Chapter 8, Problem 110P

(a)

To determine

The radius of the flywheel.

Answer to Problem 110P

The radius of the flywheel is $3.54\text{ m}$.

Explanation of Solution

Write the expression for the rotational inertia of a uniform disk.

$I=\frac{1}{2}M{R}^2$

Here, $I$ is the rotational inertia of the flywheel, $M$ is the mass of the flywheel and $R$ is its radius

Rewrite the above equation for $R$.

$R=\sqrt{\frac{2I}{M}}$ (I)

Conclusion:

Given that the rotational inertia of the flywheel is $4.55\times {10}^6\text{ kg}\cdot {\text{m}}^2$ and its mass is $7.27\times {10}^5\text{ kg}$.

Substitute $4.55\times {10}^6\text{ kg}\cdot {\text{m}}^2$ for $I$ and $7.27\times {10}^5\text{ kg}$ for $M$ in equation (I) to find the value of $R$.

$\begin{array}{c}R=\sqrt{\frac{2\left(4.55\times {10}^6\text{ kg}\cdot {\text{m}}^2\right)}{7.27\times {10}^5\text{ kg}}}\\ =3.54\text{ m}\end{array}$

Therefore, the radius of the flywheel is $3.54\text{ m}$.

(b)

To determine

The radius of the flywheel if it is a hollow cylinder with the mass concentrated at the rim.

Answer to Problem 110P

The radius of the flywheel if it is a hollow cylinder with the mass concentrated at the rim is $2.50\text{ m}$.

Explanation of Solution

Write the expression for the rotational inertia of a hollow cylinder.

$I=M{R}^2$

Rewrite the above equation for $R$.

$R=\sqrt{\frac{I}{M}}$ (II)

Conclusion:

Substitute $4.55\times {10}^6\text{ kg}\cdot {\text{m}}^2$ for $I$ and $7.27\times {10}^5\text{ kg}$ for $M$ in equation (II) to find the value of $R$.

$\begin{array}{c}R=\sqrt{\frac{\left(4.55\times {10}^6\text{ kg}\cdot {\text{m}}^2\right)}{7.27\times {10}^5\text{ kg}}}\\ =2.50\text{ m}\end{array}$

Therefore, the radius of the flywheel if it is a hollow cylinder with the mass concentrated at the rim is $2.50\text{ m}$.

(c)

To determine

The average power supplied by the flywheel during $5.00\text{ s}$.

Answer to Problem 110P

The average power supplied by the flywheel during $5.00\text{ s}$ is $4.27\times {10}^8\text{ W}$.

Explanation of Solution

Write the expression for the rate at which the energy of the flywheel is made to change.

$P_{\text{av}}=\frac{W}{\Delta t}$ (III)

Here, $P_{\text{av}}$ is the rate at which the energy of the flywheel is made to change, $W$ is the work done and $\Delta t$ is the time interval

Write the work energy theorem.

$W=\Delta K$

Here, $\Delta K$ is the change in kinetic energy of the flywheel

Put the above equation in equation (III).

$P_{\text{av}}=\frac{\Delta K}{\Delta t}$ (IV)

Write the expression for $\Delta K$.

$\begin{array}{c}\Delta K=\frac{1}{2}I{\omega }_f^2-\frac{1}{2}I{\omega }_i^2\\ =\frac{1}{2}I\left({\omega }_f^2-{\omega }_i^2\right)\end{array}$

Here, ${\omega }_f$ is the final angular speed of the flywheel and ${\omega }_i$ is the initial angular speed of the flywheel

Put the above equation in equation (IV).

$\begin{array}{c}{P}_{\text{av}}=\frac{\frac{1}{2}I\left({\omega }_f^2-{\omega }_i^2\right)}{\Delta t}\\ =\frac{I}{2\Delta t}\left({\omega }_f^2-{\omega }_i^2\right)\end{array}$

The average power supplied by the flywheel is $-P_{\text{av}}$.

$-P_{\text{av}}=\frac{I}{2\Delta t}\left({\omega }_i^2-{\omega }_f^2\right)$ (V)

Conclusion:

Given that the initial angular speed of the flywheel is $386\text{ rev/min}$ and the final angular speed is $252\text{ rev/min}$.

Substitute $4.55\times {10}^6\text{ kg}\cdot {\text{m}}^2$ for $I$. $5.00\text{ s}$ for $\Delta t$ , $386\text{ rev/min}$ for ${\omega }_i$ and $252\text{ rev/min}$ for ${\omega }_f$ in equation (V) to find the value of power supplied.

$\begin{array}{c}-P_{\text{av}}=\frac{4.55\times {10}^6\text{ kg}\cdot {\text{m}}^2}{2\left(5.00\text{ s}\right)}\left({\left(386\text{ rev}\left(\frac{2\pi \text{ rad}}{1\text{ rev}}\right)\text{/min}\left(\frac{60\text{ s}}{1\text{ min}}\right)\right)}^2-{\left(252\text{ rev}\left(\frac{2\pi \text{ rad}}{1\text{ rev}}\right)\text{/min}\left(\frac{60\text{ s}}{1\text{ min}}\right)\right)}^2\right)\\ =\frac{4.55\times {10}^6\text{ kg}\cdot {\text{m}}^2}{2\left(5.00\text{ s}\right)}\left[{\left(40.42\text{ rad/s}\right)}^2-{\left(26.39\text{ rad/s}\right)}^2\right]\\ =4.27\times {10}^8\text{ W}\end{array}$

Therefore, the average power supplied by the flywheel during $5.00\text{ s}$ is $4.27\times {10}^8\text{ W}$.