle \Delta\theta between the radii is the same as the angle between the velocities.
\dfrac{\Delta v}{v}=\dfrac{\Delta r}{r}
Rearrange this to find the change in velocity.
\Delta v=\dfrac{v}{r}\,\Delta r
Divide by \Delta t and let \Delta t\to0, so that \dfrac{\Delta r}{\Delta t}\to v.
a_c=\lim_{\Delta t\to0}\dfrac{\Delta v}{\Delta t}=\dfrac{v}{r}\,v=\dfrac{v^{2}}{r}
As \Delta t\to0 the change \Delta\vec v points towards the centre, so the acceleration does too. With v=r\omega this is also a_c=r\omega^{2}.
a_c=\dfrac{v^{2}}{r}=r\omega^{2}\quad\text{towards the centre}

Resulta_c=\dfrac{v^{2}}{r}=r\omega^{2}, \text{ directed towards the centre}

Acceleration exists even at constant speed, because the direction of the velocity keeps changing.

Applications

Tension in a string: a ball swung in a horizontal circle on a string. The tension supplies the centripetal force, T=\dfrac{mv^{2}}{r}.
Gravitation: the Moon moves around the Earth, and planets move around the Sun. Gravity supplies the centripetal force, F_g=\dfrac{mv^{2}}{r}.
Friction: a car turning on an unbanked road. The static friction between the tyres and the road supplies F_c.
Normal force on a banked road: the normal force has a horizontal component F_N\sin\theta towards the centre, giving \tan\theta=\dfrac{v^{2}}{rg} (equation 4.23).
Electric force: the force that keeps the disc moving in a CD player and the parts of a computer hard disk moving in a circle.

4.2

What is moment of inertia? Derive its relation for rigid body.

Solution

Moment of inertia

The moment of inertia I of a body about a fixed axis is the property by which it maintains its state of rest or uniform rotational motion about that axis. It is the rotational equivalent of mass: a body with a larger moment of inertia is harder to angularly accelerate. For a point mass m at perpendicular distance r from the axis:

I=mr^{2}

Its SI unit is \mathrm{kg\,m^{2}}.

Moment of inertia of a rigid body

Derivation Moment of inertia of a rigid body

Aim. Find the moment of inertia of a rigid body about a fixed axis.

Assumptions.

The body is rigid, so the particles keep fixed distances from one another.
The axis is fixed.

Each small mass m_i at distance r_i from the axis O adds m_ir_i^{2} to the moment of inertia.

Divide the body into n small masses m_1,m_2,\dots,m_n at distances r_1,r_2,\dots,r_n from the axis. Each small mass is a point mass, so its moment of inertia is I_i=m_ir_i^{2} (equation 4.24).
I_i=m_ir_i^{2}
The whole body is the sum of its parts, all about the same axis.
I=I_1+I_2+\cdots+I_n
Write the sum with the summation sign.
I=\sum_{i=1}^{n}m_ir_i^{2}
Because the body is rigid, the distances r_i stay fixed while it rotates, so I is a constant for that axis.

ResultI=\sum_{i=1}^{n}m_ir_i^{2}

The moment of inertia depends on the mass, on how the mass is spread out, and on the axis. Mass far from the axis contributes most because it enters as r^{2}.

For uniform bodies of mass M the sum gives the standard values in Fig. 4.13, for example I=\dfrac{1}{2}MR^{2} for a solid cylinder or disc, I=MR^{2} for a ring or hoop, and I=\dfrac{1}{12}ML^{2} for a thin rod about an axis through its centre.

4.3

Derive the expression for angular momentum of a body. Also deduce the relation between angular momentum and torque.

Solution

Angular momentum of a point mass

Derivation Angular momentum of a point mass

Aim. Derive L=I\omega for a point mass m moving in a circle of radius r.

For a point mass the velocity p is perpendicular to the radius r, so L = rp.

Angular momentum is the cross product of the position vector and the linear momentum, \vec L=\vec r\times\vec p, with magnitude L=rp\sin\theta.
L=rp\sin\theta
The velocity is tangent to the circle, so the angle between \vec r and \vec p is \theta=90^{\circ} and \sin90^{\circ}=1.
L=rp
Linear momentum is p=mv (equation 4.27).
p=mv
For circular motion v=r\omega (equation 4.28), so p=mr\omega (equation 4.29).
p=mr\omega
Substitute p into L=rp.
L=r\,(mr\omega)=mr^{2}\omega

ResultL=mr^{2}\omega=I\omega

Angular momentum is the product of moment of inertia and angular velocity, just as linear momentum is mass times velocity.

Angular momentum of a rigid body

Derivation Angular momentum of a rigid body

Aim. Find the angular momentum of a rigid body rotating about a fixed axis.

The net angular momentum is the sum of the angular momenta of all the small masses (equation 4.31).
L_{\text{net}}=L_1+L_2+\cdots+L_n
Each small mass has L_i=m_ir_i^{2}\omega_i (equations 4.32 to 4.35).
L_{\text{net}}=m_1r_1^{2}\omega_1+\cdots+m_nr_n^{2}\omega_n
All points of a rigid body turn with the same angular velocity, so \omega_1=\omega_2=\cdots=\omega.
\omega_i=\omega
Take \omega outside the sum.
L_{\text{net}}=\Bigl(\sum_{i=1}^{n}m_ir_i^{2}\Bigr)\omega
The bracket is the moment of inertia of the rigid body.
L_{\text{net}}=I\omega

ResultL=I\omega

The angular momentum of a rigid body has the same form as p=mv for linear motion (equation 4.37).

Relation between angular momentum and torque

Derivation Torque and the rate of change of angular momentum

Aim. Show that the torque equals the rate of change of angular momentum.

Start from the definition of angular momentum, \vec L=\vec r\times\vec p.
\vec L=\vec r\times\vec p
Divide both sides by \Delta t (equation 4.38).
\dfrac{\Delta\vec L}{\Delta t}=\vec r\times\dfrac{\Delta\vec p}{\Delta t}
Newton's second law in terms of momentum gives \vec F=\dfrac{\Delta\vec p}{\Delta t} (equation 4.39).
\dfrac{\Delta\vec p}{\Delta t}=\vec F
Substitute, and use the definition of torque \vec\tau=\vec r\times\vec F (equation 4.41).
\dfrac{\Delta\vec L}{\Delta t}=\vec r\times\vec F=\vec\tau

Result\tau=\dfrac{\Delta L}{\Delta t}

Torque is the rate of change of angular momentum. With no torque, the angular momentum is constant (section 4.4.4).

4.4

Explain conservation of angular momentum using practical life examples.

Solution

Law of conservation of angular momentum

In the absence of any external torque, the angular momentum of a system remains constant. Since \dfrac{\Delta L}{\Delta t}=\tau, no torque means \Delta L=0, so the final angular momentum equals the initial angular momentum:

L_f=L_i\quad\text{or}\quad I_f\omega_f=I_i\omega_i

For a body whose moment of inertia changes, a decrease in I must be matched by an increase in \omega, and an increase in I by a decrease in \omega.

Practical examples

Spinning ice skater: with arms extended, I is large and \omega is small. Pulling the arms in makes I smaller, so \omega increases (Fig. 4.16).
Divers and gymnasts: they leave the board with a spin given by the board's torque. Tucking the body reduces I and increases \omega; opening out increases I and reduces \omega, which controls somersaults and twists (Fig. 4.17).
Turntable with dumbbells: holding the dumbbells close to the body makes the turntable spin faster. Extending the arms reduces the angular velocity (Activity, section 4.4.4).
Flywheel: a heavy spinning wheel keeps its angular momentum, so it smooths out the power delivered by an engine.
Gyroscope: the spinning wheel keeps its orientation relative to the Earth's axis. The torque of gravity makes the axis precess rather than fall (Fig. 4.18).

4.52026 Annual I, Set B, Q.2 (x) OR2026 Annual I, Set D, Q.2 (xi) OR2025 Annual I, Set B, Q.2 (xiv)2025 Annual I, Set D, Q.2 (vi)

Derive the relation between torque and angular acceleration.

Solution

Torque on a point mass

Derivation Torque and angular acceleration

Aim. Derive \tau=I\alpha for a point mass m at distance r from the axis, driven by a tangential force F.

Assumptions.

The force acts tangentially, so it is perpendicular to the radius.
The axis is fixed.

The tangential force F has moment arm r about the axis, so its torque is \tau=rF.

Torque of a force is \tau=rF\sin\theta (equation 4.44). The force is tangential, so \theta=90^{\circ} and \sin90^{\circ}=1.
\tau=rF
Newton's second law along the tangent gives F=ma_T (equation 4.46).
F=ma_T
The tangential acceleration is related to the angular acceleration by a_T=r\alpha (equation 4.47).
F=mr\alpha
Substitute F into the expression for torque.
\tau=r(mr\alpha)=mr^{2}\alpha
The term mr^{2} is the moment of inertia of the point mass.
\tau=I\alpha

Result\tau=I\alpha

Torque is to angular acceleration what force is to linear acceleration (F=ma). A larger I needs a larger torque for the same \alpha.

The same result holds for a rigid body, because the torques on its parts add and every part has the same angular acceleration: \tau_{\text{net}}=\sum m_ir_i^{2}\,\alpha=I\alpha.

4.6

Justify how a centrifuge is used to separate materials using centripetal force.

Solution

A centrifuge separates substances suspended in a liquid by spinning the sample very quickly about an axle. The liquid in the tube applies a centripetal force on each particle, keeping it moving in a circle of radius r with angular velocity \omega. The force needed is F_c=mr\omega^{2} (equation 4.21).

A denser particle has more mass, so it needs a larger centripetal force for the same r and \omega. The liquid cannot supply that much force, so the particle does not follow the circle. By Newton's first law it moves in a straight line outwards, away from the axis, until it reaches the wall at the bottom of the test tube (Fig. 4.11).

After running at high speed, the particles are clumped at the bottom of the tube. They can then be collected and analysed. The same principle is used in cream separators, which separate milk into cream and skimmed milk, and in washing machine dryers.

4.7

Explain why the objects in orbiting satellites appear to be weightless.

Solution

A satellite and everything in it are falling freely towards the Earth under gravity. Gravity supplies the centripetal force that keeps them in orbit (F_g=\dfrac{mv^{2}}{r}), and their sideways speed keeps them moving round the Earth rather than falling into it (section 4.6).

The astronaut and the station have the same acceleration towards the Earth, so the floor does not press on the astronaut and there is no contact force. Weight is felt only through a contact force, so the apparent weight is zero.

N=0\quad\Rightarrow\quad\text{apparent weight}=0

The gravitational field at the height of most space stations is still strong, about 95\,\% of its value at the surface. Weightlessness therefore comes from free fall, not from the absence of gravity.

4.8

Describe how artificial gravity can be produced in a satellite to counter weightlessness.

Solution

Artificial gravity is produced by rotating a space station about its own axis. The occupants live on the outer rim, a tube or ring of radius R. The outer wall of the rim pushes inwards on them and provides the centripetal force that keeps them moving in a circle (Fig. 4.22). They feel this push as weight, because the centripetal acceleration acts like gravity.

Derivation Speed and angular velocity for artificial gravity

Aim. Find the speed and angular velocity needed for the centripetal acceleration at the rim to equal g.

The rim pushes the occupant inwards, and this inward acceleration is felt as gravity.

The centripetal acceleration at the rim is a_c=\dfrac{v^{2}}{R} (equation 4.52).
a_c=\dfrac{v^{2}}{R}
For the occupants to feel the same as at the Earth's surface, a_c must equal g.
a_c=g
Solve for the speed.
v=\sqrt{gR}\qquad(4.53)
Use v=\omega R (equation 4.54) to find the angular velocity.
\omega=\dfrac{v}{R}=\sqrt{\dfrac{g}{R}}\qquad(4.55)
The time period is T=\dfrac{2\pi}{\omega}=2\pi\sqrt{\dfrac{R}{g}} (equations 4.56 and 4.57).
T=2\pi\sqrt{\dfrac{R}{g}}

Result\omega=\sqrt{\dfrac{g}{R}},\qquad v=\sqrt{gR}

A larger radius needs a smaller spin rate for the same artificial gravity.

4.9

Analyse motion in curved path due to perpendicular forces.

Solution

Perpendicular force and the path

A body moves in a curved path when a force acts on it at an angle to its velocity. A force that is exactly perpendicular to the velocity changes the direction of the velocity but not its magnitude at that instant. The force therefore bends the path towards itself without speeding the body up or slowing it down.

The tension acts along the string towards O, perpendicular to the velocity v, so it bends the path without changing the speed.

Qualitative analysis

At any instant the velocity is along the tangent to the path, so the perpendicular force has no component along the velocity.
If the perpendicular force has a constant magnitude F and always stays perpendicular to the velocity, the body moves in a uniform circle. Its radius follows from F=F_c=\dfrac{mv^{2}}{r}.
r=\dfrac{mv^{2}}{F}
If the force is not perpendicular, it has two parts: a tangential part, which changes the speed (a_T=r\alpha), and a perpendicular part, which bends the path (a_c=\dfrac{v^{2}}{r}).

Examples

Stone whirled on a string: the tension acts along the string, towards the centre, perpendicular to the velocity. It keeps the stone on a circle while its speed stays nearly the same.
Car on a banked curve: the normal force has a component F_N\sin\theta perpendicular to the velocity and towards the centre, which provides the centripetal force (Fig. 4.10).
Satellites and planets: the gravitational force is perpendicular to the velocity of a body moving in a nearly circular orbit, so it bends the path into an orbit.

In every case the path curves towards the side on which the force acts, and the curvature is set by the perpendicular force through F=\dfrac{mv^{2}}{r}.

Numerical problemsp. 76

4.1

What will be the angular velocity of fly wheel of an engine if it completes 3000 revolutions in a minute?

Solution

Given

N = 3000\,\mathrm{rev\,min^{-1}}

To find

\omega in \mathrm{rad\,s^{-1}}

Convert the number of revolutions per minute to revolutions per second.
f=\dfrac{3000}{60}=50\,\mathrm{rev\,s^{-1}}
One revolution is 2\pi rad, so \omega=2\pi f.
\omega=2\pi\times50\,\mathrm{s^{-1}}\approx314\,\mathrm{rad\,s^{-1}}

\omega\approx314\,\mathrm{rad\,s^{-1}}

Book answer. 314\,\mathrm{rad\,s^{-1}}

4.2

A car is passing through a turn that is in the form of an arc of a circle of radius 14.5\,\mathrm{m}. What will be the maximum speed limit (the speed at which the car can cross the bridge without losing contact with the road) if the centre of gravity of the car is 0.5\,\mathrm{m} from the ground?

Solution

Given

R = 14.5\,\mathrm{m}
h = 0.5\,\mathrm{m} height of the centre of gravity above the road
g = 9.8\,\mathrm{m\,s^{-2}}

To find

v_{\max}

The car loses contact with the road when the normal force falls to zero. Then the weight alone provides the centripetal force: mg=\dfrac{mv^{2}}{r}, so v=\sqrt{gr}.
v=\sqrt{gr}
The centre of gravity is 0.5\,\mathrm{m} above the road, so it moves on a circle of radius r=R+h, which is 14.5\,\mathrm{m}+0.5\,\mathrm{m}=15.0\,\mathrm{m}.
r=14.5+0.5=15.0\,\mathrm{m}
Substitute.
v=\sqrt{9.8\times15.0}=\sqrt{147}\approx12.1\,\mathrm{m\,s^{-1}}

v_{\max}\approx12.1\,\mathrm{m\,s^{-1}}

Book answer. 12.1\,\mathrm{m\,s^{-1}}

Note. The answer agrees with the book when the radius of the centre-of-gravity path is taken as 14.5+0.5=15.0\,\mathrm{m}. Using 14.5\,\mathrm{m} directly would give 11.9\,\mathrm{m\,s^{-1}}. The question's wording mixes a turn and a bridge; the bridge (convex) case is the one that loses contact.

4.3

A PT teacher rotates his stick at the axis that passes through its centre. If mass of the stick is 200\,\mathrm{g} and its length is 0.8\,\mathrm{m}, find its moment of inertia.

Solution

Given

M = 200\,\mathrm{g}=0.200\,\mathrm{kg}
L = 0.8\,\mathrm{m}

To find

I about an axis through the centre

A uniform rod rotated about an axis through its centre has I=\dfrac{1}{12}ML^{2} (Fig. 4.13).
I=\dfrac{1}{12}ML^{2}
Substitute the values.
I=\dfrac{1}{12}\times0.200\times(0.8)^{2}=\dfrac{1}{12}\times0.128
Evaluate.
I\approx1.07\times10^{-2}\,\mathrm{kg\,m^{2}}

I\approx1.1\times10^{-2}\,\mathrm{kg\,m^{2}}

Book answer. 0.01\,\mathrm{kg\,m^{2}}

Note. The book's 0.01\,\mathrm{kg\,m^{2}} is the value 1.07\times10^{-2} rounded to one significant figure. The answer here keeps two significant figures, 1.1\times10^{-2}\,\mathrm{kg\,m^{2}}. This is a rounding difference, not a disagreement.

4.4

A football of mass 450\,\mathrm{g} rotates with an angular speed of 10\,\mathrm{rev\,s^{-1}}. If its radius is 11\,\mathrm{cm}, compute its angular momentum.

Solution

Given

M = 450\,\mathrm{g}=0.450\,\mathrm{kg}
R = 11\,\mathrm{cm}=0.11\,\mathrm{m}
f = 10\,\mathrm{rev\,s^{-1}}

To find

Angular velocity from the frequency: one revolution is 2\pi rad.
\omega=2\pi f=2\pi\times10=62.8\,\mathrm{rad\,s^{-1}}
Treat the football as a uniform solid sphere, whose moment of inertia is I=\dfrac{2}{5}MR^{2} (Fig. 4.13). This is the model that reproduces the book's answer.
I=\dfrac{2}{5}MR^{2}
Substitute for I.
I=0.4\times0.450\times(0.11)^{2}=2.18\times10^{-3}\,\mathrm{kg\,m^{2}}
Angular momentum is L=I\omega (equation 4.30).
L=I\omega=2.18\times10^{-3}\times62.8
Evaluate.
L\approx0.137\,\mathrm{J\,s}

L\approx0.137\,\mathrm{J\,s}

Book answer. 0.137\,\mathrm{J\,s}

Note. The book's answer is reproduced by the solid-sphere moment of inertia \dfrac{2}{5}MR^{2}. A hollow spherical shell (\dfrac{2}{3}MR^{2}) would give 0.228\,\mathrm{J\,s}, so the answer depends on the model chosen for the ball. The question does not state the model.

4.5

A merryman in a circus is standing with his arms extended on a turn table rotating with angular velocity 10\,\mathrm{rad/s}. He brings his arms closer to his body so that his moment of inertia is reduced to one third of the initial value. Find his new angular velocity.

Solution

Given

\omega_1 = 10\,\mathrm{rad\,s^{-1}}
I_2 = \dfrac{1}{3}I_1

To find

\omega_2

No external torque acts, so the angular momentum is conserved (equation 4.43).
I_1\omega_1=I_2\omega_2
Solve for the new angular velocity.
\omega_2=\dfrac{I_1}{I_2}\,\omega_1=\dfrac{I_1}{I_1/3}\,\omega_1=3\,\omega_1
Substitute.
\omega_2=3\times10=30\,\mathrm{rad\,s^{-1}}

\omega_2=30\,\mathrm{rad\,s^{-1}}

Book answer. 30\,\mathrm{rad\,s^{-1}}

4.6

A boy exerts a force of 200\,\mathrm{N} at the edge of the 30.0\,\mathrm{kg} merry-go-round, which has a 2.0\,\mathrm{m} radius. Calculate the angular acceleration produced (a) when no one is on the merry-go-round and (b) when the boy, having 20.0\,\mathrm{kg} weight, sits 1.5\,\mathrm{m} away from the centre. (Ignore friction.)

Solution

Given

F = 200\,\mathrm{N} applied at the edge
M = 30.0\,\mathrm{kg}
R = 2.0\,\mathrm{m}
m = 20.0\,\mathrm{kg} the boy, in (b)
r = 1.5\,\mathrm{m} distance of the boy from the centre, in (b)

To find

(a) \alpha with no one on it
(b) \alpha with the boy on it

(a) No one on the merry-go-round

The merry-go-round is a uniform disc, so I_{\text{disc}}=\dfrac{1}{2}MR^{2} (Fig. 4.13).
I=\dfrac{1}{2}\times30.0\times(2.0)^{2}=60.0\,\mathrm{kg\,m^{2}}
The force acts at the edge, so its torque is \tau=FR (equation 4.45).
\tau=200\times2.0=400\,\mathrm{N\,m}
Use \tau=I\alpha (equation 4.50).
\alpha=\dfrac{\tau}{I}=\dfrac{400}{60.0}\approx6.67\,\mathrm{rad\,s^{-2}}

(b) The boy sits 1.5 m from the centre

The boy is a point mass, with moment of inertia mr^{2} (equation 4.24).
I_{\text{boy}}=20.0\times(1.5)^{2}=45.0\,\mathrm{kg\,m^{2}}
The moments of inertia add, so the total is I=60.0+45.0.
I=105\,\mathrm{kg\,m^{2}}
The force is still applied at the edge, so the torque is still 400\,\mathrm{N\,m}.
\alpha=\dfrac{400}{105}\approx3.81\,\mathrm{rad\,s^{-2}}

(a)\ \alpha\approx6.67\,\mathrm{rad\,s^{-2}};\quad (b)\ \alpha\approx3.8\,\mathrm{rad\,s^{-2}}

Book answer. 6.67\,\mathrm{rad\,s^{-2}}, 3.8\,\mathrm{rad\,s^{-2}}

Note. The book gives the boy's 20.0\,\mathrm{kg} as a 'weight'; the value is a mass in kilograms and is used as mass here. Both answers agree with the book.

4.7

A wheel-shape space station provides an artificial gravity of 5.00\,\mathrm{m\,s^{-2}} to its inhabitants. If it has a diameter of 100\,\mathrm{m}, find its angular speed in rpm.

Solution

Given

a_c = 5.00\,\mathrm{m\,s^{-2}}
D = 100\,\mathrm{m}
R = 50.0\,\mathrm{m}

To find

\omega in rpm

The artificial gravity is the centripetal acceleration at the rim, a_c=R\omega^{2}.
\omega=\sqrt{\dfrac{a_c}{R}}=\sqrt{\dfrac{5.00}{50.0}}=\sqrt{0.100}\approx0.316\,\mathrm{rad\,s^{-1}}
Convert rad/s to revolutions per minute: 1\,\mathrm{rev}=2\pi\,\mathrm{rad} and 1\,\mathrm{min}=60\,\mathrm{s}.
n=\dfrac{0.316\times60}{2\pi}\approx3.0\,\mathrm{rpm}

n\approx3.0\,\mathrm{rpm}

Book answer. 3\,\mathrm{rpm}

4.8

The minute hand of a watch is 2\,\mathrm{cm} long. If it travels 9.4\,\mathrm{cm} of length, how many radians will it travel?

Solution

Given

r = 2\,\mathrm{cm}
s = 9.4\,\mathrm{cm}

To find

\theta in radians

The arc length is s=r\theta (equation 4.9), so \theta=\dfrac{s}{r}.
\theta=\dfrac{9.4\,\mathrm{cm}}{2\,\mathrm{cm}}
Evaluate. The centimetres cancel, so the angle is a pure number of radians.
\theta=4.7\,\mathrm{rad}

\theta=4.7\,\mathrm{rad}

Book answer. 4.7\,\mathrm{rad}

4.9

How many revolutions does the fidget spinner make after being flicked with an initial angular velocity of 10 revolutions per second and coming to rest in 5 seconds? (Ignore air resistance.)

Solution

Given

\omega_i = 10\,\mathrm{rev\,s^{-1}}
\omega_f = 0
t = 5\,\mathrm{s}

To find

N, the number of revolutions

The spinner slows down uniformly, so the angular acceleration is constant: \alpha=\dfrac{\omega_f-\omega_i}{t}.
\alpha=\dfrac{0-10}{5}=-2\,\mathrm{rev\,s^{-2}}
The average angular velocity is \dfrac{\omega_i+\omega_f}{2}, so the angular displacement is the average multiplied by the time.
\theta=\dfrac{\omega_i+\omega_f}{2}\,t=\dfrac{10+0}{2}\times5=25\,\mathrm{rev}
Check with \theta=\omega_i t+\dfrac{1}{2}\alpha t^{2} (Table 4.1).
\theta=10\times5+\dfrac{1}{2}(-2)(5)^{2}=50-25=25\,\mathrm{rev}

N=25\ \text{revolutions}\;(\approx157\,\mathrm{rad})

Book answer. 25 revolutions

Note. Uniform deceleration is assumed, which the question implies by asking for the time to rest.

FBISE 2025 and 2026 questions

Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2025 Annual I, Set B, MCQ 41 mark

Magnitude of centripetal force F_c =

Amr\omega^2Answer
Bmr\omega
Cm\omega^2
Dmv^2

Why

For motion in a circle the centripetal force is F_c=\dfrac{mv^{2}}{r}. Put v=r\omega: F_c=\dfrac{m(r\omega)^{2}}{r}=mr\omega^{2}. Option C (m\omega^{2}) has lost the radius, so it has the wrong dimensions; option D (mv^{2}) is missing the factor 1/r.

2025 Annual I, Set B, MCQ 51 mark

If the length of blade of a fan is increased, then its rotational speed (at same voltage and current):

AWill increase
BWill decreaseAnswer
CWill become infinite
DRemains constant

Why

The blades act like a rotating rigid body whose moment of inertia is I=\sum mr^{2}. Longer blades put the mass at larger r, so I increases a lot (it goes as r^{2}). The motor gives the same torque for the same voltage and current, and \alpha=\tau/I, so the angular acceleration and the speed reached are smaller. The tempting option A is wrong because a bigger fan does not get more torque from the motor; it only gets more rotational inertia (and more air resistance).

2026 Annual I, Set B, MCQ 71 mark

Direction of centripetal force is always:

ATangential
BPerpendicular
CRadially inwardAnswer
DOutward

Why

In circular motion the centripetal force is the force that keeps the body on its circle. It must be at right angles to the velocity, which is along the tangent, so that it changes the direction of the velocity without changing the speed. It therefore points along the radius towards the centre, that is, radially inward. Option A is the direction of the velocity (a tangential force would change the speed, not the direction). Option D, outward, is the direction of the so-called centrifugal effect, which is not a real force acting on the body. Option B is true in the sense that the force is perpendicular to the velocity, but it does not give the direction, and 'radially inward' is the complete answer.

2026 Annual I, Set B, MCQ 161 mark

If distance from axis increases, moment of inertia:

ADecreases
BIncreasesAnswer
CRemains constant
DBecomes zero

Why

The moment of inertia of a point mass is I=mr^{2}, and for a rigid body I=\sum m_ir_i^{2}. If the distance r from the axis increases, I increases, and it increases as the square of the distance. For example, a ring (MR^{2}) is harder to spin than a disc of the same mass and radius (\tfrac12MR^{2}) because its mass is farther from the axis. Option A is the tempting wrong one, because we often think that a larger distance gives a larger lever arm and so an easier turn, but that is the torque; the rotational inertia grows with distance.

2026 Annual I, Set B, MCQ 171 mark

Correct relation between torque and angular acceleration is:

A\tau = \dfrac{IR}{\alpha}
B\tau = \dfrac{\alpha}{I}
C\tau = \dfrac{I}{2}\alphaAnswer
D\tau = m\alpha

Why

The relation between torque and angular acceleration is \tau=I\alpha (torque is the rotational equivalent of F=ma). Of the four printed options only C has this form, \tau equal to a moment-of-inertia term multiplied by \alpha. Option A has \alpha in the denominator, option B is the inverse relation (\alpha=\tau/I would be the correct one for that arrangement, not \tau=\alpha/I), and option D uses the mass m instead of the moment of inertia, so it has the wrong dimensions (kg s^{-2} instead of N m).

2025 Annual I, Set B, Q.2 (iv) OR1+2 = 3 marks

Earth completes its one rotation about its axis in 24 hours. Calculate angular velocity of earth about its axis in radian per second.

Solution

Given

T = 24\,\mathrm{h}=24\times3600=86\,400\,\mathrm{s} time for one rotation
\theta = 2\pi\,\mathrm{rad} angle turned in one rotation

To find

\omega

Formula (1 mark). Angular velocity is the angle turned per unit time.
\omega=\dfrac{\Delta\theta}{\Delta t}=\dfrac{2\pi}{T}
Substitution (1 mark).
\omega=\dfrac{2\times3.14}{86\,400}
Answer with unit (1 mark).
\omega=7.27\times10^{-5}\,\mathrm{rad\,s^{-1}}

\omega=7.27\times10^{-5}\,\mathrm{rad\,s^{-1}}

2026 Annual I, Set B, Q.2 (ix) OR1+1+1 = 3 marks

The front wheel of a tractor travels 700 revolutions while the rear wheel 280 in a time interval of 40 seconds. Find their angular velocities.

Solution

Given

N_f = 700\ \text{rev} front wheel
N_r = 280\ \text{rev} rear wheel
t = 40\,\mathrm{s} time interval

To find

\omega_f (front)
\omega_r (rear)

Formula (1 mark). One revolution is 2\pi rad, so the angular displacement is \theta=2\pi N, and the angular velocity is the angular displacement per unit time.
\omega=\dfrac{\theta}{t}=\dfrac{2\pi N}{t}
Front wheel (1 mark).
\omega_f=\dfrac{2\pi\times700}{40}=35\pi=110\,\mathrm{rad\,s^{-1}}
Rear wheel (1 mark).
\omega_r=\dfrac{2\pi\times280}{40}=14\pi=44\,\mathrm{rad\,s^{-1}}

\omega_f=110\,\mathrm{rad\,s^{-1}},\quad\omega_r=44\,\mathrm{rad\,s^{-1}}

2026 Annual I, Set B, Q.2 (x) OR3 marks

Drive a relationship between torque and angular acceleration.

Solution

Torque on a point mass

Derivation Torque and angular acceleration

Aim. Derive \tau=I\alpha for a point mass m at distance r from the axis, driven by a tangential force F.

Assumptions.

The force acts tangentially, so it is perpendicular to the radius.
The axis is fixed.

The tangential force F has moment arm r about the axis, so its torque is \tau=rF.

Torque of a force is \tau=rF\sin\theta (equation 4.44). The force is tangential, so \theta=90^{\circ} and \sin90^{\circ}=1.
\tau=rF
Newton's second law along the tangent gives F=ma_T (equation 4.46).
F=ma_T
The tangential acceleration is related to the angular acceleration by a_T=r\alpha (equation 4.47).
F=mr\alpha
Substitute F into the expression for torque.
\tau=r(mr\alpha)=mr^{2}\alpha
The term mr^{2} is the moment of inertia of the point mass.
\tau=I\alpha

Result\tau=I\alpha

Torque is to angular acceleration what force is to linear acceleration (F=ma). A larger I needs a larger torque for the same \alpha.

The same result holds for a rigid body, because the torques on its parts add and every part has the same angular acceleration: \tau_{\text{net}}=\sum m_ir_i^{2}\,\alpha=I\alpha.

2025 Annual I, Set B, Q.2 (xiv)1x3 = 3 marks

Derive the relation between torque, moment of inertia and angular acceleration.

Solution

Torque on a point mass

Derivation Torque and angular acceleration

Aim. Derive \tau=I\alpha for a point mass m at distance r from the axis, driven by a tangential force F.

Assumptions.

The force acts tangentially, so it is perpendicular to the radius.
The axis is fixed.

The tangential force F has moment arm r about the axis, so its torque is \tau=rF.

Torque of a force is \tau=rF\sin\theta (equation 4.44). The force is tangential, so \theta=90^{\circ} and \sin90^{\circ}=1.
\tau=rF
Newton's second law along the tangent gives F=ma_T (equation 4.46).
F=ma_T
The tangential acceleration is related to the angular acceleration by a_T=r\alpha (equation 4.47).
F=mr\alpha
Substitute F into the expression for torque.
\tau=r(mr\alpha)=mr^{2}\alpha
The term mr^{2} is the moment of inertia of the point mass.
\tau=I\alpha

Result\tau=I\alpha

Torque is to angular acceleration what force is to linear acceleration (F=ma). A larger I needs a larger torque for the same \alpha.

The same result holds for a rigid body, because the torques on its parts add and every part has the same angular acceleration: \tau_{\text{net}}=\sum m_ir_i^{2}\,\alpha=I\alpha.

2025 Annual I, Set B, Q.61+1+4 = 6 marks

Why does an astronaut feel weightless in space? How can artificial gravity be produced in a satellite? For a satellite derive expression for angular velocity, time period and frequency of satellite.

Solution

Why an astronaut feels weightless (1 mark)

A satellite and everything inside it are in free fall towards the Earth. Gravity supplies the centripetal force that keeps the satellite in its orbit, and the satellite and the astronaut have the same acceleration. The floor therefore does not push on the astronaut: the contact force (apparent weight) is zero, and he floats. Gravity has not disappeared; at the height of a space station it is still about 90\,\% of its surface value.

N=0\ \Rightarrow\ \text{apparent weight}=0

Artificial gravity (1 mark)

Artificial gravity is produced by rotating the satellite (space station) about its own axis. The outer wall of the rotating rim pushes the occupants inwards and provides the centripetal force that keeps them on a circle. They feel this contact force as weight, so the centripetal acceleration plays the part of g.

Derivation (4 marks)

Derivation Angular velocity, time period and frequency of the rotating satellite

Aim. Find \omega, T and f for a satellite of radius R whose rim has centripetal acceleration equal to g.

Assumptions.

The occupants stand on the outer rim, at distance R from the axis.
The centripetal acceleration at the rim is made equal to g.

The rim pushes the occupant inwards, and this inward acceleration is felt as gravity.

The centripetal acceleration at the rim is a_c=\dfrac{v^{2}}{R}. For the same effect as gravity, a_c=g.
g=\dfrac{v^{2}}{R}\ \Rightarrow\ v=\sqrt{gR}
Use v=\omega R and compare it with v=\sqrt{gR} to get the angular velocity.
\omega R=\sqrt{gR}\ \Rightarrow\ \omega=\sqrt{\dfrac{g}{R}}
The time period is the time for one rotation, T=\dfrac{2\pi}{\omega}.
T=2\pi\sqrt{\dfrac{R}{g}}
Frequency is the reciprocal of the time period.
f=\dfrac1T=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{R}}

Result\omega=\sqrt{\dfrac{g}{R}},\qquad T=2\pi\sqrt{\dfrac{R}{g}},\qquad f=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{R}}

A larger station needs a slower spin to give the same artificial gravity. The mass of the occupant does not appear.

2026 Annual I, Set D, MCQ 11 mark

Which equation correctly relates torque (\tau), moment of inertia (I), and angular acceleration (\alpha)?

A\tau = I^2\alpha
B\tau = \dfrac{I}{\alpha}
C\tau = I + \alpha
D\tau = I\alphaAnswer

Why

Torque equals the moment of inertia times the angular acceleration, \tau=I\alpha, the rotational form of F=ma (Section 4.9 of the book). Option A squares I without reason, so the dimensions are wrong. Option B divides instead of multiplying, which would make the angular acceleration larger for a larger I; it is the other way round. Option C adds two quantities of different kinds, which is not allowed.

2025 Annual I, Set D, MCQ 41 mark

Moment of inertia of a body spinning about its axis DOES NOT depend on the:

AAngular velocity of bodyAnswer
BMass of body
COrientation of the axis
DDistribution of mass around axis

Why

The moment of inertia I=\sum mr^{2} depends on the mass of the body, on how that mass is distributed about the axis, and on where the axis is and how it is oriented (the same body has different I about different axes). It does not depend on how fast the body spins. Option C looks tempting, but moving the axis changes every distance r, so I does change with the orientation of the axis.

2026 Annual I, Set D, MCQ 81 mark

If velocity of a body moving in circular path doubles, centripetal force becomes:

AFour timesAnswer
BSame
CDouble
DHalf

Why

The centripetal force is F_c=\dfrac{mv^{2}}{r}. It is proportional to the square of the speed when m and r stay the same. If v\to2v, then F_c\to\dfrac{m(2v)^{2}}{r}=4\dfrac{mv^{2}}{r}, so the force becomes four times. Option C (double) is the tempting mistake of treating F_c as proportional to v.

2026 Annual I, Set D, MCQ 171 mark

A solid cylinder with mass M=5kg and radius R=0.3m rotates about its central axis. Calculate the moment of inertia of the cylinder.

A0.23\,\mathrm{kgm^2}Answer
B5.3\,\mathrm{kgm^2}
C4.7\,\mathrm{kgm^2}
D1.5\,\mathrm{kgm^2}

Why

For a solid cylinder (disc) about its central axis the moment of inertia is I=\tfrac12MR^{2} (Figure 4.13 of the book). Substituting,

I=\tfrac12\times5\times(0.3)^{2}=\tfrac12\times5\times0.09=0.225\approx0.23\,\mathrm{kg\,m^{2}}

The wrong options come from slips: 1.5=MR forgets to square R, and MR^{2}=0.45 forgets the factor \tfrac12.

2025 Annual I, Set D, Q.2 (iv)3 marks

What is meant by Gyroscope? Explain briefly.

Solution

Meaning (1 mark). A gyroscope is a device that uses the principle of angular momentum to maintain its orientation relative to the Earth's axis, or to resist changes in its orientation. It consists of a wheel mounted on an axle that can rotate freely, secured in a metal frame.

Explanation (2 marks). When the wheel is spun, it has a large angular momentum \vec L along its axle, and a spinning gyroscope can be balanced on a flat surface; when the wheel stops, it falls. If it is tilted it does not fall but turns slowly (precesses) about the vertical. A change in the direction of \vec L needs a torque. Here the torque comes from the weight, and the change in angular momentum is in the direction of the torque:

\Delta\vec L=\vec\tau\,\Delta t

So the size of \vec L stays the same but its direction turns round the vertical, which is the precession. Because the axle keeps pointing in a fixed direction unless a torque acts, gyroscopes are used in navigation systems (for example in aeroplanes) to hold their orientation.

2025 Annual I, Set D, Q.2 (v)3 marks

A space ship has diameter of 8\,\mathrm{m} rotates about its axis to produce artificial gravity of magnitude 17.47\,\mathrm{m\,s^{-2}} in it. Calculate with which angular speed the satellite rotates?

Solution

Given

D = 8\,\mathrm{m}\Rightarrow R=4\,\mathrm{m}
a_c=g' = 17.47\,\mathrm{m\,s^{-2}} artificial gravity at the rim

To find

\omega

Formula (1 mark). The artificial gravity is the centripetal acceleration at the rim, a_c=\omega^{2}R, so
\omega=\sqrt{\dfrac{a_c}{R}}
Substitution (1 mark).
\omega=\sqrt{\dfrac{17.47}{4}}=\sqrt{4.37}
Answer with unit (1 mark).
\omega=2.09\,\mathrm{rad\,s^{-1}}

\omega=2.09\,\mathrm{rad\,s^{-1}}

2025 Annual I, Set D, Q.2 (vi)3 marks

Derive the relation between torque, moment of inertia and angular acceleration.

Solution

Torque on a point mass

Derivation Torque and angular acceleration

Aim. Derive \tau=I\alpha for a point mass m at distance r from the axis, driven by a tangential force F.

Assumptions.

The force acts tangentially, so it is perpendicular to the radius.
The axis is fixed.

The tangential force F has moment arm r about the axis, so its torque is \tau=rF.

Torque of a force is \tau=rF\sin\theta (equation 4.44). The force is tangential, so \theta=90^{\circ} and \sin90^{\circ}=1.
\tau=rF
Newton's second law along the tangent gives F=ma_T (equation 4.46).
F=ma_T
The tangential acceleration is related to the angular acceleration by a_T=r\alpha (equation 4.47).
F=mr\alpha
Substitute F into the expression for torque.
\tau=r(mr\alpha)=mr^{2}\alpha
The term mr^{2} is the moment of inertia of the point mass.
\tau=I\alpha

Result\tau=I\alpha

Torque is to angular acceleration what force is to linear acceleration (F=ma). A larger I needs a larger torque for the same \alpha.

The same result holds for a rigid body, because the torques on its parts add and every part has the same angular acceleration: \tau_{\text{net}}=\sum m_ir_i^{2}\,\alpha=I\alpha.

2026 Annual I, Set D, Q.2 (x) OR1+1+1 = 3 marks

Determine the equivalent value of 1 radian in degrees.

Solution

Relation (1 mark). For one complete revolution the arc is the circumference, S=2\pi r, so the angle in radians is \theta=\dfrac{S}{r}=2\pi\,\mathrm{rad}. The same angle in degrees is 360^{\circ}.
2\pi\ \mathrm{rad}=360^{\circ}
**Divide by 2\pi (1 mark).**
1\ \mathrm{rad}=\dfrac{360^{\circ}}{2\pi}=\dfrac{180^{\circ}}{\pi}
Value (1 mark).
1\ \mathrm{rad}=\dfrac{180^{\circ}}{3.1416}=57.3^{\circ}

1\ \mathrm{rad}=57.3^{\circ}

2026 Annual I, Set D, Q.2 (xi) OR3 marks

Derive the relation between Torque and angular acceleration.

Solution

Torque on a point mass

Derivation Torque and angular acceleration

Aim. Derive \tau=I\alpha for a point mass m at distance r from the axis, driven by a tangential force F.

Assumptions.

The force acts tangentially, so it is perpendicular to the radius.
The axis is fixed.

The tangential force F has moment arm r about the axis, so its torque is \tau=rF.

Torque of a force is \tau=rF\sin\theta (equation 4.44). The force is tangential, so \theta=90^{\circ} and \sin90^{\circ}=1.
\tau=rF
Newton's second law along the tangent gives F=ma_T (equation 4.46).
F=ma_T
The tangential acceleration is related to the angular acceleration by a_T=r\alpha (equation 4.47).
F=mr\alpha
Substitute F into the expression for torque.
\tau=r(mr\alpha)=mr^{2}\alpha
The term mr^{2} is the moment of inertia of the point mass.
\tau=I\alpha

Result\tau=I\alpha

Torque is to angular acceleration what force is to linear acceleration (F=ma). A larger I needs a larger torque for the same \alpha.

The same result holds for a rigid body, because the torques on its parts add and every part has the same angular acceleration: \tau_{\text{net}}=\sum m_ir_i^{2}\,\alpha=I\alpha.

2026 Annual I, Set D, Q.2 (xiii)1+1+1 = 3 marks

The orbit of an electron revolving around the nucleus of an atoms is considered to be circular of radius 50\,\mathrm{pm} and the period of motion is 0.2\,\mathrm{fs}. Find out angular velocity of electron (\omega).

Solution

Given

r = 50\,\mathrm{pm}=50\times10^{-12}\,\mathrm{m} radius of the orbit (not needed for \omega)
T = 0.2\,\mathrm{fs}=0.2\times10^{-15}\,\mathrm{s}=2\times10^{-16}\,\mathrm{s} period

To find

\omega

Formula (1 mark). In one period the electron turns through 2\pi rad, so
\omega=\dfrac{2\pi}{T}
Substitution (1 mark).
\omega=\dfrac{2\times3.14}{2\times10^{-16}}
Answer with unit (1 mark).
\omega=3.14\times10^{16}\,\mathrm{rad\,s^{-1}}

\omega=3.14\times10^{16}\,\mathrm{rad\,s^{-1}}

2025 Annual I, Set D, Q.5 OR1+6 = 7 marks

What is centripetal acceleration? Derive an expression for centripetal acceleration in terms of angular velocity '\omega'.

Solution

Centripetal acceleration (1 mark)

Centripetal acceleration is the acceleration of a body moving in a circle (with constant speed) that is directed towards the centre of the circle. It arises because the direction of the velocity keeps changing, even though the speed does not.

Derivation (6 marks)

Derivation Centripetal acceleration in terms of angular velocity

Aim. Find the acceleration of a particle moving with constant speed v on a circle of radius r.

Assumptions.

The speed v is constant (uniform circular motion).
A and B are so close that \Delta t\to0.

In Figure 4.9(a) the particle is at A at time t_i with velocity \vec v_i and at B at a later time t_f with velocity \vec v_f. For uniform circular motion \vec v_i and \vec v_f differ only in direction, so their magnitudes are equal.
|\vec v_i|=|\vec v_f|=|\vec v|\qquad |\vec r_i|=|\vec r_f|=|\vec r|
Redraw the velocity vectors tail to tail (Figure 4.9(b)). The vector \Delta\vec v joining their heads is the vector change of velocity.
\vec v_f=\vec v_i+\Delta\vec v
The velocity is perpendicular to the position vector at each point. So the angle \Delta\theta between \vec v_i and \vec v_f equals the angle between \vec r_i and \vec r_f. The two triangles (position triangle with side \Delta r, velocity triangle with side \Delta v) are similar, so their sides are in proportion.
\dfrac{\Delta v}{v}=\dfrac{\Delta r}{r}\qquad\text{or}\qquad\Delta v=v\,\dfrac{\Delta r}{r}
Divide both sides by \Delta t to get the average acceleration.
a_{av}=\dfrac{\Delta v}{\Delta t}=\dfrac{v}{r}\,\dfrac{\Delta r}{\Delta t}
Now let A and B come extremely close together, so \Delta t\to0 and the acceleration becomes instantaneous.
a=\dfrac{v}{r}\lim_{\Delta t\to0}\dfrac{\Delta r}{\Delta t}
As \Delta t\to0 the chord \Delta r becomes the arc, so \lim\dfrac{\Delta r}{\Delta t}=v.
a=\dfrac{v}{r}\times v=\dfrac{v^{2}}{r}
This is the magnitude of the acceleration towards the centre: a_c=\dfrac{v^{2}}{r}. The negative sign in vector form only shows that it points towards the centre, opposite to the outward radial unit vector \hat r.
a_c=\dfrac{v^{2}}{r}
Now bring in the angular velocity. The linear speed and the angular velocity are related by v=r\omega. Put it in a_c=v^{2}/r.
a_c=\dfrac{(r\omega)^{2}}{r}=r\omega^{2}

Resulta_c=\dfrac{v^{2}}{r}=r\omega^{2}\quad\left(\text{in vector form } \vec a_c=-r\omega^{2}\,\hat r\right)

Magnitude a_c=v^{2}/r=r\omega^{2}, directed towards the centre of the circle at every instant. Because its direction keeps changing, it is not a constant vector even when the speed is constant.

Exam focus

Derive the relation between torque, moment of inertia and angular acceleration, \tau=I\alpha (point mass, with the figure). (short)
Asked every time: Q.2 (x) OR in 2026 Set B, Q.2 (xi) OR in 2026 Set D, Q.2 (xiv) in 2025 Set B and Q.2 (vi) in 2025 Set D, 3 marks each. Same as long question 5 of the book. MCQ 17 of 2026 Set B and MCQ 1 of Set D ask the formula itself.
Go to answer
Moment of inertia: I=\sum mr^{2}, what it depends on, and I=\tfrac12MR^{2} for a solid cylinder. (mcq)
2026 Set D MCQ 17 (numerical on the solid cylinder), 2026 Set B MCQ 16 (distance from axis), 2025 Set B MCQ 5 and Set D MCQ 4. Book exercise MCQs 3 and 5.
Go to answer
Centripetal force F_c=mv^{2}/r=mr\omega^{2}: direction (radially inward) and how it changes when the speed doubles. (mcq)
2026 Set B MCQ 7 and Set D MCQ 8, and 2025 Set B MCQ 4. Book long questions 1 and 6.
Go to answer
Angular velocity numericals: \omega=2\pi N/t=2\pi/T (tractor wheels, electron orbit, Earth's rotation). (numerical)
2026 Set B Q.2 (ix) OR (tractor wheels, Assignment 4.1), 2026 Set D Q.2 (xiii) (electron, T=0.2 fs) and 2025 Set B Q.2 (iv) OR (Earth). Book numerical 1.
Go to answer
Radian and degree: 1\ \mathrm{rad}=180^{\circ}/\pi=57.3^{\circ} from 2\pi\ \mathrm{rad}=360^{\circ}. (short)
Q.2 (x) OR in 2026 Set D, 3 marks; also numerical 8 and short question 12 of the book.
Go to answer
Derive a_c=v^{2}/r=r\omega^{2} by similar triangles, with both parts of the figure. (long)
Q.5 OR (7 marks) in 2025 Set D; not asked in 2026, but a standard long derivation of the unit.
Go to answer
Weightlessness and artificial gravity; derive \omega=\sqrt{g/R}, T and f of the rotating satellite. (long)
Q.6 (6 marks) in 2025 Set B and numerical Q.2 (v) in 2025 Set D; not asked in 2026. Book long questions 7 and 8 and numerical 7.
Go to answer
Gyroscope: meaning, working and use, with Figure 4.18. (short)
Q.2 (iv) in 2025 Set D (not in 2026); also short question 10 of the book.
Go to answer
State and apply conservation of angular momentum (I_f\omega_f=I_i\omega_i): skater, diver, merry-go-round. (long)
Not asked in 2025 or in 2026, but long question 4 and numerical 5 of the book and a stated SLO.
Go to answer

Rotational motion

4.1.1Angular position (θ)

4.1.2Angular displacement (Δθ)

Units of angular displacement

Relation between radian and degree

Direction of angular displacement

4.1.3Angular velocity (ω)Often examined

Average angular velocity (ωav​)

Instantaneous angular velocity (ωinst​)

4.1.4Angular acceleration (α)Often examined

Average angular acceleration (αav​)

Instantaneous angular acceleration (αinst​)

4.1.5Relationship between linear and angular kinematic quantitiesOften examined

Derivation Relation between linear and angular displacement, S=rθ

Derivation Relation between linear and angular velocity, v=rω

Derivation Relation between linear and angular acceleration, aT​=rα

Tangential and radial components of acceleration

Kinematic equations of rotational motion

4.2Centripetal acceleration and centripetal forceOften examined

Derivation Centripetal acceleration a_c=\dfrac{v^{2}}{r}

4.2.1Centripetal forceOften examined

Frictional force and normal force as centripetal force

Derivation Banked curve: \tan\theta=\dfrac{v^{2}}{rg}

Tension force as centripetal force

Gravitational force as centripetal force

Centrifuge

4.3Moment of inertia

Derivation Moment of inertia of a rigid body, I=\sum m_ir_i^{2}

Moments of inertia of uniform objects

4.4Angular momentumOften examined

For a point mass

Derivation Angular momentum of a point mass, L=I\omega

For a rigid body

Derivation Angular momentum of a rigid body, L=I\omega

4.4.3Relation between torque and angular momentumOften examined

Derivation Torque equals the rate of change of angular momentum

4.4.4Conservation of angular momentumOften examined

Spinning ice skater

Gymnasts and divers

Gyroscope

Flywheel

4.5Torque and angular acceleration

For a point mass

Derivation Torque and angular acceleration, \tau=I\alpha

4.6Weightlessness in satellites

Effects on astronauts

4.7Artificial gravity

Derivation Speed, angular velocity, period and frequency for artificial gravity

Formula sheet

Key terms

Multiple choice questionsp. 74–75

Short questionsp. 75

Comprehensive questionsp. 75–76

Centripetal force

Why a centripetal acceleration exists

Derivation Centripetal acceleration

Applications

Moment of inertia

Moment of inertia of a rigid body

Derivation Moment of inertia of a rigid body

Angular momentum of a point mass

Derivation Angular momentum of a point mass

Angular momentum of a rigid body

Derivation Angular momentum of a rigid body

Relation between angular momentum and torque

Derivation Torque and the rate of change of angular momentum

Law of conservation of angular momentum

Practical examples

Torque on a point mass

Derivation Torque and angular acceleration

Derivation Speed and angular velocity for artificial gravity

Perpendicular force and the path

Qualitative analysis

Examples

Numerical problemsp. 76

(a) No one on the merry-go-round

(b) The boy sits 1.5 m from the centre

FBISE 2025 and 2026 questions

Torque on a point mass

Derivation Torque and angular acceleration

Average angular velocity ( $ω_{a v}$ )

Instantaneous angular velocity ( $ω_{in s t}$ )

Average angular acceleration ( $α_{a v}$ )

Instantaneous angular acceleration ( $α_{in s t}$ )

Derivation Relation between linear and angular displacement, $S = r θ$

Derivation Relation between linear and angular velocity, $v = r ω$

Derivation Relation between linear and angular acceleration, $a_{T} = r α$