Search any question & find its solution
Question:
Answered & Verified by Expert
A hollow sphere is rolling on a plane surface about its axis of symmetry. The ratio of rotational kinetic energy to its total kinetic energy is $\frac{x}{5}$. The value of $x$ is _____ .
Solution:
1938 Upvotes
Verified Answer
The correct answer is:
2
$\begin{aligned} & \frac{\frac{1}{2} \mathrm{I} \omega^2}{\frac{1}{2} \mathrm{I} \omega^2+\frac{1}{2} \mathrm{mv}^2}=\frac{\left(\frac{1}{2}\right)\left(\frac{2}{3} \mathrm{mR}^2\right) \omega^2}{\left(\frac{1}{2}\right)\left(\frac{2}{3} \mathrm{mR}^2\right) \omega^2+\frac{1}{2} \mathrm{~m}(\mathrm{R} \omega)^2} \\ & =\frac{\frac{2}{3}}{\frac{2}{3}+1}=\frac{2}{5} \\ & \mathrm{x}=2\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.