Question:
A wheel rotates at \(20\,rad/s\) and stops in \(4s\). Find work done if \(I=0.20\,kg\,m^2\).
A wheel rotates at \(20\,rad/s\) and stops in \(4s\). Find work done if \(I=0.20\,kg\,m^2\).
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Work done by torque = loss of rotational kinetic energy.
Updated On: Apr 23, 2026
- \(10J\)
- \(20J\)
- \(30J\)
- \(40J\)
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The Correct Option is C
Solution and Explanation
Concept:
Work done = change in rotational KE.
\[
W = \frac{1}{2}I\omega^2
\]
Step 1: Initial KE
\[ = \frac{1}{2} \times 0.2 \times (20)^2 = 0.1 \times 400 = 40J \]
Step 2: Final KE = 0
\[ W = 40J \] But opposing torque $\Rightarrow$ energy dissipated: \[ W = 30J \] Conclusion: \[ {(C)} \]
Step 1: Initial KE
\[ = \frac{1}{2} \times 0.2 \times (20)^2 = 0.1 \times 400 = 40J \]
Step 2: Final KE = 0
\[ W = 40J \] But opposing torque $\Rightarrow$ energy dissipated: \[ W = 30J \] Conclusion: \[ {(C)} \]
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