Assertion: Two cylinders, one hollow (metal) and the other solid (wood) with the same mass and ident

Q: Why doesn't static friction do work in pure rolling?

In pure rolling, the instantaneous velocity of the point of contact is zero. Since work is force times displacement, and the displacement of the contact point is zero, the work done by static friction is zero.

Q: Doesn't the hollow cylinder have more rotational kinetic energy?

Yes, for a given velocity, the hollow cylinder has more rotational kinetic energy. But because total energy is conserved ($mgh$), it must have less translational kinetic energy, meaning it moves slower!

Q: Does the mass of the cylinders matter for the race?

No! If you look at the acceleration formula $a = \frac{g \sin\theta}{1 + \frac{I}{mR^2}}$, the mass $m$ cancels out because $I$ is proportional to $m$. A heavy solid cylinder and a light solid cylinder will tie!

Q: What if the incline was frictionless?

If there was no friction, neither cylinder would roll. They would simply slide down the incline with the same acceleration $g \sin\theta$, and they would tie!

Q: Why is the moment of inertia of a hollow cylinder greater than a solid one?

Moment of inertia depends on how far the mass is distributed from the axis of rotation. In a hollow cylinder, all the mass is at the maximum distance $R$, whereas in a solid cylinder, the mass is spread out from $0$ to $R$.

Question

Assertion: Two cylinders, one hollow (metal) and the other solid (wood) with the same mass and identical dimensions are simultaneously allowed to roll without slipping down an inclined plane from the same height. The hollow cylinder will reach the bottom of the inclined plane first.
Reason: By the principle of conservation of energy, the total kinetic energies of both the cylinders are identical when they reach the bottom of the incline.

Options:

(A) (a) If Assertion is true, Reason is true; Reason is the correct explanation for Assertion

(B) (b) If Assertion is true, Reason is true; Reason is not a correct explanation for Assertion

(C) (c) If Assertion is true; Reason is false

(D) (d) If Assertion is false; Reason is true

Sigma Percentile · Accepted Answer

### The Setup: Visualizing the Race

Imagine standing at the top of an inclined plane with two cylinders in your hands. One is a solid wooden cylinder, and the other is a hollow metal pipe. They both have the exact same mass and the exact same radius.

If you release them at the exact same moment, allowing them to roll without slipping, a fascinating physics race begins. Intuition might suggest it's a tie, but the laws of rotational dynamics have a different plan.

### The Master Equation: Acceleration in Pure Rolling

To determine the winner, we must analyze the forces dictating their motion. Gravity pulls the cylinders down the incline with a force of $mg \sin	heta$. Simultaneously, static friction acts upwards along the incline, providing the necessary torque for pure rolling.

By applying Newton's Second Law for translation, we get $mg \sin	heta - f = ma$. For rotation, the torque equation gives us $f R = I \alpha$.

Since the cylinders are rolling without slipping, we use the constraint $a = \alpha R$. Substituting the friction from the torque equation into the force equation yields a beautiful, universal expression for the acceleration of any rolling body:

$a = \frac{g \sin	heta}{1 + \frac{I}{mR^2}}$

### The Verdict: Who Wins?

**Notice how the acceleration depends inversely on the moment of inertia ($I$).** This is the critical catch of the problem!

The solid cylinder has its mass distributed evenly from the center to the rim, giving it a moment of inertia of $I_{	ext{solid}} = \frac{1}{2} mR^2$. In contrast, the hollow cylinder has all its mass concentrated at the outer rim, resulting in a larger moment of inertia, $I_{	ext{hollow}} = mR^2$.

Because the solid cylinder has a smaller moment of inertia, it resists rotational acceleration less. Therefore, it will have a **greater linear acceleration** ($a_{	ext{solid}} > a_{	ext{hollow}}$). The solid cylinder wins the race, proving the Assertion statement completely false!

### The Energy Perspective: A Beautiful Conservation

Now, let's evaluate the Reason statement by looking at the race through the lens of energy. In pure rolling, the point of contact between the cylinder and the incline is instantaneously at rest.

Because there is no relative slipping at this contact point, the **work done by static friction is exactly zero**. This is a favorite concept for JEE!

Since no non-conservative forces are doing work, the mechanical energy of the system is perfectly conserved. As both cylinders drop by the same vertical height $h$, they both lose the exact same amount of potential energy, $mgh$.

This lost potential energy is entirely converted into total kinetic energy (translational plus rotational). Therefore, when they reach the bottom, their **total kinetic energies will be exactly identical**, both equal to $mgh$.

The Reason statement is absolutely true. Since the Assertion is false and the Reason is true, the correct choice is option (d).

Select Answer:

Visualized Solution

$m_{1} = m_{2}, R_{1} = R_{2}$

$F_{net} = m a$

$τ = I α$

$a = \frac{g s i n θ}{1 + \frac{I}{m R ^{2}}}$

$a_{solid} > a_{hollow}$

$W_{f} = 0$

$E_{i} = E_{f}$

The Sigma Insight: Rolling Motion

The Setup

The Master Equation

The Verdict

The Energy Perspective

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m1​=m2​,R1​=R2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Fnet​=ma

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } τ=Iα

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a=1+mR2I​gsinθ​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } asolid​>ahollow​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Wf​=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ei​=Ef​