Question 395: Centre of Mass and Rotational Motion

Question

A circular ring and a solid sphere having same radius roll down on an inclined plane from rest without slipping. The ratio of their velocities when reached at the bottom of the plane is $\sqrt{\frac{x}{5}}$ where $x=$ ________.

Accepted Answer

To determine the ratio of velocities for a circular ring and a solid sphere rolling down an inclined plane without slipping, we apply the principle of mechanical energy conservation:
Conservation of Mechanical Energy:
$ k_i + U_i = k_f + U_f $
Initially (at the top), we have:
Initial kinetic energy, $k_i = 0$ (since they start from rest)
Initial potential energy, $U_i = Mgh$
Finally (at the bottom), we have:
Final potential energy, $U_f = 0$
Final kinetic energy, $k_f = \frac{1}{2}mv^2\left(1 + \frac{k^2}{R^2}ight)$
This gives:
$ Mgh = \frac{1}{2}mv^2\left(1 + \frac{k^2}{R^2}ight) $
Solving for the velocity $V$:
$ V = \sqrt{\frac{2gh}{1 + \frac{k^2}{R^2}}} $
Ratio of Velocities:
For the circular ring (moment of inertia $I = mR^2$), $\frac{k^2}{R^2} = 1$.
For the solid sphere (moment of inertia $I = \frac{2}{5}mR^2$), $\frac{k^2}{R^2} = \frac{2}{5}$.
The ratio is:
$ \frac{V_{	ext{Ring}}}{V_{	ext{Solid Sphere}}} = \sqrt{\frac{1 + \frac{2}{5}}{1 + 1}} = \sqrt{\frac{7}{10}} $
Thus, $x = 3.5$. Rounding off, $x = 4$.

Centre of Mass and Rotational Motion

A circular ring and a solid sphere having same radius roll down on an inclined plane from rest without slipping. The ratio of their velocities when reached at the bottom of the plane is \sqrt{\frac{x}{5}} where x= ________.

Integer / Numeric Answer Type