Chapter 9, Problem 84P

(a)

To determine

The percent of total kinetic energy of a sphere is its translational kinetic energy.

Answer to Problem 84P

  $71.4\%$

Explanation of Solution

Given:

A uniform sphere is rolling without friction.

Formula used:

The total kinetic energy is sum of translational kinetic energy and rotational kinetic energy is given as:

  $\begin{array}{l}KE=K_{translational}+K_{rotational}\\ =\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2\end{array}$

Where, m is the mass, v is the velocity, I is the moment of inertia and $\omega$ is the angular velocity.

Calculation:

Translational Kinetic energy of sphere

  $K_{translational}=\frac{1}{2}m{v}^2$

Kinetic energy of an object is the sum of its rotational kinetic energy and translational kinetic energy which can be determined as follows

  $\begin{array}{l}KE=K_{translational}+K_{rotational}\\ =\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2\\ =\frac{1}{2}m{v}^2+\frac{1}{2}km{r}^2{v}^2/r^2\\ =\frac{1}{2}m{v}^2(1+k)\end{array}$

The ratio of translational kinetic energy and rotational kinetic energy is given as

  $\frac{K_{translational}}{KE}=\frac{\frac{1}{2}m{v}^2}{\frac{1}{2}m{v}^2(1+k)}\mathrm{...}(1)$

For sphere, $k=2/5$

Therefore,

  $\begin{array}{l}\frac{K_{translational}}{KE}=\frac{\frac{1}{2}m{v}^2}{\frac{1}{2}m{v}^2(1+2/5)}\\ \frac{K_{translational}}{KE}=\frac{1}{1+.4}\\ \frac{K_{translational}}{KE}=0.714\\ \frac{K_{translational}}{KE}=71.4\%\end{array}$

Conclusion:

71.4% percentage of a sphere total kinetic energy is its translational kinetic energy

(b)

To determine

The percent of kinetic energy is translational kinetic energy of a uniform cylinder.

Answer to Problem 84P

66.7%

Explanation of Solution

Given:

Uniform Cylinder is rolling without friction.

Formula used:

The total kinetic energy is sum of translational kinetic energy and rotational kinetic energy is given as:

  $\begin{array}{l}KE=K_{translational}+K_{rotational}\\ =\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2\end{array}$

Where, m is the mass, v is the velocity, I is the moment of inertia and $\omega$ is the angular velocity.

Calculation:

Translational Kinetic energy of uniform cylinder

  $K_{translational}=\frac{1}{2}m{v}^2$

Kinetic energy of an object is the sum of its rotational kinetic energy and translational kinetic energy which can be determined as follows

  $\begin{array}{l}KE=K_{translational}+K_{rotational}\\ =\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2\\ =\frac{1}{2}m{v}^2+\frac{1}{2}km{r}^2{v}^2/r^2\\ =\frac{1}{2}m{v}^2(1+k)\end{array}$

The ratio of translational kinetic energy and rotational kinetic energy is given as

  $\frac{K_{translational}}{KE}=\frac{\frac{1}{2}m{v}^2}{\frac{1}{2}m{v}^2(1+k)}\mathrm{...}(1)$

For uniform cylinder $k=1/2$

Therefore,

  $\begin{array}{l}\frac{K_{translational}}{KE}=\frac{\frac{1}{2}m{v}^2}{\frac{1}{2}m{v}^2(1+1/2)}\\ \frac{K_{translational}}{KE}=\frac{1}{1+.5}\\ \frac{K_{translational}}{KE}=0.667\\ \frac{K_{translational}}{KE}=66.7\%\end{array}$

Conclusion:

66.7% percentage of a uniform cylinder total kinetic energy is its translational kinetic energy

(c)

To determine

The percent of total kinetic energy of a hoop is its translational kinetic energy.

Answer to Problem 84P

50%

Explanation of Solution

Given:

Hoop is rolling without friction

Formula used:

The total kinetic energy is sum of translational kinetic energy and rotational kinetic energy is given as:

  $\begin{array}{l}KE=K_{translational}+K_{rotational}\\ =\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2\end{array}$

Where, m is the mass, v is the velocity, I is the moment of inertia and $\omega$ is the angular velocity.

Calculation:

Translational Kinetic energy of a hoop

  $K_{translational}=\frac{1}{2}m{v}^2$

Kinetic energy of an object is the sum of its rotational kinetic energy and translational kinetic energy which can be determined as follows

  $\begin{array}{l}KE=K_{translational}+K_{rotational}\\ =\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2\\ =\frac{1}{2}m{v}^2+\frac{1}{2}km{r}^2{v}^2/r^2\\ =\frac{1}{2}m{v}^2(1+k)\end{array}$

The ratio of translational kinetic energy and rotational kinetic energy is given as

  $\frac{K_{translational}}{KE}=\frac{\frac{1}{2}m{v}^2}{\frac{1}{2}m{v}^2(1+k)}\mathrm{...}(1)$

For hoop $k=1$

Therefore,

  $\begin{array}{l}\frac{K_{translational}}{KE}=\frac{\frac{1}{2}m{v}^2}{\frac{1}{2}m{v}^2(1+1)}\\ \frac{K_{translational}}{KE}=\frac{1}{1+1}\\ \frac{K_{translational}}{KE}=0.50\\ \frac{K_{translational}}{KE}=50.0\%\end{array}$

Conclusion:

50.0% percentage of a sphere total kinetic energy is its translational kinetic energy