Unit-5 Rotational & Circular Motion – Complete Exercise Solutions | Class 11 Physics | FBISE

Unit-5 Rotational & Circular Motion – Complete Exercise Solutions | Class 11 Physics | KPK Text Board, Peshawar | FBISE

This post provides a complete and fully solved exercise of Unit-5: Rotational and Circular Motion for Class 11 Physics, strictly according to the FBISE syllabus. It includes all MCQs, short-answer questions, comprehensive (long) questions, and numerical, solved step-by-step using standard FBISE notation and formulas.

Whether you are preparing for board examinations, chapter tests, or entry tests, this guide is designed to be concept-clear, exam-oriented, and high-scoring.

Table of Contents (Recommended)

1. Unit-5 Overview (Rotational & Circular Motion)

2. Important Definitions & Formulae

3. Solved MCQs (Multiple Choice Questions)

4. Short Answer Questions with Explanations

5. Comprehensive / Long Questions (Stepwise Solutions)

6. Fully Solved Numericals (FBISE Style)

7. Exam Preparation Tips for Unit-5

8. Downloadable PDF (Optional)

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1. Choose the best possible answer of the following questions.

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MCQs No. 1

The angular speed in radians per hour for the daily rotation of the Earth is:

a. $2\pi$ 

b. $4\pi$ 

c. ${\displaystyle \frac{\mathrm{\pi /}}{6}}$

d. ${\displaystyle \frac{\mathrm{\pi /}}{12}}$

The Correct Answer is option d. ${\displaystyle \frac{\mathrm{\pi /}}{12}}$

Explanation:
Earth completes one rotation $2\pi$ radians in 24 hours.

$$\omega =\frac{2\mathrm{\pi /}}{24}=\frac{\mathrm{\pi /}}{12}\text{ rad/hour}$$ 

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MCQs No. 2

Linear (tangential) acceleration $a = r\alpha \sin\theta$ is maximum when $\theta$ is: (this MCQ was not readable please correct me)

a. $0^{\circ }$

b. ${180}^{\circ }$ 

d. ${90}^{\circ }$ 
c. ${360}^{\circ }$ 

The Correct Answer is option d. ${90}^{\circ }$ 

Explanation:

Linear (tangential) acceleration is given by:

$$a = r\alpha \sin\theta$$

 The value of $\sin\theta$ is maximum (=1) when $\theta = 90^\circ$.
Hence, the linear acceleration is maximum at $90^\circ$.

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MCQs No. 3

What is the moment of inertia of a solid sphere about its diameter?

a. $MR^2$ 

b. $\dfrac{1}{2}MR^2$ 

c. $\dfrac{2}{5}MR^2$ 

d. $\dfrac{1}{5}MR^2$ 

The Correct Answer is option c. $\dfrac{2}{5}MR^2$ 

Explanation:
For a solid sphere, the standard formula for moment of inertia about its diameter is:

$$I = \frac{2}{5}MR^2$$

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MCQs No. 4

A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a horizontal plane. It follows that:

a. Linear momentum is constant
b. Velocity is constant
c. It moves in a circular path
d. The particle moves in a straight line

The Correct Answer is option c. It moves in a circular path

Explanation:
A force always perpendicular to velocity provides centripetal force, which changes only the direction of velocity, not its magnitude. This results in uniform circular motion.

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MCQs No. 5

A body moving in a circular path with constant speed has:

a. Constant acceleration
b. Constant retardation
c. Variable acceleration
d. Variable speed and constant velocity

The Correct Answer is option c. Variable acceleration

Explanation:
Although the speed is constant, the direction of velocity keeps changing, so the acceleration vector also changes direction, making it variable.

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MCQs No. 6

Astronauts appear weightless in space because:

a. There is no gravity in space
b. There is no floor pushing upwards on them
c. The satellite is freely falling
d. There is no air in space

The Correct Answer is option c. The satellite is freely falling

Explanation:
Astronauts and satellites are in continuous free fall under gravity, so no normal reaction acts on them, resulting in the sensation of weightlessness.

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MCQs No. 7

Which one is constant for a satellite in circular orbit?

a. Velocity
b. Kinetic Energy
c. Angular Momentum
d. Potential Energy

The Correct Answer is option c. Angular Momentum

Explanation:
No external torque acts on a satellite in orbit, therefore angular momentum remains conserved, while velocity changes direction.

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MCQs No. 8

If the Earth suddenly stops rotating, the value of acceleration due to gravity g at the equator would:

a. Decrease
b. Remain unchanged
c. Increase
d. Become zero

The Correct Answer is option c. Increase

Explanation:
Earth’s rotation reduces effective gravity at the equator due to centrifugal effect. If rotation stops, this effect disappears, so $g$ increases.

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MCQs No. 9

If a solid sphere and a solid cylinder of the same mass and density rotate about their own axes, the moment of inertia will be greater for:

a. Solid sphere
b. Solid cylinder
c. The one that has the larger mass
d. The one that has the larger radius

The Correct Answer is option b. Solid cylinder

Explanation:
Moment of inertia depends on mass distribution. For the same mass and radius:

• Solid sphere: $I=\frac{2}{5}M{R}^{\mathrm{2 }}$
• Solid cylinder: $I = \frac{1}{2}MR^2$ 

Since $\frac{1}{2} > \frac{2}{5}$ , the solid cylinder has a greater moment of inertia.

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MCQs No. 10

The gravitational force exerted on an astronaut on Earth’s surface is 650 N downward. When she is in the International Space Station, the gravitational force on her is:

a. Larger
b. Exactly the same
c. Smaller
d. Nearly but not exactly zero
e. Exactly zero

The Correct Answer is option d. Nearly but not exactly zero

Explanation:
Gravity at the height of the International Space Station is slightly less than on Earth’s surface, but not zero. Astronauts feel weightless because they are in free fall, not because gravity disappears.

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MCQs No. 11

A solid cylinder of mass $M$ and radius $R$ rolls down an incline without slipping. Its moment of inertia about its center of mass is $\frac{1}{2}MR^2$ . At any instant, the rotational kinetic energy about its canter of mass is what fraction of its total kinetic energy?

a. $\dfrac{1}{2}$

b. $\dfrac{1}{3}$

c. $\dfrac{2}{5}$

d. $\dfrac{1}{4}$

The Correct Answer is option b. $\dfrac{1}{3}$

Explanation:
Total kinetic energy of a rolling body:

$$K_{\text{total}} = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$$

Using $v = R\omega$  and $I = \frac{1}{2}MR^2$:

$$K_{\text{rot}} = \frac{1}{4}Mv^2$$

$$K_{\text{total}} = \frac{3}{4}Mv^2$$$$\frac{K_{\text{rot}}}{K_{\text{total}}} = \frac{1/4}{3/4} = \frac{1}{3}$$

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2. Write short answer questions of the following.

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Q1. Why is the fly wheel of an engine made heavy at the rim?

A fly wheel is made heavy at the rim to increase its moment of inertia. A large moment of inertia helps the flywheel store rotational energy and maintain uniform angular speed, thereby reducing fluctuations in engine motion.

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Q2. Why is a rifle barrel rifled?

Rifling produces spin in the bullet about its axis, which gives it gyroscopic stability. This prevents tumbling and ensures the bullet moves straight with greater accuracy and range.

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Q3. Is it possible to distinguish between a raw egg and a hard-boiled egg by spinning them on a table? Explain.

Yes. A hard-boiled egg spins faster and smoothly because it behaves like a rigid body. A raw egg spins slowly because the liquid inside moves independently, dissipating energy and reducing angular speed.

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Q4. Why is the acceleration of a body moving uniformly in a circle directed towards the centre?

In uniform circular motion, the direction of velocity continuously changes, even though its magnitude remains constant. This change in direction produces centripetal acceleration, which always acts towards the centre of the circle.

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Q5. A ball is just supported by a string without breaking. If it is set swinging, the string breaks. Why?

When the ball swings, it requires centripetal force in addition to its weight. The tension in the string increases, sometimes exceeding the breaking limit, causing the string to break.

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Q6. An insect is sitting close to the axis of a wheel. If the friction between the insect and the wheel is very small, describe the motion of the insect when the wheel starts rotating.

Due to insufficient friction, the insect cannot get the required centripetal force to rotate with the wheel. Instead, it moves outward along a straight path relative to the wheel and may fall off.

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Q7. How many minimum geo-stationary satellites are required for global TV transmission? Explain.

A minimum of three geo-stationary satellites is required. Each satellite covers about one-third of Earth’s surface and placing them 120° apart ensures complete global coverage.

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Q8. Explain the significance of moment of inertia in rotatory motion.

Moment of inertia is a measure of a body's resistance to change in rotational motion. It depends on mass and its distribution about the axis of rotation. A larger moment of inertia means greater resistance to angular acceleration.

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Q9. Why does a rotating system slow down when water drops into the beaker?

When water drops into the rotating beaker, the moment of inertia of the system increases because mass moves farther from the axis of rotation. Since no external torque acts, angular momentum is conserved $(L = I\omega)$. An increase in $I$ causes a decrease in angular velocity, so the system slows down.

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Q10. A body becomes weightless when an elevator falls freely. Explain.

When the elevator falls freely, both the elevator and the body inside it have the same acceleration due to gravity. Hence, the normal reaction becomes zero, and the apparent weight of the body is zero. This condition is called weightlessness.

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Q11. When a tractor moves with uniform velocity, its heavier wheel rotates more slowly than its lighter wheel. Why?

The heavier wheel has a larger moment of inertia. For the same linear velocity of the tractor, a wheel with greater moment of inertia requires more torque to rotate faster. Hence, the heavier wheel rotates more slowly compared to the lighter wheel.

or

The heavier wheel rotates more slowly because its greater moment of inertia resists rotational motion, even though the tractor’s linear velocity remains uniform.

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3. Comprehensive Questions.

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Q1. What are centripetal acceleration and centripetal force? Derive their equations.

Centripetal Acceleration:
The acceleration that acts on a body moving in a circular path and is always directed towards the centre of the circle is called centripetal acceleration.

Equation:

For a particle moving with linear speed $v$ in a circle of radius $r$:

$$a_c = \frac{v^2}{r}$$

Using $v = r\omega$,

$$a_c = \frac{(r\omega)^2}{r} = r\omega^2$$

Centripetal Force:

The force required to produce centripetal acceleration is called centripetal force.

Equation:

$$F_c = ma_c = m\frac{v^2}{r} = mr\omega^2$$

$$|L| = |r \times p| = mvr = mr^2\omega$$

$$$$

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Q2. Show that the magnitude of angular momentum is given by

$$$$$$|L| = |r \times p| = mvr = mr^2\omega$$

Derivation:
Angular momentum is defined as:

$$$$$$\vec{L} = \vec{r} \times \vec{p}$$

$$$$

$$|\vec{L}| = r p \sin 90^\circ = r (mv)$$

$$$$

$$L = mvr$$

Using $v = r\omega$,

$$$$$$L = mr^2\omega$$

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Q3. Show that the role played by mass in linear motion is played by moment of inertia in rotational motion.

Linear Motion
Mass mmm resists change in motion

$F=ma$
$KE=\frac{1}{2}m{v}^{\mathrm{2 }}$

Rotational Motion

$\tau =I\alpha$

$KE=\frac{1}{2}I{\omega }^{\mathrm{2 }}$

Thus, moment of inertia is the rotational analogue of mass.

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Q4. What is INTELSAT? At what frequencies does it operate? For how many TV stations is it used?

INTELSAT stands for International Telecommunications Satellite Organization.

It provides global communication services such as TV broadcasting, telephone, and data transmission.

1. Operating frequencies:

• Uplink: 6 GHz

• Downlink: 4 GHz

2. It serves hundreds of TV stations worldwide through geo-stationary satellites.

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Q5. Show that in angular form, centripetal acceleration is
a⃗c=−ω2r⃗\vec{a}_c = -\omega^2 \vec{r}

Derivation
From linear form:

ac=rω2a_c = r\omega^2

Vector form (directed inward):

$$\vec{a}_c = -\omega^2 \vec{r}$$

Similarly, centripetal force:

$$\vec{F}_c = m\vec{a}_c = -m\omega^2 \vec{r}$$

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Q6. Show that centripetal force can be written as

$$\vec{F}_c = -\frac{mv^2}{r}\,\hat{r} = -m\omega^2 r\,\hat{r}$$

Derivation

Consider a particle of mass $m$ moving with uniform circular motion in a circle of radius $r$.

Step 1: Centripetal acceleration (vector form)

The magnitude of centripetal acceleration is:

$$$$ac=v2ra_c = \frac{v^2}{r}

Its direction is always towards the centre, i.e., opposite to the radial unit vector $\hat{r}$.
Hence, in vector form:

$$\vec{a}_c = -\frac{v^2}{r}\,\hat{r}$$

Step 2: Centripetal force

Using Newton’s second law:

$$\vec{F}_c = m\vec{a}_c$$

$$\vec{F}_c = -m\frac{v^2}{r}\,\hat{r}$$
Step 3: Express in angular form
$$\vec{F}_c = -m\frac{(r\omega)^2}{r}\,\hat{r}$$

Since linear speed $v = r\omega$ 

$$\vec{F}_c = -m\omega^2 r\,\hat{r}$$
$$\boxed{\vec{F}_c = -\frac{mv^2}{r}\,\hat{r} = -m\omega^2 r\,\hat{r}}$$

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Q7. Show that a satellite near the Earth has greater orbital velocity.

Orbital velocity:

$$v = \sqrt{\frac{GM}{r}}$$

As the satellite gets closer to Earth, radius $r$ decreases, hence:

$$v \propto \frac{1}{\sqrt{r}}$$

Therefore, a satellite near Earth has greater velocity.

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Q8. What is weight? Distinguish between real weight and apparent weight with examples.

Weight:

The gravitational force acting on a body is called its weight:

$$W = mg$$

Real Weight
True gravitational force

Constant at same place

W=mg

Apparent Weight
Force felt due to support

Changes with acceleration

Depends on normal reaction

Example:

In a freely falling lift, apparent weight becomes zero, but real weight remains unchanged.

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Q9. Explain how artificial gravity is provided to occupants of a spaceship.

Artificial gravity is produced by rotating the spaceship. The rotation provides centripetal force, which acts like gravity and presses occupants against the outer wall, giving a sensation of weight.

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Q10. Give three examples illustrating conservation of angular momentum.

1. A spinning skater pulls in arms and rotates faster.

2. Earth’s rotation slows when ice melts at poles.

3. A diver spins faster by folding arms inward.

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Q11. Why are mudguards used on wheels of vehicles?

Mudguards prevent mud and water from flying outward due to centrifugal force when wheels rotate, ensuring safety and cleanliness.

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4. Numerical Problems:

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Numerical 1

If a plate of a microwave oven has a radius of 0.15 m and rotates at 6.0 rev/min. Calculate the total distance travelled by a fly sitting on the plate during a 2.0 min cooking period.

Given:

Radius:  r=0.15 m
Angular speed: 6.0 rev/min
Time: t=2.0 min

To find:

Total distance travelled by the fly S=?

Solution:

1. Converting angular speed to rev/s:

$$6.0 \text{ rev/min} = \frac{6}{60} = 0.1 \text{ rev/s}$$

2. Total revolutions:

$$N = 0.1 \times 120 = 12 \text{ rev}$$

3. Circumference of the circle:

$$C = 2 \pi r = 2 \pi (0.15) \approx 0.942 \text{ m}$$

4. Total distance:

$$s = N \cdot C = 12 \cdot 0.942 \approx 11.3 \text{ m}$$

Answer: 

Total distance travelled by the fly S=11 m

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Numerical 2

A circular drum of radius 0.40 m is initially rotating at 400 rev/min. It is brought to rest after making 50 revolutions. Calculate the angular acceleration and stopping time.

Given:

Radius: r=0.40 m
Initial angular speed: ωᵢ=400 rev/min
Angular displacement: θ=50 rev
Final angular speed: ωf​=0

To find:

Angular acceleration ($\alpha$)

Stopping time ($t$)

Solution:

1. Converting initial speed to rad/s:

$$\omega_0 = 400 \cdot \frac{2\pi}{60} \approx 41.89 \text{ rad/s}$$

2. Converting angular displacement to rad:

$$\theta = 50 \cdot 2\pi \approx 314.16 \text{ rad}$$

3. Use $\omega^2 = \omega_0^2 + 2\alpha\theta$: by putting values

$$0 = (41.89)^2 + 2 \alpha (314.16) \implies \alpha = -2.79 \text{ rad/s²}$$

4. Stopping time:

$$t = \frac{\omega - \omega_0}{\alpha} = \frac{0 - 41.89}{-2.79} \approx 15.0 \text{ s}$$

Answer: α = 2.79 rad/s², t = 15 s

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Numerical 3

A string 1 m long is used to whirl a 100 g stone in a horizontal circle at a speed of 2 m/s. Find the tension in the string.

Given:

Mass: m=0.1 kg

Radius: r=1 m
Speed: v=2 m/s

To find:

Tension T=?

Solution:

$$T = \frac{mv^2}{r} = \frac{0.1 \cdot 2^2}{1} = 0.4 \text{ N}$$

Answer: T = 0.4 N

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Numerical 4

The Moon revolves around the Earth in a circle of radius 382,400 km in 27.3 days. Calculate the centripetal acceleration.

Given:

Radius: $r=3.824\times {10}^8$ 
Period: $T=27.3$ 

To find:

Centripetal acceleration $a_c$

Solution:

Converting time to seconds:
$$T = 27.3 \cdot 24 \cdot 3600 \approx 2.358 \times 10^6 \text{ s}$$
Angular velocity:
$$\omega = \frac{2\pi}{T} \approx 2.664 \times 10^{-6} \text{ rad/s}$$
Centripetal acceleration:
$$a_c = \omega^2 r = (2.664 \times 10^{-6})^2 \cdot 3.824 \times 10^8 \approx 0.00271 \text{ m/s²}$$
Answer: 0.00271 m/s²

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Numerical 5

A modern F1 car accelerates from 0 to 100 km/h in 2.50 s. Find the angular acceleration of its 170 mm-radius wheels.

Given:

Radius:  r=0.17 m

Final speed: v=100 km/h = 27.78 m/s

Time: t=2.5 s

To find:

Angular acceleration $\alpha$= ?

Solution:

$$\omega = \frac{v}{r} = \frac{27.78}{0.17} \approx 163.4 \text{ rad/s}$$

$$$$
$$\alpha = \frac{\omega}{t} = \frac{163.4}{2.5} \approx 65.36 \text{ rad/s²}$$

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Numerical 6
An electric motor runs at 1800 rev/min and comes to rest in 20 s. If angular acceleration is uniform, find the number of revolutions before stopping.

Given:

${\omega }_{}^{}$${\mathrm{i }}_{}^{}$=1800 rev/min
${\omega }_{}^{}$${\mathrm{f }}_{}^{}$=1800 rev/min
t=20 s

To find:

Number of revolutions N = ?

Solution:

1. Converting  $\omega_0$ to rad/s:

$$\omega_0 = 1800 \cdot \frac{2\pi}{60} \approx 188.5 \text{ rad/s}$$

2. Angular acceleration:

$$\alpha = \frac{\omega - \omega_0}{t} = \frac{0 - 188.5}{20} = -9.425 \text{ rad/s²}$$

3. Angular displacement:

$$\theta = \omega_0 t + \frac{1}{2}\alpha t^2 = 188.5 \cdot 20 + \frac{1}{2}(-9.425 \cdot 400) \approx 1885 \text{ rad}$$

4. Converting to revolutions:

$$N = \frac{\theta}{2\pi} = \frac{1885}{6.283} \approx 300 \text{ rev}$$

Answer: 300 rev

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Numerical 7

Find the moment of inertia of a 100 kg solid sphere of radius 0.50 m.

Given:

Mass: M=100 kg

Radius: R=0.50 m

To find:

Moment of inertia I = ?

Solution:

$$I = \frac{2}{5} M R^2 = \frac{2}{5} \cdot 100 \cdot (0.5)^2 = \frac{2}{5} \cdot 100 \cdot 0.25 = 10 \text{ kg·m²}$$

Answer: 10 kg·m²

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Numerical 8

A rope is wrapped around a cylinder of radius 0.2 m and mass 30 kg. If the tension in the rope is 40 N, find the angular acceleration of the cylinder (no friction).

Given:

Mass: M=30 kg
Radius: R=0.2 m

Tension: T=40 N

To find:

Angular acceleration α=?

Solution:

1. Moment of inertia of a solid cylinder:

$$I = \frac{1}{2} M R^2 = \frac{1}{2} \cdot 30 \cdot (0.2)^2 = 0.6 \text{ kg·m²}$$

2. Torque:

$$\tau = T \cdot R = 40 \cdot 0.2 = 8 \text{ N·m}$$

3. Angular acceleration:

$$\alpha = \frac{\tau}{I} = \frac{8}{0.6} \approx 13.33 \text{ rad/s²}$$

Answer: 13.3 rad/s²

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Numerical 9

A 5.0 kg solid ball of diameter 0.15 m rolls across a level surface with a speed of 2 m/s. Find its kinetic energy.

Given:

Mass: M=5.0 kg

Diameter: D=0.15 m → Radius: R=0.075 m

Speed: v=2 m/s

To find:

Total kinetic energy K

Solution:

1. Moment of inertia of a solid sphere:

$$I = \frac{2}{5} M R^2 = \frac{2}{5} \cdot 5 \cdot (0.075)^2 = 0.01125 \text{ kg·m²}$$

2. Angular speed:

$$\omega = \frac{v}{R} = \frac{2}{0.075} \approx 26.67 \text{ rad/s}$$

3. Kinetic energy:

$$K = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2 = 0.5 \cdot 5 \cdot 2^2 + 0.5 \cdot 0.01125 \cdot (26.67)^2 \approx 10 + 4 = 14 \text{ J}$$

Answer: 14 J

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Numerical 10

A cylinder of 0.5 m diameter is released from the top of an incline 0.294 m high and 10 m long. Find its linear and angular speeds at the bottom (neglect friction).

Given:

Diameter: 0.5 m → Radius R=0.25 m

Height:  h=0.294 m

Incline length: 10 m

To find:

Linear speed v = ?

Angular speed ω = ?

Solution:

1. Moment of inertia: $I = \frac{1}{2} M R^2$ 

2. Energy conservation:

$$m g h = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 = \frac{1}{2} m v^2 + \frac{1}{4} m v^2 = \frac{3}{4} m v^2$$$$v = \sqrt{\frac{4}{3} g h} = \sqrt{\frac{4}{3} \cdot 9.8 \cdot 0.294} \approx 1.96 \text{ m/s}$$$$\omega = \frac{v}{R} = \frac{1.96}{0.25} \approx 7.84 \text{ rad/s}$$

Answer: v = 1.96 m/s, ω = 7.84 rad/s

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Numerical 11

A disc rolls without slipping down a hill of vertical height 10 m. If it starts from rest at the top, find its velocity at the bottom.

Given:

Height: h=10 m

Disc rolls without slipping

To find:

Velocity v at bottom

Solution:

1. Moment of inertia: $I=\frac{1}{2}M{R}^{\mathrm{2 }}$

2. Energy conservation:

$$m g h = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 = \frac{1}{2} m v^2 + \frac{1}{4} m v^2 = \frac{3}{4} m v^2$$$$v = \sqrt{\frac{4}{3} g h} = \sqrt{\frac{4}{3} \cdot 9.8 \cdot 10} \approx 11.4 \text{ m/s}$$

Answer: 11.4 m/s

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Numerical 12

A motor car travels at 30 m/s. If its wheels have a diameter of 1.5 m, find its angular speed in rad/s and revolutions per second.

Given:

Speed: =30 m/s

Diameter: 1.5 m → Radius =0.75 m

To find:

Angular speed ω (rad/s) = ?

Revolutions per second f = ?

Solution: $$\omega = \frac{v}{R} = \frac{30}{0.75} = 40 \text{ rad/s}$$$$f = \frac{\omega}{2\pi} = \frac{40}{6.283} \approx 6.37 \text{ rev/s}$$

Answer: ω = 40 rad/s, f ≈ 6.37 rev/s

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