Question

Two wheels $A$ and $C$ are connected by a belt $B$ as shown in the figure. The radius of $C$ is three times the radius of $A$. What would be the ratio of the rotational inertias $\left( \dfrac{{{I}_{C}}}{{{I}_{A}}} \right)$ if both the wheels have the same kinetic energy?

Answer 1

Hint: For finding the relation between the rotational inertias of wheels $A$ and $C$, we will apply the conservation of total energy throughout the motion rolling without slipping. For finding the relation between angular velocities of the two wheels, we will use the formula of relating linear and angular velocity of a body with the radius of the body or the radius of rotation, depending upon the situation.

Formula used:
$v=r\omega $
$K{{E}_{\text{Rotational}}}=\dfrac{I{{\omega }^{2}}}{2}$

Complete step by step answer:
Moment of inertia is a measure of the body's tendency to resist angular acceleration. It is described as the sum of the products of the mass of each mass particle in the body with the square of its distance from the axis of rotation.
Angular or rotational moment of inertia of a rigid body is a quantity that determines the value of torque needed to achieve a desired value of angular acceleration about a rotational axis. Rotational inertia is the property of any object or body which can rotate. It is a scalar quantity which tells us how difficult it is to change the value of rotational velocity of the object about a fixed rotational axis. Rotational inertia in rotational mechanics is analogous to mass in linear mechanics.
Rotational inertia is given by the symbol $I$. For a single object, such as a ball of mass$m$, rotating in the radius of $r$ from the axis of rotation,
Rotational inertia is given as,
$I=m{{r}^{2}}$
The SI unit of Moment of Inertia is $kg{{m}^{2}}$.
In rotational motion, when there is a situation of rolling without slipping, it signifies that the point of contact has no tendency to slip against the surface and hence there is no relative sliding between the object and the surface. We can also say that rolling without slipping is a combination of translation and rotation when the point of contact is instantaneously at rest.
We are given that two wheels $A$ and $C$ are connected by a belt $B$. Since there is no slipping between the belts, the velocity at the contact surface of the belt and wheel would be the same for both.
${{v}_{A}}={{v}_{C}}$
${{v}_{A}}={{v}_{C}}={{\omega }_{A}}\times {{r}_{A}}={{\omega }_{C}}\times {{r}_{C}}$
Given that,
${{r}_{C}}=3{{r}_{A}}$
Therefore, ${{\omega }_{A}}=3{{\omega }_{C}}$
In rolling without slipping condition, total energy of the rolling body remains the same. As it is given that the kinetic energies of the wheels are same; therefore, rotational energies for wheels $A$ and $C$ are same, so,
$\dfrac{{{I}_{A}}\omega _{A}^{2}}{2}=\dfrac{{{I}_{C}}\omega _{C}^{2}}{2}$
$\dfrac{{{I}_{A}}{{\left( 3{{\omega }_{C}} \right)}^{2}}}{2}=\dfrac{{{I}_{C}}\omega _{C}^{2}}{2}$
$\dfrac{{{I}_{A}}}{{{I}_{C}}}=\dfrac{1}{9}$
Therefore,
$\dfrac{{{I}_{C}}}{{{I}_{A}}}=9$

Note: In rolling without slipping motion, because of the absence of any non-conservative force that would take energy of the system in the form of heat, the total energy of a rolling object without slipping is conserved and is constant throughout the motion. In the above question, it was given that the kinetic energy of the wheels is the same. Therefore, we equate the rotational energy of the wheels.