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

${\text{K.E}}_R=\frac{1}{2}{\rm I}{\omega }^2$ …(1)

Here, $I$ 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,

$K.E_T=\frac{1}{2}M{v}^2$ …(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,

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

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

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

${K.E}_R=\frac{1}{2}\times \frac{2}{5}M{R}^2\times {\left(\frac{v}{d}\right)}^2$

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

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).

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

$=\frac{1}{2}M{d}^2$

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 ${\text{K.E}}_R=\frac{1}{2}\left(\frac{2}{5}M{R}^2\right){\left(\frac{v}{d}\right)}^2$.

role="math" localid="1654147660681" $K.E_{}=\frac{1}{2}M{d}^2$role="math" localid="1654147958776" $+\frac{1}{2}\left(\frac{2}{5}M{R}^2\right){\left(\frac{v}{d}\right)}^2$

role="math" localid="1654148067935" $=\frac{1}{2}\left(M{d}^2+\frac{2}{5}M{R}^2\right){\left(\frac{v}{d}\right)}^2$

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