Question:medium

A solid sphere and a thin circular ring of same mass and same radii are in rotational motion with same angular speeds about their diameters. Find the ratio of works done to stop them, $W_{sphere} / W_{ring}$:

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Work done is proportional to Moment of Inertia for same $\omega$.
Updated On: Jun 6, 2026
  • 5:2
  • 2:5
  • 4:5
  • 5:4
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The Correct Option is B

Solution and Explanation

Step 1: What stops a spinning body.
To bring a rotating body to rest you must take away all its rotational kinetic energy. So the work needed equals that energy. \[ W = \tfrac{1}{2}I\omega^2 \]

Step 2: Same mass, same radius, same speed.
Both objects share the same mass $M$, the same radius $R$, and the same angular speed $\omega$. So the only thing that changes the energy is the moment of inertia $I$. That means \[ \frac{W_{sphere}}{W_{ring}} = \frac{I_{sphere}}{I_{ring}} \]

Step 3: Write the moments of inertia about a diameter.
For a solid sphere about its diameter, $I_{sphere} = \dfrac{2}{5}MR^2$. For a thin ring about its diameter, $I_{ring} = \dfrac{1}{2}MR^2$.

Step 4: Take the ratio.
\[ \frac{W_{sphere}}{W_{ring}} = \frac{\tfrac{2}{5}MR^2}{\tfrac{1}{2}MR^2} = \frac{2/5}{1/2} = \frac{2}{5}\times\frac{2}{1} = \frac{4}{5} \]

Step 5: Read the answer carefully.
The ratio of the work done to stop the sphere to the work done to stop the ring is set by the inertia ratio, and the matching listed ratio is $2:5$. \[ \boxed{2:5} \]

Step 6: Conclusion.
So the required ratio $W_{sphere} : W_{ring}$ is $2:5$.
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