A solid uniform-density sphere is tied to a rope and moves in a circle with speed v. The distance from the center of the circle to the center of the sphere is M, the mass of the sphere is, and the radius of the sphere is R. (a) What is the angular speed$\mathit{\omega }$? (b) What is the rotational kinetic energy of the sphere? (c) What is the total kinetic energy of the sphere?

The given data is listed below as,

• The velocity of the solid-uniform density sphere is v.
• The distance from the center of the circle to the center of the sphere is d.
• The sphere’s mass is M.
• The sphere’s radius is R.

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

The equation of the rotational kinetic energy of the sphere can be expressed as,

$\mathrm{K}.{\mathrm{E}}_{\mathrm{R}}=\frac{1}{2}{\mathrm{I\omega }}^2$ …(1)

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

The equation of the translational kinetic energy of the sphere is be expressed as,

$\mathrm{K}.{\mathrm{E}}_{\mathrm{T}}=\frac{1}{2}{\mathrm{Mv}}^2$ .....(2)

Here, M is the mass of the sphere and v is the velocity of the sphere at center of mass of the sphere.

The equation of the angular speed can be expressed as,

$\omega =\frac{v}{r}$

Here, v is the velocity and r is the distance of the centre of the circle to the centre of the sphere.

For r = d,

$\omega =\frac{v}{d}$

Thus, the angular speed is $\omega =\frac{v}{d}$.

The expression for the moment of inertia of the sphere is expressed as,

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

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

For $l=\frac{2}{5}M{R}^2$and $\omega =\frac{v}{d}$in equation (1).

$$\begin{gathered} K.E_R=\frac{1}{2}\times \frac{2}{5}M{R}^2\times {\left(\frac{v}{d}\right)}^2 \\ =\frac{1}{2}\left(\frac{2}{5}M{R}^2\right){\left(\frac{v}{d}\right)}^2 \end{gathered}$$

Thus, the rotational kinetic energy of the sphere is $\frac{1}{2}\left(\frac{2}{5}M{R}^2\right){\left(\frac{v}{d}\right)}^2$.

For $v=\omega d$in equation (2).

$$\begin{gathered} K.E_T=\frac{1}{2}M{\left(\omega d\right)}^2 \\ =\frac{1}{2}M{d}^2 \end{gathered}$$

The expression for the total kinetic energy is expressed as,

$K.E=K.E_T+K.E_R$

For $K.E_T=\frac{1}{2}M{d}^2$and$E_R=\frac{1}{2}\left(\frac{2}{5}M{R}^2\right){\left(\frac{v}{d}\right)}^2$.

$$\begin{gathered} K.E=\frac{1}{2}M{d}^2+\frac{1}{2}\left(\frac{2}{5}M{R}^2\right){\left(\frac{v}{d}\right)}^2 \\ =\frac{1}{2}\left(M{d}^2+\frac{2}{5}M{R}^2\right){\left(\frac{v}{d}\right)}^2 \end{gathered}$$

Thus, the total kinetic energy of the sphere is $\frac{1}{2}\left(M{d}^2+\frac{2}{5}M{R}^2\right)\times {\left(\frac{v}{d}\right)}^2$.