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
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Hint: The quantity that marks the distinction between the two spheres i.e., a solid sphere and a hollow sphere is their moments of inertia. The moment of inertia of a particle around an axis is just the product of the square of its distance from that axis to its mass.
Formula used:
Moment of inertia of a solid sphere:
$I=\dfrac{2}{5}MR^2$
Moment of inertia of a hollow sphere:
$I=\dfrac{2}{3}MR^2$
Complete step-by-step solution:
First, we compare the moment of inertia of a solid sphere and a hollow sphere. Clearly the moment of inertia of a solid sphere which is $\dfrac{2}{5}MR^2$ is lower than the moment of inertia of a hollow sphere $\dfrac{2}{3}MR^2$ (as the denominator is more for solid sphere).
The analog of force F is just torque in rotational motion and the analogue of mass is I inertia. We find $a = \dfrac{F}{m}$ in linear motion. So, in case of rotational motion when $\tau = I.\alpha$ the acceleration will be given by
$\alpha = \dfrac{\tau}{I}$
This formula is quite enough to tell us that more the moment of inertialessness will be the acceleration. In our case the moment of inertia for a hollow sphere is more, therefore, it will have lesser acceleration. We, therefore, can say that a solid sphere will have more acceleration as compared to the hollow sphere.
Therefore, the correct answer seems to be an option (B). The solid sphere reaches the bottom first with greater speed as its acceleration is more.
Note: The question mentions a rolling without slipping condition as the inclined surface is rough. Also, the formula for acceleration had an alpha in it which is angular acceleration which is the rate of change of angular velocity. It is not the same as the regular linear acceleration, which is the rate of the change of linear velocity.
Formula used:
Moment of inertia of a solid sphere:
$I=\dfrac{2}{5}MR^2$
Moment of inertia of a hollow sphere:
$I=\dfrac{2}{3}MR^2$
Complete step-by-step solution:
First, we compare the moment of inertia of a solid sphere and a hollow sphere. Clearly the moment of inertia of a solid sphere which is $\dfrac{2}{5}MR^2$ is lower than the moment of inertia of a hollow sphere $\dfrac{2}{3}MR^2$ (as the denominator is more for solid sphere).
The analog of force F is just torque in rotational motion and the analogue of mass is I inertia. We find $a = \dfrac{F}{m}$ in linear motion. So, in case of rotational motion when $\tau = I.\alpha$ the acceleration will be given by
$\alpha = \dfrac{\tau}{I}$
This formula is quite enough to tell us that more the moment of inertialessness will be the acceleration. In our case the moment of inertia for a hollow sphere is more, therefore, it will have lesser acceleration. We, therefore, can say that a solid sphere will have more acceleration as compared to the hollow sphere.
Therefore, the correct answer seems to be an option (B). The solid sphere reaches the bottom first with greater speed as its acceleration is more.
Note: The question mentions a rolling without slipping condition as the inclined surface is rough. Also, the formula for acceleration had an alpha in it which is angular acceleration which is the rate of change of angular velocity. It is not the same as the regular linear acceleration, which is the rate of the change of linear velocity.
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