
A hollow sphere and a solid sphere having the 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 We are given two spheres, one of them is hollow and the other one is solid and are said that they are rolled down a rough inclined plane at the same time and are asked to evaluate the motion of the two spheres. Thus, we will use the formula for moment of inertia for both the spheres and then will compare the resistance and the velocity of the spheres.
Formulae Used:
${I_H} = \dfrac{2}{3}{M_H}{R_H}^2$
Where,${I_H}$ is the moment of inertia of a hollow sphere,${M_H}$ is the mass of the hollow sphere and${R_H}$ is the radius of the hollow sphere.
${I_S} = \dfrac{2}{5}{M_S}{R_S}^2$
Where,${I_S}$ is the moment of inertia of a solid sphere,${M_S}$ is the mass of a solid sphere and ${R_S}$ is the radius of the solid sphere.
Complete step by step answer
Here,
Given,
The mass of the hollow and the solid sphere are equal.
Let us assume their mass to be $M$.
That is,
${M_H} = {M_S} = M$
Also,
It is given that their radii are also equal.
Let us assume that the radius is $R$.
That is,
${R_H} = {R_S} = R$
Thus,
The moment of inertia of the hollow sphere is,
${I_H} = \dfrac{2}{3}M{R^2}$
And,
The moment of inertia of solid sphere is,
${I_S} = \dfrac{2}{5}M{R^2}$
Now,
Clearly,
Moment of inertia of the hollow sphere is greater than the moment of inertia of the solid sphere.
That is,
${I_H} > {I_S}$
Thus,
The resistance on the hollow sphere will be more than that on the solid sphere.
Thus,
The solid sphere will have a faster motion than that of the hollow sphere.
Hence,
The solid sphere will reach the bottom first. Also, as the velocity of the solid sphere is higher than that of the hollow sphere and thus will reach the bottom with a greater kinetic energy.
Hence, The correct options are (B) and (C).
Note We have compared the motion of the two spheres with respect to their moment of inertia as the bodies are rolling as their motion proceeds. Also, the only parameter which defines the motion of the spheres is its moment of inertia.
Formulae Used:
${I_H} = \dfrac{2}{3}{M_H}{R_H}^2$
Where,${I_H}$ is the moment of inertia of a hollow sphere,${M_H}$ is the mass of the hollow sphere and${R_H}$ is the radius of the hollow sphere.
${I_S} = \dfrac{2}{5}{M_S}{R_S}^2$
Where,${I_S}$ is the moment of inertia of a solid sphere,${M_S}$ is the mass of a solid sphere and ${R_S}$ is the radius of the solid sphere.
Complete step by step answer
Here,
Given,
The mass of the hollow and the solid sphere are equal.
Let us assume their mass to be $M$.
That is,
${M_H} = {M_S} = M$
Also,
It is given that their radii are also equal.
Let us assume that the radius is $R$.
That is,
${R_H} = {R_S} = R$
Thus,
The moment of inertia of the hollow sphere is,
${I_H} = \dfrac{2}{3}M{R^2}$
And,
The moment of inertia of solid sphere is,
${I_S} = \dfrac{2}{5}M{R^2}$
Now,
Clearly,
Moment of inertia of the hollow sphere is greater than the moment of inertia of the solid sphere.
That is,
${I_H} > {I_S}$
Thus,
The resistance on the hollow sphere will be more than that on the solid sphere.
Thus,
The solid sphere will have a faster motion than that of the hollow sphere.
Hence,
The solid sphere will reach the bottom first. Also, as the velocity of the solid sphere is higher than that of the hollow sphere and thus will reach the bottom with a greater kinetic energy.
Hence, The correct options are (B) and (C).
Note We have compared the motion of the two spheres with respect to their moment of inertia as the bodies are rolling as their motion proceeds. Also, the only parameter which defines the motion of the spheres is its moment of inertia.
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