Chapter 10, Problem 35P

(a)

To determine

The moment of inertia of the wheel.

Answer to Problem 35P

The moment of inertia for the wheel is $21.6\text{ kg}\cdot {\text{m}}^2$.

Explanation of Solution

Write the expression for torque for wheel.

    $\tau =I\alpha$                                                                                                           (I)

Here, $\tau$ is the torque, $I$ is the moment of inertia and $\alpha$ is the angular acceleration.

Write the expression for angular acceleration.

    $\alpha =\frac{{\omega }_f-{\omega }_i}{\Delta t}$

Here, ${\omega }_f$ is final angular velocity, ${\omega }_i$ is initial angular velocity and $\Delta t$ is time interval.

Substitute $\frac{{\omega }_f-{\omega }_i}{\Delta t}$ for $\alpha$ in equation (I) and rearrange for moment of inertia.

    $I=\frac{\tau \Delta t}{{\omega }_f-{\omega }_i}$                                                                                                  (II)

Conclusion:

Substitute $36.0\text{ N}\cdot \text{m}$ for $\tau$, $10.0\text{ rad/s}$ for ${\omega }_f$, $0\text{ rad/s}$ for ${\omega }_i$ and $6.00\text{ s}$ for $\Delta t$ in equation (II).

    $\begin{array}{c}I=\frac{\left(36.0\text{ N}\cdot \text{m}\right)\left(6.00\text{ s}\right)}{\left(10.0\text{ rad/s}\right)-\left(0\text{ rad/s}\right)}\\ =\frac{216\text{ N-m s}}{10.0\text{ rad/s}}\\ =21.6\text{ kg}\cdot {\text{m}}^2\end{array}$

Thus, the moment of inertia for the wheel is $21.6\text{ kg}\cdot {\text{m}}^2$.

(b)

To determine

The magnitude of the torque due to friction.

Answer to Problem 35P

The magnitude of the frictional torque is $3.60\text{ N}\cdot \text{m}$.

Explanation of Solution

Frictional force can only act between the wheel and the rope on the wheel. Therefore torque due to friction is produced due to wheel.

Write the expression for frictional torque.

    ${\tau }_f=I\alpha$                                                                                                      (III)

Here, ${\tau }_f$ is the frictional torque and $\alpha$ is angular acceleration.

Substitute $\frac{{\omega }_f-{\omega }_i}{\Delta t}$ for $\alpha$ in equation (III).

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

Conclusion:

Substitute $21.6{\text{ kg-m}}^2$ for $I$, $60.0\text{ s}$ for $\Delta t$, $0$ for ${\omega }_i$ and $10.0\text{ rad/s}$ for ${\omega }_f$ in equation (III).

    $\begin{array}{c}{\tau }_f=21.6\text{ kg}\cdot {\text{m}}^2\left(\frac{\left(10.0\text{ rad/s}\right)-\left(0\text{rad/s}\right)}{60.0\text{ s}}\right)\\ =3.60\text{ N}\cdot \text{m }\end{array}$

Thus, the magnitude of the frictional torque is $3.60\text{ N}\cdot \text{m}$.

(c)

To determine

The totalnumber of revolutions of the wheelduring the entire interval of $66.0\text{s}$.

Answer to Problem 35P

Thetotalnumber of revolutions of the wheel is $52.5\text{ revolutions}$.

Explanation of Solution

The total number of revolutions can be calculated by using angular displacement.

Write the expression for angular displacement at first time interval.

    $\Delta {\theta }_1={\omega }_{avg1}\Delta t_1$                                                                                              (IV)

Here, $\Delta {\theta }_1$ is angular displacement in first time, ${\omega }_{avg1}$ is the average angular velocity in first interval and $\Delta t_1$ is the first time interval.

Write the expression for the average angular velocity.

    $\Delta {\omega }_{avg1}=\left(\frac{{\omega }_f+{\omega }_i}{2}\right)$   

Substitute $\frac{{\omega }_f+{\omega }_i}{2}$ for $\Delta {\omega }_{avg1}$ in equation (IV).

    $\Delta {\theta }_1=\left(\frac{{\omega }_f+{\omega }_i}{2}\right)\Delta t_1$                                                                                    (V)

Write the expression for angular displacement at second time interval.

    $\Delta {\theta }_2={\omega }_{avg2}^{}\Delta t_2$                                                                                            (VI)

Here, $\Delta {\theta }_2$ is angular displacement in second time, ${\omega }_{avg2}$ is the average angular velocity in second interval and $\Delta t_2$ is the second time interval.

Write the expression for the average angular velocity.

    $\Delta {\omega }_{avg2}=\left(\frac{{\omega }_{f2}+{\omega }_{i2}}{2}\right)$   

Here, ${\omega }_{f2}$ is the final angular velocity after second interval and ${\omega }_{i2}$ is the initial angular velocity in second interval.

Substitute $\frac{{\omega }_{f2}+{\omega }_{i2}}{2}$ for $\Delta {\omega }_{avg2}$ in equation (VI).

    $\Delta {\theta }_2=\left(\frac{{\omega }_{f2}+{\omega }_{i2}}{2}\right)\Delta t_2$                                                                              (VII)

Write the expression for total angular displacement.

    $\Delta \theta =\Delta {\theta }_1+\Delta {\theta }_2$

Substitute $\left(\frac{{\omega }_f+{\omega }_i}{2}\right)\Delta t_1$ for $\Delta {\theta }_1$ and $\left(\frac{{\omega }_{f2}+{\omega }_{i2}}{2}\right)\Delta t_2$ for $\Delta {\theta }_2$ in above equation.

  $\Delta \theta =\left(\frac{{\omega }_f+{\omega }_i}{2}\right)\Delta t_1+\left(\frac{{\omega }_{f2}+{\omega }_{i2}}{2}\right)\Delta t_2$                                                      (VIII)

Here, $\Delta \theta$ is total angular displacement.

Write the expression for the total number of revolutions of the wheel.

    $N=\frac{\Delta \theta }{2\pi }$                                                                                                      (IX)

Here, $N$ is the revolution per second.

Substitute $\left(\frac{{\omega }_f+{\omega }_i}{2}\right)\Delta t_1+\left(\frac{{\omega }_{f2}+{\omega }_{i2}}{2}\right)\Delta t_2$ for $\Delta \theta$.

    $N=\frac{1}{2\pi }\left[\left(\frac{{\omega }_f+{\omega }_i}{2}\right)\Delta t_1+\left(\frac{{\omega }_{f2}+{\omega }_{i2}}{2}\right)\Delta t_2\right]$                                                   (X)

Conclusion:

Substitute $0$ for ${\omega }_i$, $10.0\text{ rad/s}$ for ${\omega }_f$, $6.0\text{ s}$ for $\Delta t_1$, $0$ for ${\omega }_{i2}$, $10.0\text{ rad/s}$ for ${\omega }_{f2}$ and $60.0\text{ s}$ for $\Delta t_2$ in equation (X).

    $\begin{array}{c}N=\frac{1}{2\pi }\left[\left(\frac{10\text{rad/s}+0}{2}\right)\left(6\text{s}\right)+\left(\frac{0+10\text{rad/s}}{2}\right)\left(60\text{s}\right)\right]\\ =\frac{1}{2\pi }\left[\left(30\text{rad}\right)+\left(300\text{rad}\right)\right]\\ =52.5\end{array}$

Thus, the totalnumber of revolutions of the wheel is $52.5\text{ revolutions}$.