Question

Derive an expression for kinetic energy, when a rigid body is rolling on a horizontal surface without slipping. Hence find the kinetic energy of a solid sphere.

Answer 1

Hint: Define what is kinetic energy. Study how we can define the kinetic energy of a rolling body. Obtain the equation using suitable format. Then find the value of kinetic energy for a solid sphere using the physical quantities for a solid sphere.

Complete Step-by-Step solution:
The total kinetic energy of the rolling body can be defined as the sum of both translational kinetic energy and rotational kinetic energy.
So,
$\text{total kinetic energy = translational kinetic energy + rotational kinetic energy}$
Translational kinetic energy is the kinetic energy of a body moving in linear direction. Rotational kinetic energy is due to the angular velocity of the object.
$\text{total kinetic energy = }\dfrac{1}{2}M{{V}^{2}}+\dfrac{1}{2}I{{\omega }^{2}}$
Where, M is the mass of the rolling body, V is the linear velocity of the rolling body, I is the moment of inertia of the body and w is the angular velocity of the body.
Now, we can relate angular velocity with the linear velocity as,
$\omega =\dfrac{V}{R}$
So, putting this value on the above equation we can write,
$\text{total kinetic energy = }\dfrac{1}{2}M{{V}^{2}}+\dfrac{1}{2}I{{\left( \dfrac{V}{R} \right)}^{2}}$
Now consider a solid sphere of mass M and radius R. Now we can use the equation found above to find out the kinetic energy of the rolling sphere.
Let the linear velocity of the sphere is V and the angular velocity is $\omega$.
$\text{total kinetic energy = }\dfrac{1}{2}M{{V}^{2}}+\dfrac{1}{2}I{{\left( \dfrac{V}{R} \right)}^{2}}$
Now a moment of inertia of a rolling sphere can be found. The sphere is moving around an axis passing through its centre.
So, the moment of inertia of the rolling solid sphere is,
$I=\dfrac{2}{5}M{{R}^{2}}$
Putting this value on the above equation, we get,
$\begin{align}
  & \text{total kinetic energy = }\dfrac{1}{2}M{{V}^{2}}+\dfrac{1}{2}\times \dfrac{2}{5}M{{R}^{2}}{{\left( \dfrac{V}{R} \right)}^{2}} \\
 & \text{total kinetic energy = }\dfrac{1}{2}M{{V}^{2}}+\dfrac{1}{5}M{{V}^{2}} \\
 & \text{total kinetic energy = }\dfrac{7}{10}M{{V}^{2}} \\
\end{align}$
So, the kinetic energy of a rolling solid sphere is $\dfrac{7}{10}M{{V}^{2}}$

Note: When a body is moving in a straight line, we only have translational kinetic energy. Consider a car. It only has the forward motion. It doesn’t have rolling motion. Now consider a ball rolling in a straight path. It has one translational kinetic energy because it is moving forward and will have another rotational kinetic energy because it is also rolling in its motion. So, in this case we need to add these two kinds of energies.