A size-5 soccer ball of diameter $\mathbf{22}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{cm}$and mass $\mathbf{426}\mathbf{}\mathbf{g}$rolls up a hill without slipping, reaching a maximum height of $\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}$above the base of the hill. We can model this ball as a thin-walled hollow sphere. (a) At what rate was it rotating at the base of the hill? (b) How much rotational kinetic energy did it have then?

Mass of the ball $\left(m\right)=426g$.

Diameter of the ball $\left(D\right)=22.6cm$.

The radius of the ball $\left(R\right)=11.3cm$.

Height $\left(h\right)=5.00m$.

Determine the formula for the potential energy:

$\mathbf{U}\mathbf{=}\mathbf{mgh}$ …… (1)

Determine the formula for the kinetic energy:

$$\begin{gathered} {\mathbf{K}}_{\mathbf{tot}}\mathbf{=}{\mathbf{K}}_{\mathbf{tr}}\mathbf{+}{\mathbf{K}}_{\mathbf{rot}} \\ {\mathbf{K}}_{\mathbf{tot}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{mv}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{I\omega }}^{\mathbf{2}} \end{gathered}$$ …… (2)

Consider the formula for the speed at the center of mass:

$v=R\omega$ …… (3)

Here, $\omega$is the angular speed of the ball about its center of mass and R is radius of the ball.

Consider the formula for the moment of inertia:

$I=\frac{2}{3}m{R}^2$ …… (4)

Thus, using $\left(3\right), \left(4\right)$in equation $\left(2\right)$solve as:

$$\begin{gathered} K_{tot}=\frac{1}{2}m{R}^2{\omega }^2+\frac{1}{2}·\frac{2}{3}m{R}^2{\omega }^2 \\ {K}_{tot}=\frac{1}{2}m{R}^2{\omega }^2+\frac{1}{3}m{R}^2{\omega }^2 \\ {K}_{tot}=\frac{5}{6}m{R}^2{\omega }^2 \end{gathered}$$ …… (5)

Now, equating equations $\left(1\right)$and $\left(5\right)$.

localid="1667972804314" $$\begin{gathered} mgh=\frac{5}{6}m{R}^2{\omega }^2 \\ \omega =\sqrt{\frac{6gh}{5R^2}} \end{gathered}$$ …… (6)

Now, substituting the values in $\left(6\right)$and solve:

$$\begin{gathered} \omega =\sqrt{\frac{6gh}{5R^2}} \\ \Rightarrow \omega =\sqrt{\frac{6\left(9.80\right)\left(5.00\right)}{5{\left(0.113\right)}^2}} \\ \Rightarrow \omega =67.9\frac{rad}{\text{s}} \end{gathered}$$

Hence, the angular speed is, $\omega =67.9\frac{rad}{\text{s}}$.

The rotational kinetic energy from the base is,

$K_{rot}=\frac{1}{2}I{\omega }^2=\frac{1}{3}m{R}^2{\omega }^2$.

Substituting values, we get,

Hence, the rotational kinetic energy is, $K_{rot}=8.35 J$.