A rope is wound around a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of th

I. Multiple Choice Questions · Q4

Q. A rope is wound around a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force 30 N?

(a)

0.25\ \text{rad s}^{-2}

(b)

25\ \text{rad s}^{-2}

(c)

5\ \text{rad s}^{-2}

(d)

250\ \text{rad s}^{-2}

Tamil Nadu DgeTextbookSubjectiveImportance★★★★★

18% · 16/87 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Rotational Dynamics

Rotational Dynamics: The Physics of Spinning Things

Imagine you're trying to open a heavy door. You push near the hinge — it barely moves. Push near the handle — it swings open easily. Same force, different result. That's the first clue: rotation isn't just about how much you push, but where and in what direction.

Now think about a spinning bicycle wheel. Why is it so hard to tilt it sideways when it's spinning fast? And why does a figure skater spin faster when she pulls her arms in? These are the questions rotational dynamics answers.

The Core Idea

Rotational dynamics is the study of why things rotate and how their rotation changes. It's the spinning-world equivalent of Newton's laws for straight-line motion.

In linear motion, you have:

Force ( $F$ ) causes acceleration ( $a$ )
Mass ( $m$ ) resists acceleration

In rotational motion, you have:

Torque ( $\tau$ ) causes angular acceleration ( $\alpha$ )
Moment of inertia ( $I$ ) resists angular acceleration

The master equation is:

\tau_{\text{net}} = I \alpha

This is the rotational version of $F = ma$ . Every term has a direct parallel.

Breaking It Down

Torque — The Rotational "Push"

Torque isn't just force — it's force multiplied by the distance from the pivot point (the lever arm). That's why the door handle works better than the hinge.

\tau = r F \sin\theta

Where $r$ is the distance from the axis, $F$ is the force, and $\theta$ is the angle between them. Maximum torque happens when you push perpendicular to the lever arm ( $\theta = 90^\circ$ ).

Tip

Think of torque as "twisting effectiveness." A wrench works because the handle gives you a long lever arm. A short wrench needs more force to do the same job.

Moment of Inertia — The Rotational "Mass"

Mass resists linear acceleration. Moment of inertia resists angular acceleration. But unlike mass, moment of inertia depends on how the mass is distributed relative to the axis of rotation.

For a point mass $m$ at distance $r$ from the axis:

I = mr^2

For extended objects, you sum (or integrate) over all mass elements:

I = \sum m_i r_i^2

Note

| Object | Axis | Moment of Inertia |

|--------|------|-------------------|

| Thin hoop | Through center, perpendicular to plane | $MR^2$ |

| Solid disk | Through center, perpendicular to plane | $\frac{1}{2}MR^2$ |

| Solid sphere | Through center | $\frac{2}{5}MR^2$ |

| Thin rod | Through center, perpendicular to rod | $\frac{1}{12}ML^2$ |

Notice: a hoop has more moment of inertia than a disk of the same mass and radius because its mass is farther from the axis. That's why a hoop is harder to start spinning.

Angular Acceleration — How Fast Rotation Changes

Just as acceleration is the rate of change of velocity, angular acceleration $\alpha$ is the rate of change of angular velocity $\omega$ :

\alpha = \frac{d\omega}{dt}

And angular velocity is the rate of change of angular displacement $\theta$ :

\omega = \frac{d\theta}{dt}

The Complete Picture: Rotational Analogues

Linear Quantity	Rotational Analogue
Displacement $x$	Angular displacement $\theta$
Velocity $v$	Angular velocity $\omega$
Acceleration $a$	Angular acceleration $\alpha$
Mass $m$	Moment of inertia $I$
Force $F$	Torque $\tau$
Newton's 2nd law: $F = ma$	$\tau = I\alpha$
Kinetic energy: $\frac{1}{2}mv^2$	$\frac{1}{2}I\omega^2$
Momentum: $p = mv$	Angular momentum: $L = I\omega$

The Key Insight: Conservation of Angular Momentum

This is where rotational dynamics gets beautiful. Just as linear momentum is conserved when no external force acts, angular momentum is conserved when no external torque acts:

$L = I\omega = \text{constant}$ …

Start your 14-day free trial to unlock the full solution →

Step 1. Note the given values.

Mass of hollow cylinder $M = 3\ \text{kg}$ , radius $R = 40\ \text{cm} = 0.4\ \text{m}$ , pulling force $F = 30\ \text{N}$ applied via a rope wound around the cylinder (so the force acts tangentially at the rim, at distance $R$ from the axis).

Step 2. Compute the moment of inertia of the hollow cylinder.

For a hollow (thin-walled) cylinder about its central axis, all the mass sits at radius $R$ , so

I = MR^2 = 3 \times (0.4)^2 = 3 \times 0.16 = 0.48\ \text{kg m}^2.

Step 3. Compute the torque produced by the rope tension.

Since the rope is wound around the cylinder, the force $F$ acts tangentially at the rim, so the full force contributes to torque (moment arm $=R$ ):

\tau = FR = 30 \times 0.4 = 12\ \text{N m}.

Step 4. Apply the rotational equation of motion.

\tau = I\alpha \quad\Rightarrow\quad \alpha = \dfrac{\tau}{I} = \dfrac{12}{0.48} = 25\ \text{rad s}^{-2}.

Step 5. Rule out the other options.

(a) $0.25\ \text{rad s}^{-2}$ — off by two orders of magnitude, the kind of error you get from mixing up $R$ in cm vs. m without converting. …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 25 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. Case study: Moment of Inertia of a body about an axis of rotation of the body is defined as the sum of the products of the masses of the constituent particles of the body and square of the distances of the respective particles from the axis of rotation. Its value depends upon the axis of rotation and the distribution of mass of the body. M.I. plays the same role in rotational motion as the force plays in the translation motion. (A) S.I. unit of moment of Inertia is :$
(a) Kg^2 m^2
(b) Kg m^2
(c) Kg m^-2
(d) Kg m.
›Reveal solutionSolution

Since I = Σ m r², its unit is (unit of mass) × (unit of distance)² = kg·m².

Moment of Inertia is defined as:

I = Σ mi ri²

where mi is the mass of each constituent particle and ri is its perpendicular distance from the axis of rotation.

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. (B) Moment of Inertia is the rotational analogue of :$
(a) Torque
(b) Mass
(c) Force
(d) Angular Momentum.
›Reveal solutionSolution

Moment of inertia is the rotational counterpart of mass — both measure a body's inertia (resistance to a change in its state of motion).

In translational motion, Newton's second law is F = ma; mass m measures the body's inertia — its resistance to a change in linear velocity.

In rotational motion, the analogous equation is τ = Iα (torque = moment of inertia × angular acceleration); moment of inertia I measures the body's rotational inertia — its resistance to a change in angular velocity.

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set sz1 markMCQ
$Q. Radius of gyration is denoted by:$
(a) R
(b) G
(c) K
(d) I
›Reveal solutionSolution

The radius of gyration is conventionally denoted by K, defined through I = MK^2.

The radius of gyration K of a rigid body about an axis is defined by the relation I = M K^2, where I is the moment of inertia about that axis and M is the total mass. Physically, K is the distance from the axis at which the entire mass of the body could be concentrated without changing its moment of inertia. It …

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. On which of the following quantity does the moment of inertia of a body not depend?$
(a) Angular acceleration
(b) Axis of rotation
(c) Mass of body
(d) Distribution of mass
›Reveal solutionSolution
Moment of inertia is a purely geometric/mass property of a body about a given axis; it does not depend on the body's angular acceleration.

Moment of inertia is defined as I = Σ mi ri², the sum over all mass elements of (mass) × (perpendicular distance from the axis)². This definition shows I depends on:

(b) Axis of rotation — yes, changing the axis changes every ri, hence I.

(c) Mass of the body — yes, directly appears in the formula.

(d) Distribution of mass — yes, mass spread farther from the axis increases I (this is why a hollow cylinder has a larger I than a solid cylinder of the same mass and radius). …
Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. What will be the moment of inertia of a circular disc about its diameter, if its mass and radius is unity? (A) 1 unit (B) 1/2 unit (C) 1/4 unit (D) 1/8 unit$
›Reveal solutionSolution

With unit mass and unit radius, the disc's moment of inertia about a diameter is 1/4 unit.

The moment of inertia of a uniform circular disc of mass $M$ and radius $R$ about an axis through its centre and perpendicular to its plane is $I_z = \tfrac{1}{2}MR^2$ . By the perpendicular axis theorem, for two perpendicular diameters $x$ and $y$ in the plane of the disc, $I_z = I_x + I_y$ . By sym …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. The unit of moment of inertia of a solid cylinder having mass M and radius R about its geometrical axis is -$
(a) Nm^2
(b) kgm^-1
(c) kgm^3
(d) kgm^2
›Reveal solutionSolution

Moment of inertia has the SI unit kg m^2, since I is mass multiplied by the square of a distance.

Moment of inertia of a rigid body about an axis is defined as I = summation(m_i r_i^2), where m_i is the mass of each particle and r_i is its perpendicular distance from the axis of rotation. For a solid cylinder of mass M and radius R rotating about its geometrical (central) axis, I = (1/2)MR^2. …

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$Q. The mass and radius of a ring are M and r respectively. The moment of inertia of the ring about an axis passing through its centre and perpendicular to its plane is (A) ½Mr^2 (B) Mr^4 (C) M^2r^2 (D) Mr^2$
›Reveal solutionSolution

$I=Mr^2$ for a ring about the axis through its centre, perpendicular to its plane.

Moment of inertia is $I=\sum m_i r_i^2$ . In a ring, the entire mass $M$ is distributed at the same perpendicular distance $r$ (the ring's radius) from …

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$Q. A solid sphere is rolling on a horizontal plane surface. The ratio of its angular kinetic energy to linear kinetic energy will be equal to (A) 2/5 (B) 5/2 (C) 5/7 (D) 7/5$
›Reveal solutionSolution

For a rolling solid sphere, $KE_{rot}/KE_{trans}=2/5$ .

For a solid sphere, $I=\tfrac25mr^2$ . Rolling without slipping means $v=r\omega$ , so $\omega=v/r$ .

$KE_{rot}=\tfrac12I\omega^2=\tfrac12\left(\tfrac25mr^2\right)\left(\dfrac{v}{r}\right)^2=\tfrac15mv^2$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$Q. Rotational kinetic energy is equal to$
(a) I omega^2
(b) I^2 omega^2
(c) (3/2) I omega^2
(d) (1/2) I omega^2
›Reveal solutionSolution

Rotational kinetic energy = (1/2) I ω^2, the rotational analogue of (1/2) m v^2.

For a rigid body rotating with angular velocity ω, each mass element dm at distance r from the axis has speed v = rω and kinetic energy (1/2) dm v^2 = (1/2) dm r^2 ω^2. Summing (integrating) ove …

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$Q. Match the Column given below: Column-I (I) Movement of the tip of the minute hand of a clock (II) A stone falling from a height in straight line path (III) Moving wheel of a sewing machine (IV) Rotational analogue of force Column-II (A) Torque (B) Rational motion (C) Circular motion (D) Linear motion$
(a) (I)-A (II)-B (III)-C (IV)-D
(b) (I)-D (II)-A (III)-B (IV)-C
(c) (I)-C (II)-D (III)-B (IV)-A
(d) (I)-B (II)-C (III)-D (IV)-A
›Reveal solutionSolution

(I) tip of minute hand -> Circular motion (C); (II) falling stone -> Linear motion (D); (III) sewing-machine wheel -> Rotational motion (B); (IV) rotational analogue of force -> Torque (A). This is option (c).

Matching each item in Column-I to Column-II:

(I) The tip of the minute hand of a clock moves along a fixed circular path of constant radius about the clock's centre -- this is Circular motion -> matches (C).

(II) A stone falling from a height along a straight-line path is a case of one-dimensional, straight-line motion -- this is Linear motion -> matches (D).

(III) The wheel of a sewing machine spins about its own fixed axis -- every point in the wheel (other than the axis) revolves about that axis, which is Rotational motion -> matches (B).

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 mark
$Q. Fill in the blank: The quantity equivalent to mass in rotational motion is ______.$
›Reveal solutionSolution

Moment of inertia is the rotational analogue of mass — it measures a body's inertia (resistance to change) in rotational motion.

In translational motion, mass measures a body's inertia — its resistance to a change in linear velocity under an applied force (F = ma). …

Start your 14-day free trial to see the full solution →
CBSE 2023Set ANNUAL1 markMCQ
$Q. The relation between rotational kinetic energy K, moment of inertia I and angular velocity w is:$
(a) K = I²w/2
(b) K = Iw/2
(c) K = I²w²/2
(d) K = Iw²/2
›Reveal solutionSolution

The rotational kinetic energy of a rigid body is K = (1/2) I ω², directly analogous to translational KE = (1/2) m v².

A rotating rigid body can be thought of as made of many small mass elements mi, each at distance ri from the axis, moving with speed vi = ri ω. The kinetic energy of each element is (1/2) mi vi² = (1/2) mi ri² ω². Summing over all elements:

K = Σ (1/2) mi ri² ω² = (1/2) ω² Σ mi ri² = (1/2) I ω²

…

Start your 14-day free trial to see the full solution →

←Previous question · 5.7Q3 — A particle is moving with a constant velocity along a line parallel to…→Next question · 5.7Q5 — A closed cylindrical container is partially filled with water. As the …

Q.A rope is wound around a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force 30 N?

🎓Unlock everything free for 14 days

Unlock everything free for 14 days