Question

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 ω about their own symmetry axes. The amounts of work (W) required to bring them to rest, would satisfy the relation

• A. W_C>W_B>W_A
• B. W_B>W_A>W_C
• C. W_A>W_B>W_C
• D. W_A>W_C>W_B

Answer 1

The Correct Option is A

 To solve this problem, we need to calculate the work done to bring each of the three objects to rest. The work done on a rotating object to bring it to rest is equal to its rotational kinetic energy, given by the formula:

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

where \(I\) is the moment of inertia of the object and \(\omega\) is the angular speed.

Let's find the moment of inertia for each object:

• For a solid sphere: \(I_A = \frac{2}{5}MR^2\)
• For a thin circular disk: \(I_B = \frac{1}{2}MR^2\)
• For a circular ring: \(I_C = MR^2\)

Using the formula for rotational kinetic energy, the work done \(W\) for each object to bring it to rest is:

• Work done for the solid sphere: \(W_A = \frac{1}{2}\left(\frac{2}{5}MR^2\right)\omega^2 = \frac{1}{5}MR^2\omega^2\)
• Work done for the thin circular disk: \(W_B = \frac{1}{2}\left(\frac{1}{2}MR^2\right)\omega^2 = \frac{1}{4}MR^2\omega^2\)
• Work done for the circular ring: \(W_C = \frac{1}{2}(MR^2)\omega^2 = \frac{1}{2}MR^2\omega^2\)

Comparing these expressions, we find:

• \(W_C = \frac{1}{2}MR^2\omega^2\)
• \(W_B = \frac{1}{4}MR^2\omega^2\)
• \(W_A = \frac{1}{5}MR^2\omega^2\)

Therefore, the relationship among the works needed to bring each object to rest is:

\(W_C > W_B > W_A\)

Thus, the correct option is: W_C>W_B>W_A