A solid circular disc of mass $50 \mathrm{~kg}$ rolls along a horizontal floor so that its center of mass has a speed of $0.4 \mathrm{~m} / \mathrm{s}$. The absolute value of work done on the disc to stop it is ________ J.
Answer (integer)
6
Solution
<p>Using work energy theorem</p>
<p>$$\begin{aligned}
& \mathrm{W}=\Delta \mathrm{KE}=0-\left(\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{I} \omega^2\right) \\
& \mathrm{W}=0-\frac{1}{2} \mathrm{mv}^2\left(1+\frac{\mathrm{K}^2}{\mathrm{R}^2}\right) \\
& =-\frac{1}{2} \times 50 \times 0.4^2\left(1+\frac{1}{2}\right)=-6 \mathrm{~J}
\end{aligned}$$</p>
<p>Absolute work $=+6 \mathrm{~J}$</p>
<p>$W=-6 J \quad|W|=6 J$</p>
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
This question is part of PrepWiser's free JEE Main question bank. 158 more solved questions on Rotational Motion are available — start with the harder ones if your accuracy is >70%.