Question

A soccer ball (k = 2/3) with a radius of 0.109 meters and a mass of 0.407 kilograms rolls on a field with an angular speed of 98 radians per second. a. What is the translational kinetic energy of the ball?b. What is the rotational kinetic energy of the ball?c. The ball then rolls up a hill. What height will the ball reach before coming to rest?

Answer 1

Given,

The value of k is

$k=\frac{2}{3}$

The radius is

$r=0.109\text{ m}$

The mass is

$m=0.407\text{ kg}$

The angular speed is

$\omega=98\text{ radians per second}$

To find:

(a) What is the translational kinetic energy of the ball.

$\begin{gathered} v=\omega r \\ v=98\times0.109 \\ v=10.682\text{ m/s} \end{gathered}$

Now, kinetic energy is

$\begin{gathered} KE=\frac{1}{2}\times0.407\times(10.682)^2 \\ KE=23.22\text{ J} \end{gathered}$

(b) The rotational kinetic energy is

$\begin{gathered} KE=\frac{1}{2}I\omega^2 \\ KE=\frac{1}{2}\times mk^2\times\omega^2 \\ KE=\frac{1}{2}\times0.407\times(\frac{2}{3})^2\times(98)^2 \\ KE=8.86\text{ J} \end{gathered}$

(c) The ball then rolls up a hill.

What height will the ball reach before coming to rest?

$\begin{gathered} KE=PE \\ \frac{1}{2}mv^2=mgh \\ \frac{v^2}{2}=gh \end{gathered}$

The value of the acceleration due to gravity is 9.8 m/s^2.

Put the given values.

$\begin{gathered} h=\frac{v^2}{2g} \\ h=\frac{(10.682)^2}{2\times9.8} \\ h=5.82\text{ m} \end{gathered}$

Thus, the value of h is 5.82 m.