Question:medium

Three objects, $A$ : (a solid sphere), $B$ : (a thin circular disk) and $C$ : (a circular ring), each have the same mass $M$ and radius $R$. They all spin with the same angular speed to about their own symmetry axes. The amounts of work $(W)$ required to bring them to rest, would satisfy the relation

Updated On: May 7, 2026
  • $W_A > W_C > W_B$
  • $W_C > W_B > W_A $
  • $ W_B > W_A > W_C $
  • $W_A > W_B > W_C $
Show Solution

The Correct Option is B

Solution and Explanation

 To solve this problem, we need to compare the work done to bring each object (a solid sphere, a thin circular disk, and a circular ring) to rest. The work done is related to the rotational kinetic energy of each object, which depends on the moment of inertia and the angular speed.

The formula for rotational kinetic energy \( K \) is given by:

\(K = \frac{1}{2} I \omega^2\)

where:

  • \(I\) is the moment of inertia of the object
  • \(\omega\) is the angular speed

 

Since all objects have the same mass \( M \) and radius \( R \), and they spin with the same angular speed \( \omega \), the difference in the work done to stop them is purely due to their differing moments of inertia.

The moment of inertia for each object is as follows:

  • Solid sphere \( (A) \): \(I_A = \frac{2}{5} MR^2\)
  • Thin circular disk \( (B) \): \(I_B = \frac{1}{2} MR^2\)
  • Circular ring \( (C) \): \(I_C = MR^2\)

 

Now, substituting these moments of inertia in the kinetic energy formula, we compare their kinetic energies:

  • For solid sphere \( (A) \): \(K_A = \frac{1}{2} \cdot \frac{2}{5} MR^2 \cdot \omega^2 = \frac{1}{5} MR^2 \omega^2\)
  • For thin circular disk \( (B) \): \(K_B = \frac{1}{2} \cdot \frac{1}{2} MR^2 \cdot \omega^2 = \frac{1}{4} MR^2 \omega^2\)
  • For circular ring \( (C) \): \(K_C = \frac{1}{2} \cdot MR^2 \cdot \omega^2 = \frac{1}{2} MR^2 \omega^2\)

 

From these calculations, we can see the relation:

\(K_C > K_B > K_A\)

Since the work done to bring an object to rest is equal to its kinetic energy, the same relation holds for the work done:

\(W_C > W_B > W_A\)

Hence, the correct answer is: \(W_C > W_B > W_A\)

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