A uniform-density sphere whose mass is $\mathbf{10}\mathbf{kg}$and radius is $\mathbf{0}.\mathbf{4}\mathbf{m}$makes one complete rotation every$\mathbf{0}.\mathbf{2}\mathbf{s}$. What is the rotational kinetic energy of the sphere?

The given data can be listed below as-

• The mass of the uniform density sphere is, $10\text{kg}$.

• The radius of the sphere is,$0.4\text{m}$.
• The time taken by the sphere in making one complete rotation is, localid="1662468578481" $0.2\text{s}$

The moment of inertia for a body state that the force that a body exhibits while altering an axis is due to the application of a turning force which is torque.

The rotational kinetic energy can be determined by taking half of the product of the moment of inertia and angular velocity of the object. The expression is given as follows,

$K.E=\frac{1}{2}{\rm I}{\omega }^2$

…(1)

Here, $I$is the moment of inertia and $\omega$is the angular velocity.

The expression for the moment of inertia of a sphere can be expressed as,

$I=\frac{2}{5}M{R}^2$

Here, $M$is the mass of the sphere and R is the radius of the sphere.

For $M=10\text{kg}$and $R=0.4\text{m}$.

$$\begin{gathered} I=\frac{2}{5}\times \left(10kg\right)\times {\left(0.4m\right)}^2 \\ =0·64kg·m \end{gathered}$$

The expression for the angular velocity can be expressed as,

$\omega =2\pi f$

Here, $f$is the frequency of revolution that is assumed to belocalid="1662468686058" $5\text{rad/s}$Substituting the values in the above equation.

localid="1662468692118" $$\begin{gathered} \omega =2\mathrm{\pi }\times 5\mathrm{rad}/\mathrm{s} \\ =10\mathrm{\pi }\mathrm{rad}/\mathrm{s} \end{gathered}$$

Substituting all the values in equation (1).

$$\begin{gathered} K·E=\frac{1}{2}\times \left(0.64kg.m^2\right)\times {\left(10\mathrm{\pi }\mathrm{rad}/\mathrm{s}\right)}^2 \\ =315.827kg.m^2/s^2 \\ =315.827kg.m^2/s^2\times \frac{1J}{1kg.m^2/s^2} \\ =316J \end{gathered}$$

Thus, the rotational kinetic energy of the sphere is $316\text{J}$.