Question

A symmetrical body of mass $M$, radius $R$ and radius of gyration $k$ is rolling on a horizontal surface without slipping. If the linear velocity of the centre of mass is ${v_c}$ and the angular velocity is $\omega $ then which of the following options are correct?
A. The total K.E. of the body is $\dfrac{1}{2}M{v_c}^2\left( {1 + \dfrac{{{k^2}}}{{{R^2}}}} \right)$ .
B. The rotational K.E. of the body is $\dfrac{1}{2}M{R^2}{\omega ^2}$ .
C. The translational K.E. of the body is $\dfrac{1}{2}M{v_c}^2$ .
D. The moment of inertia of the body is $M{k^2}$ .

Answer 1

Hint:The rolling body possesses rotational kinetic energy and translational kinetic energy. The total kinetic energy of the body will be the sum of these two kinetic energies. The moment of inertia of the rolling body is expressed in terms of the mass of the body and its radius of gyration.

Formulae used:
->The translational kinetic energy of a rolling body is given by, $K{E_{transl}} = \dfrac{1}{2}m{v^2}$ where $m$ is the mass of the body and $v$ is its linear velocity.
->The rotational kinetic energy of a rolling body is given by, $K{E_{rot}} = \dfrac{1}{2}I{\omega ^2}$ where $I$ is the moment of inertia of the body and $\omega $ is its angular velocity.
->The angular velocity of a rotating body is given by, $\omega = \dfrac{v}{r}$ where $r$ is the radius of the circle described by the rotating body and $v$ is its linear velocity.
->The moment of inertia of a body is given by, $I = m{r^2}$ where $m$ is the mass of the body and $r$ is the distance from the axis of rotation.

Complete step-by-step solution:
->Step 1: Mention the features of the given body.
The mass of the rolling body is $M$ and its radius is given to be $R$ .
The radius of gyration of the body is $k$ .
It is also given that the linear velocity of the centre of mass of the body is ${v_c}$ while its angular velocity is given to be $\omega $ .
We have to determine its moment of inertia, translational kinetic energy, rotational kinetic energy and total kinetic energy.
->Step 2: Express the moment of inertia of the given body.
The moment of inertia of the given body can be expressed as $I = M{r^2}$ ; $r$ is the distance from the axis of rotation to the centre of mass.
The distance from the axis of rotation is often referred to as the radius of gyration i.e., $r = k$
So the moment of inertia of the given body will be $I = M{k^2}$ .
This is in agreement with option D.
->Step 3: Express the translational kinetic energy of the rolling body.
The translational kinetic energy of a rolling body is given by, $K{E_{transl}} = \dfrac{1}{2}m{v^2}$ where $m$ is the mass of the body and $v$ is its linear velocity.
Here we have the mass of the body as $M$ and its linear velocity is given to be ${v_c}$ .
So the translational kinetic energy of the given body will be $K{E_{transl}} = \dfrac{1}{2}M{v_c}^2$
This is in agreement with option C.
->Step 4: Express the rotational kinetic energy of the rolling body.
The rotational kinetic energy of a rolling body is given by, $K{E_{rot}} = \dfrac{1}{2}I{\omega ^2}$ where $I$ is the moment of inertia of the body and $\omega $ is its angular velocity.
Substituting for $I = M{k^2}$ in the above relation we get, $K{E_{rot}} = \dfrac{1}{2}M{k^2}{\omega ^2}$.
This is not in agreement with option B.
->Step 5: Express the total kinetic energy of the given body.
The total kinetic energy of the given body will be the sum of its translational kinetic energy and its rotational kinetic energy.
i.e., $K{E_{total}} = K{E_{transl}} + K{E_{rot}}$ ------- (1)
Substituting for $K{E_{rot}} = \dfrac{1}{2}M{k^2}{\omega ^2}$ and $K{E_{transl}} = \dfrac{1}{2}M{v_c}^2$ in equation (1) we get, $K{E_{total}} = \dfrac{1}{2}M{v_c}^2 + \dfrac{1}{2}M{k^2}{\omega ^2}$
Substituting for $\omega = \dfrac{{{v_c}}}{R}$ in the above equation we get, $K{E_{total}} = \dfrac{1}{2}M{v_c}^2 + \dfrac{1}{2}M{k^2}\dfrac{{{v_c}^2}}{{{R^2}}}$
$ \Rightarrow K{E_{total}} = \dfrac{1}{2}M{v_c}^2\left( {1 + \dfrac{{{k^2}}}{{{R^2}}}} \right)$
This is in agreement with option A.
So the correct options are A, C and D.

Note:- As the body rolls along the horizontal path, the body undergoes a translational motion as the body moves forward and a rotational motion as it rotates about an axis. The radius of gyration only exists for a body undergoing rotation. The radius of gyration refers to the distance at a point of mass from the axis of rotation where the entire mass of the body will be equal to the mass at that point. The moment of inertia of the whole body will also be the same at the distance equal to the radius of gyration.