Chapter 13, Problem 32PQ

(a)

To determine

The ratio of the rotational kinetic energy to the translational kinetic energy for each toy.

Answer to Problem 32PQ

The ratio of the rotational kinetic energy to the translational kinetic energy for spherical toy is $\frac{2}{5}$ and ratio of the rotational kinetic energy to the translational kinetic energy for cylindrical toy is $\frac{1}{2}$

Explanation of Solution

Take $R$ to represent the radius of both cylindrical toy as well as spherical toy, since radius will be cancelled in kinetic energy expression of each one. Given that translational speed is same for spherical and cylindrical toy.

Write the relation between linear and angular speed of rotation.

  $v=R\omega$                                                                                                                    (I)

Here, $R$ is the radius, $v$ is the linear speed and $\omega$ is the angular speed.

Rearrange the equation for $\omega$.

  $\omega =\frac{v}{R}$                                                                                                                     (II)

Write the expression for rotational kinetic energy of the spherical shaped toy

  $K_{r,sphere}=\frac{1}{2}{I}_{\text{sphere}}{\omega }^2$                                                                                                (III)

Here, $K_{r,sphere}$ is the rotational kinetic energy of the sphere spherical shaped toy and $I$ is the rotational inertia of the spherical shaped toy.

Write the expression for rotational inertia of the spherical shaped toy.

  $I_{\text{sphere}}=\frac{2}{5}M{R}^2$                                                                                                        (IV)

Here, $M$ is the mass of the spherical shaped toy

Write the expression for rotational kinetic energy of the cylindrical shaped toy.

  $K_{r,cyl}=\frac{1}{2}{I}_{\text{cyl}}{\omega }^2$                                                                                                       (V)

Here, $K_{r,cyl}$ is the rotational kinetic energy of the small cylindrical shape toys and $I_{\text{cyl}}$ is the rotational inertia of the cylinder.

Write the expression for rotational inertia of the cylinder.

  $I_{\text{cyl}}=\frac{1}{2}M{R}^2$                                                                                                           (VI)

Here, $M$ is the mass of the toys in cylindrical shaped toy

Write the expression for the translational kinetic energy of spherical shaped toy.

  $K_{t,sphere}=\frac{1}{2}M{v}^2$                                                                                                    (VII)

Write the expression for the translational kinetic energy of cylindrical shaped toy.

  $K_{t,cyl}=\frac{1}{2}M{v}^2$                                                                                                      (VIII)

Conclusion:

Substitute equations (II) and (IV) in equation (III).

  $\begin{array}{c}{K}_{r,sphere}=\frac{1}{2}\left(\frac{2}{5}M{R}^2\right){\left(\frac{v}{R}\right)}^2\\ =\frac{1}{2}\left(\frac{2}{5}M{R}^2\right)\frac{v^2}{R^2}\\ =\frac{2}{10}M{v}^2\\ =\frac{1}{5}M{v}^2\end{array}$                                                                                    (IX)

Substitute equations (II) and (VI) in equation (V).

  $\begin{array}{c}{K}_{r,cyl}=\frac{1}{2}\left(\frac{1}{2}M{R}^2\right)\frac{v^2}{R^2}\\ =\frac{1}{4}M{v}^2\end{array}$                                                                                            (X)

Divide equation (IX) by (VI) to get ratio of rotational kinetic energy to the translational kinetic energy for spherical shaped toy.

  $\begin{array}{c}\frac{K_{r,sphere}}{K_{t,sphere}}=\frac{\frac{1}{5}M{v}^2}{\frac{1}{2}m{v}^2}\\ =\frac{2}{5}\end{array}$

Divide equation (X) by (VII) to get ratio of rotational kinetic energy to the translational kinetic energy for cylindrical shaped toy.

  $\begin{array}{c}\frac{K_{r,cyl}}{K_{t,cyl}}=\frac{\frac{1}{4}M{v}^2}{\frac{1}{2}M{v}^2}\\ =\frac{1}{2}\end{array}$

Therefore, the ratio of the rotational kinetic energy to the translational kinetic energy for spherical toy is $\frac{2}{5}$ and ratio of the rotational kinetic energy to the translational kinetic energy for cylindrical toy is $\frac{1}{2}$ .

(b)

To determine

Comparison of the translational speeds of sphere and cylinder if they same angular speed instead of same translational speed.

Answer to Problem 32PQ

The linear speed of cylinder is $0.65$ times the linear speed of sphere.

Explanation of Solution

Write the expression for angular speed of sphere.

  ${\omega }_{sphere}=\frac{v_{sphere}}{R_{sphere}}$                                                                                                       (XI)

Here, ${\omega }_{sphere}$ is the angular speed of the sphere, $v_{sphere}$ is the linear speed of sphere and $R_{sphere}$ is the radius of sphere.

Write the expression for angular speed of cylinder.

  ${\omega }_{cyl}=\frac{v_{cyl}}{R_{cyl}}$                                                                                                          (XII)

Here, ${\omega }_{cyl}$ is the angular speed of cylinder, $v_{cyl}$ is the linear speed of cylinder and $R_{cyl}$ is the radius of cylinder.

Conclusion:

In problem 31, it is given that radius of cylinder is $0.013\text{m}$ and radius of sphere is $0.020\text{m}$ . In problem, it is given that angular speed of sphere and cylinder are same.

Write condition given in question.

  ${\omega }_{sphere}={\omega }_{cyl}$

Equate equation (XI) and (XII) to get $\frac{v_{cyl}}{v_{sphere}}$ .

  $\begin{array}{l}\frac{v_{sphere}}{R_{sphere}}=\frac{v_{cyl}}{R_{cyl}}\\ \frac{v_{cyl}}{v_{sphere}}=\frac{R_{cyl}}{R_{sphere}}\end{array}$

Substitute $0.013\text{m}$ for $R_{cyl}$ and $0.020\text{m}$ for $R_{sphere}$ in above equation to get $\frac{v_{cyl}}{v_{sphere}}$ .

  $\begin{array}{c}\frac{v_{cyl}}{v_{sphere}}=\frac{0.013\text{m}}{0.020\text{m}}\\ v_{cyl}=0.65v_{sphere}\end{array}$

Therefore, the linear speed of cylinder is $0.65$ times the linear speed of sphere.