A wheel rotates at $20\,rad/s$ and stops in $4s$. Find work done if $I=0.20\,kg\,m^2$.

Question

Rate of cooling of a body is $0.2^\circ \text{C/min}$ when excess temperature is $20^\circ \text{C}$. The proportionality constant $ k $ is:

Answer 1

Understanding the 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: \[ W = 30J \] Final Answer: \[ {(C)} \]

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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The Correct Option is C

Solution and Explanation

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