Chapter 8, Problem 45P

To determine

To find: The total kinetic energy.

Answer to Problem 45P

The total kinetic energy of the bowling ball is $56\text{J}$ .

Explanation of Solution

Given information:

A bowling ball of mass $7.3\text{kg}$ , radius $9.0\text{cm}$ rolls without slipping down a lane at $3.3\text{m/s}$ .

Formula used:

Write the expression for the energy conversion.

  $\begin{array}{l}KE=K{E}_{trans}+K{E}_{rot}\\ KE=\frac{1}{2}m{v}_{CM}^2+\frac{1}{2}{I}_{CM}{\omega }^2\end{array}$

Here,

  $KE$ is the total kinetic energy,

  $K{E}_{trans}$ is the translational kinetic energy,

  $K{E}_{rot}$ is the rotational kinetic energy,

  $m$ is the mass,

  $v_{CM}$ is the velocity of center of mass,

  $I_{CM}$ is the moment of inertia at center of mass,

  $\omega$ is the angular velocity.

Calculation:

The mass of ball $m=7.3\text{kg}$ and velocity of the ball is $v=3.3\text{m/s}$ . The total kinetic energy is given by

  $\begin{array}{l}KE=K{E}_{trans}+K{E}_{rot}\\ KE=\frac{1}{2}m{v}_{CM}^2+\frac{1}{2}{I}_{CM}{\omega }^2\end{array}$

Since the ball is rolling without slipping so the angular velocity is given by $\omega =\frac{v}{R}$ and the rotational inertia of a sphere about an axis through its center is

  $I=\frac{2}{5}m{R}^2$

So,

  $\begin{array}{l}KE=\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{2}{5}m{R}^2\right){\left(\frac{v}{R}\right)}^2\\ =\frac{1}{2}m{v}^2+\frac{1}{5}m{v}^2\\ KE=\frac{7}{10}m{v}^2\end{array}$

Substituting the values we get

  $\begin{array}{l}KE=\frac{7}{10}\left[\left(\text{7}\text{.3}\text{kg}\right)\text{×}{\left(\text{3}\text{.3}\text{m/s}\right)}^{\text{2}}\right]\text{J}\\ KE=55.6\text{J}\end{array}$

The total kinetic energy of the bowing ball $56\text{J}$ .

Conclusion:

The total kinetic energy of the bowing ball $56\text{J}$ .