Question

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 $\frac{v^2_r}{v^2_s}$ is

• A. 0.2
• B. 0.5
• C. 0.7
• D. 0.9

Answer 1

Answer: (C)

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, $\frac{k^{2}}{R^{2}}=\frac{2}{5}$
$\therefore \frac{v_{\text {bottom ring }}^{2}}{v_{\text {bottom sphere }}^{2}}=\frac{1+\frac{2}{5}}{1+1}=0 \cdot 7$