Chapter 11, Problem 61CP

(a)

To determine

The angular speed of the disk once pure rolling takes place.

Answer to Problem 61CP

The angular speed of the disk once pure rolling takes place is $\underset{\_}{\frac{1}{3}{\omega }_i}$.

Explanation of Solution

For the particle under the net force model the net force will be the rate of change of linear momentum of the body.

Write the expression for net force on the disc .

  $f_k=\frac{m\left(v_f-v_i\right)}{\Delta t}$

Here, $m$ is mass of the disc, $v_f$ is final linear velocity of the disc, $\Delta t$ is time of contact, $v_i$ is initial linear velocity of the disc and $f_k$ is for net force on the disc.

The initial velocity of the disc is zero.

Substitute $0$ for $v_i$ in above expression as.

  $f_k=\frac{m{v}_f}{\Delta t}$                                                                                                       (I)

For the particle under the net torque model the net torque will be the rate of change of angular momentum of the body.

Write the expression for net torque on the disc as.

  $-f_{k}R=\frac{I\left({\omega }_f-{\omega }_i\right)}{\Delta t}$

Simplify the above expression for $f_k$ as.

  $-f_k=\frac{I\left({\omega }_f-{\omega }_i\right)}{R\Delta t}$                                                                                         (II)

Here, ${\omega }_i$ is the initial angular speed of the disc, ${\omega }_f$ is the final angular speed of the disc, $I$ is a moment of inertia of the disc about a rotational axis and $R$ is the radius of the disc.

Substitute $\frac{m{v}_f}{\Delta t}$ for $f_k$ in equation (II).

  $-\frac{m{v}_f}{\Delta t}=\frac{I\left({\omega }_f-{\omega }_i\right)}{R\Delta t}$

Simplify the above-obtained expression for $R$,

  $R=\frac{I\left({\omega }_i-{\omega }_f\right)}{m{v}_f}$                                                                                           (III)

Write the expression for final linear velocity of the disc.

  $v_f=R{\omega }_f$                                                                                                    (IV)

Substitute $R{\omega }_f$ for $v_f$ in equation (IV).

  $R=\frac{I\left({\omega }_i-{\omega }_f\right)}{mR{\omega }_f}$                                                                                           (V)

Simplify equation (V) for ${\omega }_f$ .

  $R=\frac{I\left({\omega }_i-{\omega }_f\right)}{mR{\omega }_f}$

Re-arrange the terms.

  $\begin{array}{l}m{R}^2{\omega }_f=I{\omega }_i-I{\omega }_f\\ {\omega }_f\left(I+m{R}^2\right)=I{\omega }_i\end{array}$

Simplify the above expression for ${\omega }_f$ .

  ${\omega }_f=\frac{I{\omega }_i}{I+m{R}^2}$                                                                                            (VI)

Write the expression for moment of inertia of the disc as.

  $I=\frac{1}{2}m{R}^2$                                                                                                 (VII)

Conclusion:

Substitute $\frac{1}{2}m{R}^2$ for $I$ in equation (VII).

  ${\omega }_f=\frac{\left(\frac{1}{2}m{R}^2\right){\omega }_i}{\left(\frac{1}{2}m{R}^2\right)+m{R}^2}$

Simplify the above expression for ${\omega }_f$ it gives.

  $\begin{array}{c}{\omega }_f=\frac{\frac{1}{2}m{R}^2{\omega }_i}{\frac{3}{2}m{R}^2}\\ =\frac{1}{3}{\omega }_i\end{array}$

Thus, the angular speed of the disk once pure rolling takes place is $\underset{\_}{\frac{1}{3}{\omega }_i}$.

(b)

To determine

The fractional change in kinetic energy from the moment the disk is set down until pure rolling occurs.

Answer to Problem 61CP

The fractional change in kinetic energy from the moment the disk is set down until pure rolling occurs $\underset{\_}{-\frac{2}{3}}$.

Explanation of Solution

Write the expression for the rotational kinetic energy of the disc as.

  ${\left(KE\right)}_R=\frac{1}{2}I{\omega }^2$                                                                                        (VIII)

Here, ${\left(KE\right)}_R$ is the rotational kinetic energy of the disc and $\omega$ is the corresponding angular speed of the disc.

Write the expression for kinetic energy due to the linear velocity of the disc as

  $E_L=\frac{1}{2}m{v}_{CM}{}^2$                                                                                              (IX)

Here, $E_L$ is kinetic energy due to the linear velocity of the disc and $v_{CM}$ linear velocity of the center of mass of the disc.

The change in kinetic energy is the difference between initial and final kinetic energy.

Initially, the disc is only rotating so initial kinetic energy is rotational kinetic energy. In the final condition, the disc has linear and angular velocities so the kinetic energy becomes the sum of linear and rotational kinetic energy.

Write the expression for the change in kinetic energy of the disc as.

  $\Delta E=\frac{1}{2}I{\omega }_f{}^2+\frac{1}{2}m{v}_{CM}{}^2-\frac{1}{2}I{\omega }_i{}^2$                                                                  (X)

Here, $\Delta E$ is the change in kinetic energy.

Divide equation (X) by equation (IX) it gives.

  $\frac{\Delta E}{E_L}=\frac{\frac{1}{2}I{\omega }_f{}^2+\frac{1}{2}m{v}_{CM}{}^2-\frac{1}{2}I{\omega }_i{}^2}{\frac{1}{2}I{\omega }_i{}^2}$                                                              (XI)

Conclusion:

Substitute $\frac{{\omega }_i}{3}$ for ${\omega }_f$, $\frac{1}{2}m{R}^2$ for $I$ and $\frac{R{\omega }_i}{3}$ for $v_{CM}$ in equation (XI)

  $\begin{array}{c}\frac{\Delta E}{E_L}=\frac{\frac{1}{2}\left(\frac{1}{2}m{R}^2\right){\left(\frac{{\omega }_i}{3}\right)}^2+\frac{1}{2}m{\left(\frac{R{\omega }_i}{3}\right)}^2-\frac{1}{2}\left(\frac{1}{2}m{R}^2\right){\omega }_i{}^2}{\frac{1}{2}\left(\frac{1}{2}m{R}^2\right){\omega }_i{}^2}\\ =\frac{\left(\frac{1}{2}m{R}^2{\omega }_i{}^2\right)\left[\left(\frac{1}{2}\right)\left(\frac{1}{9}\right)+\left(\frac{1}{9}\right)-\left(\frac{1}{2}\right)\right]}{\frac{1}{2}\left(\frac{1}{2}m{R}^2\right){\omega }_i{}^2}\\ =\frac{\left(-\frac{1}{3}\right)}{\left(\frac{1}{2}\right)}\end{array}$

Simplify the above expression for $\frac{\Delta E}{E_L}$ it gives.

  $\frac{\Delta E}{E_L}=-\frac{2}{3}$

Thus, the fractional change in kinetic energy from the moment the disk is set down until pure rolling occurs $\underset{\_}{-\frac{2}{3}}$.

(c)

To determine

The time interval after setting the disk down before pure rolling motion begins.

Answer to Problem 61CP

The time interval after setting the disk down before pure rolling motion begins is $\underset{\_}{\frac{R{\omega }_f}{\mu g}}$.

Explanation of Solution

Write the expression for time interval after setting the disk down before pure rolling motion begins as.

  $\Delta t=\frac{\Delta p}{f}$                                                                                                   (XII)

Here, $\Delta p$ change in momentum, $f$ is friction force between disc and plane surface.

Initially, the linear velocity of the disc is zero therefore the initial linear momentum of the disc becomes zero. As the initial momentum is zero, the change in linear momentum equal to the final linear momentum of the disc.

Write the expression for change in linear momentum as.

  $\Delta p=m{v}_f$                                                                                                  (XIII)

Write the expression for friction force between disc and plane surface as.

  $f=\mu mg$                                                                                                  (XIV)

Here, $\mu$ is coefficient of friction between disc and plane surface and $g$ gravitational acceleration.

Substitute $m{v}_f$ for $\Delta p$, $\mu mg$ for $f$ and in equation (XII).

  $\Delta t=\frac{m{v}_f}{\mu mg}$                                                                                                (XV)

Conclusion:

Substitute $R{\omega }_f$ for $v_f$ in equation (XV).

  $\Delta t=\frac{m\left(R{\omega }_f\right)}{\mu mg}$

Simplify the above obtained expression for $\Delta t$ as.

  $\Delta t=\frac{R{\omega }_f}{\mu g}$

Thus, the time interval after setting the disk down before pure rolling motion begins is $\underset{\_}{\frac{R{\omega }_f}{\mu g}}$.

(d)

To determine

The distance travel by the disc before pure rolling begins.

Answer to Problem 61CP

The distance travel by the disc before pure rolling begins is $\underset{\_}{\frac{R^2{\omega }_i{}^2}{18\mu g}}$.

Explanation of Solution

Write the expression for average velocity of the disc as.

  $v_{avg}=\frac{v_i-v_f}{2}$

Initial linear velocity of the disc is zero.

Substitute $0$ for $v_i$ in above expression as.

  $v_{avg}=\frac{v_f}{2}$

Substitute $R{\omega }_f$ for $v_f$ in above expression.

  $v_{avg}=\frac{R{\omega }_f}{2}$                                                                                               (XVI)

Here, $v_{avg}$ average speed of the disc.

Write the expression for the disc under constant acceleration as.

  $\Delta x=v_{avg}\Delta t$                                                                                              (XVII)

Here, $\Delta x$ is distance travelled by the disc before pure rolling begins.

Conclusion:

Substitute $\frac{R{\omega }_f}{2}$ for $v_{avg}$ and $\frac{R{\omega }_f}{\mu g}$ for $\Delta t$ in equation (XVII).

  $\Delta x=\frac{R{\omega }_f}{2}\left(\frac{R{\omega }_f}{\mu g}\right)$

Substitute $\frac{{\omega }_i}{3}$ for ${\omega }_f$ in above expression as.

  $\begin{array}{c}\Delta x=\frac{R\left(\frac{{\omega }_i}{3}\right)}{2}\left(\frac{R\left(\frac{{\omega }_i}{3}\right)}{\mu g}\right)\\ =\frac{R{\omega }_i}{6}\left(\frac{R{\omega }_i}{3\mu g}\right)\end{array}$

Simplify the above expression for $\Delta x$ as.

  $\Delta x=\frac{R^2{\omega }_i{}^2}{18\mu g}$

Thus, the distance travel by the disc before pure rolling begins is $\underset{\_}{\frac{R^2{\omega }_i{}^2}{18\mu g}}$.