Question

A uniform sphere of mass $500\mathrm{g}$ rolls without slipping on a plane surface so that its center moves at a speed of $0.02\mathrm{m}/\mathrm{s}$. What is the total kinetic energy of the rolling sphere in joules $\left(\mathrm{J}\right)$.

Answer 1

Step 1: Given information

Mass of sphere $=500\mathrm{g}$

Speed $=0.02\mathrm{m}/\mathrm{s}$

Step 2: Calculate the total kinetic energy of the rolling sphere

The total kinetic energy of the sphere $\left(\mathrm{E}\right)$ will be the sum of translational kinetic energy and rotational kinetic energy.

$\Rightarrow \mathrm{E}=\frac{1}{2}{\mathrm{mv}}^2+\frac{1}{2}{\mathrm{I\omega }}^2$

Here,

$\mathrm{m}$ is the mass of the sphere.

$\mathrm{v}$ is the velocity.

$\mathrm{\omega }$ is the angular velocity.

$\mathrm{I}$ is the moment of inertia of the sphere around the axis of rotation.

The moment of inertia of a sphere about the diameter is

$\mathrm{I}=\frac{2{\mathrm{MR}}^2}{5}$

Now,

$$\begin{gathered} \Rightarrow \mathrm{E}=\frac{1}{2}{\mathrm{mv}}^2+\frac{1}{2}(2{\mathrm{MR}}^2/5){\mathrm{\omega }}^2 \\ =\frac{1}{2}{\mathrm{mv}}^2+\frac{1}{5}{\mathrm{mv}}^2(v=R\omega ) \\ =\frac{7}{10}{\mathrm{mv}}^2 \end{gathered}$$

Substitute the given values in the above equation.

$$\begin{gathered} =\frac{7}{10}\times 0.5\times {(0.02)}^2 \\ =1.4\times {10}^{-4}\mathrm{J} \end{gathered}$$

Therefore, the total kinetic energy is $1.4\times {10}^{-4}\mathrm{J}$.