Answer to Question #138924 in Classical Mechanics for dein

Question

A uniform solid cylinder, a hollow cylinder, and a uniform solid sphere are rolling at the same speed on a horizontal table without slipping. If they all have the same mass and radius, which statement is true?
A. They all have the same angular momentum.
B. The hollow cylinder has the most angular momentum.
C. The solid sphere has the most total kinetic energy.
D. The solid cylinder has the greatest rate of rotation.
E. They all have the same kinetic energy of rotation

Answer 1

As per the given question,

Masses and Radius of all the given shapes are same. Let it is M and radius is R.

We know that the moment of inertia of the shapes-

Moment of inertia of the solid cylinder, about the axis $=\frac{MR^2}{2}$

moment of inertia of the solid cylinder about the point of rotation (about the point of contact at the ground)$(I_1)=\frac{3MR^2}{2}$

Moment of inertia of the hollow cylinder about the point of contact at the ground $(I_2)=2MR^2$

Moment of inertia of the uniform solid sphere $(I_3)=\frac{7MR^2}{2}$

Now,

Let the angular momentum of all the given shapes are $L_1, L_2$ and $L_3$ and angular speed of the shapes are $\omega_1, \omega_2$ and $\omega_3$. As per the question, all are moving with the same speed so $\omega_1=\omega_2=\omega_3=\omega$

$L_1=I_1\omega=\frac{3MR^2}{2}\omega$

$L_2=I_2\omega=2MR^2\omega$

$L_3=I_3\omega=\frac{7MR^2}{2}\omega$

from the above comparison, we can conclude that $L_3>L_2>L1$ So, statement A is incorrect.

Kinetic energy of the given shapes are $K_1=\frac{I_1\omega^2}{2}$

$K_2=\frac{I_2\omega^2}{2}$ and $K_3=\frac{I_3\omega^2}{2}$

As here, $I_3>I_2>I1$ , hence $K_3 > K_2>K_1$

From the above explanation, we can conclude that option (c) is the correct answer.