Question

A hollow sphere and a solid sphere having the same mass and same radii are rolled down a rough inclined plane.
(A) The hollow sphere reaches the bottom first.
(B) The solid sphere reaches the bottom with greater speed.
(C) The solid sphere reaches the bottom with greater kinetic energy.
(D) The two spheres will reach the bottom with the same linear momentum.

Answer 1

Hint : The moment of inertia of a solid sphere is different from that of a hollow sphere. We need to specify the moment of inertia with respect to a chosen axis of rotation and using this value of moment of inertia, we need to find which sphere has a greater rotational acceleration.

Formula Used: The formulae used in the solution are given here.
Moment of inertia of a solid sphere $\dfrac{{2m{r^2}}}{5}$ where m is the mass and r is the radius.
Moment of inertia of a hollow sphere $\dfrac{{2m{r^2}}}{3}$ where m is the mass and r is the radius.
$\Rightarrow \tau = I \cdot \alpha$ where $\tau$ is the torque in rotational motion, $I$ is the inertia and $\alpha$ is the rotational acceleration.

Complete step by step answer
The moment of inertia of a point mass with respect to an axis is defined as the product of the mass times the distance from the axis squared.
For a point mass, the moment of inertia is just the mass times the square of perpendicular distance to the rotation axis, $I = m{r^2}$ where $m$ is the mass and $r$ is the radius.
It is known to us that the analogue of mass is inertia, and the analogue of force is torque in rotational motion.
So, $\tau = I \cdot \alpha$ where $\tau$ is the torque in rotational motion, $I$ is the inertia and $\alpha$ is the rotational acceleration.
Acceleration in linear motion is given by $a = \dfrac{F}{m}$ where $a$ is the linear acceleration, $F$ is the force and $m$ is the mass of the body.
So the rotational acceleration is given similarly by, $\alpha = \dfrac{\tau }{I}$ .
Thus, greater the moment of inertia, less is the rotational acceleration of the body.
Moment of inertia of a solid sphere $\dfrac{{2m{r^2}}}{5}$ where $m$ is the mass and $r$ is the radius.
Moment of inertia of a hollow sphere $\dfrac{{2m{r^2}}}{3}$ where $m$ is the mass and $r$ is the radius.
Since the moment of inertia of a hollow sphere is more than the moment of inertia of a solid sphere, hence we can say that the acceleration is greater for solid sphere than the hollow sphere. So the solid sphere reaches the bottom with greater speed.
Hence the correct answer is option B.

Note
The spheres are released from the same position, thus the initial potential energy of both the hollow and solid sphere is $mgh$ where $m$ is the mass, $g$ is the acceleration due to gravity and $h$ is the vertical height of the inclined plane from the ground.
So when they reach the bottom they lose the same amount of potential energy and this loss of potential energy will be the gain in the kinetic energy. So both the spheres will have the same kinetic energy at the bottom. Hence option C is not correct.
Since the solid sphere reaches the bottom with a greater speed and both have the same mass, they cannot have the same linear momentum. Thus option D is also not correct.