Search any question & find its solution
Question:
Answered & Verified by Expert
A solid sphere and a ring of same radius roll down an inclined plane without slipping. Both start from rest from the top of the inclined plane. If the sphere and the ring reach the bottom of the inclined plane with velocities $v_s$ and $v_r$ respectively, then $\frac{v_r^2}{v_s^2}$ is
Options:
Solution:
1190 Upvotes
Verified Answer
The correct answer is:
0.7
When a body rolls down an inclined plane, energy conservation gives
$$
m g h=\frac{1}{2} m v^2+\frac{1}{2} I \omega^2=\frac{1}{2} m v^2\left(1+\frac{k^2}{R^2}\right)
$$
where, $I=m k^2$ and $k=$ radius of gyration.
So,
$$
v_{\text {bottom }}=\sqrt{\left(\frac{2 g h}{1+\frac{k^2}{R^2}}\right)}
$$
Now, for ring, $\frac{k^2}{R^2}=1$ and for sphere, $\frac{k^2}{R^2}=\frac{2}{5}$
$$
\therefore \frac{v_{\text {bottom ring }}^2}{v_{\text {bottom sphere }}^2}=\frac{1+\frac{2}{5}}{1+1}=0 \cdot 7
$$
$$
m g h=\frac{1}{2} m v^2+\frac{1}{2} I \omega^2=\frac{1}{2} m v^2\left(1+\frac{k^2}{R^2}\right)
$$
where, $I=m k^2$ and $k=$ radius of gyration.
So,
$$
v_{\text {bottom }}=\sqrt{\left(\frac{2 g h}{1+\frac{k^2}{R^2}}\right)}
$$
Now, for ring, $\frac{k^2}{R^2}=1$ and for sphere, $\frac{k^2}{R^2}=\frac{2}{5}$
$$
\therefore \frac{v_{\text {bottom ring }}^2}{v_{\text {bottom sphere }}^2}=\frac{1+\frac{2}{5}}{1+1}=0 \cdot 7
$$
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.