Search any question & find its solution
Question:
Answered & Verified by Expert
A spherical ball rolls on a table without slipping. Then the fraction of its total energy associated with rotation is
Options:
Solution:
1239 Upvotes
Verified Answer
The correct answer is:
$\frac{2}{7}$
$\begin{aligned} & \mathrm{KE}_{\mathrm{Total}}=\frac{1}{2} \mathrm{IW}^2+\frac{1}{2} \mathrm{MV}^2 \\ & =\frac{1}{2} \times \frac{2}{5} \mathrm{MR}^2 \mathrm{~W}^2+\frac{1}{2} \mathrm{MV}^2 \\ & =\frac{1}{5} \mathrm{MR}^2 \mathrm{~W}^2+\frac{1}{2} \mathrm{MV}^2=\frac{1}{5} \mathrm{MV}^2+\frac{1}{2} \mathrm{MV}^2 \\ & =\frac{7}{10} \mathrm{MV}{ }^2\end{aligned}$
$\begin{aligned} & \mathrm{KE}_{\text {rot }}=\frac{1}{2} \mathrm{IW}^2=\frac{1}{5} \mathrm{MV}^2 \\ & \frac{\mathrm{KE}_{\text {rot }}}{\mathrm{KE}_{\text {Total }}}=\frac{\frac{1}{5} \mathrm{MV}^2}{\frac{7}{10} \mathrm{MV}^2} \\ & =\frac{2}{7}\end{aligned}$
$\begin{aligned} & \mathrm{KE}_{\text {rot }}=\frac{1}{2} \mathrm{IW}^2=\frac{1}{5} \mathrm{MV}^2 \\ & \frac{\mathrm{KE}_{\text {rot }}}{\mathrm{KE}_{\text {Total }}}=\frac{\frac{1}{5} \mathrm{MV}^2}{\frac{7}{10} \mathrm{MV}^2} \\ & =\frac{2}{7}\end{aligned}$
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.