91Ó°ÊÓ

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,

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

Here, l is the moment of inertia of the sphere and $\mathrm{Ó\neg }$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,

$\mathrm{Ó\neg }=\frac{v}{r}$

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,

$\mathrm{Ó\neg }=\frac{v}{d}$

Thus, the angular speed is $\mathrm{Ó\neg }=\frac{v}{d}$.

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 $\mathrm{Ó\neg }=\frac{v}{d}$in equation (1).

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

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 $v=\mathrm{Ó\neg }d$in equation (2).

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

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 $\frac{1}{2}\left(M{d}^2+\frac{2}{5}M{R}^2\right)×{\left(\frac{v}{d}\right)}^2$.