Chapter 8, Problem 131P

(a)

To determine

The angular acceleration of disc during $0.20\text{s}$ time interval.

Answer to Problem 131P

Angular acceleration is $55\text{rad}/{\text{s}}^{\text{2}}$.

Explanation of Solution

The final angular speed is $11\text{rad}/\text{s}$, rotational inertia of disc is $1.5\text{kg}\cdot {\text{m}}^{\text{2}}$, and the radius of disc is $11.5\text{cm}$.

Write the equation to find the angular acceleration.

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

Here, $\alpha$ is the angular acceleration, ${\omega }_f$ is the final angular velocity, ${\omega }_i$ is the initial angular velocity, and $\Delta t$ is the time taken for the change of velocity.

Conclusion:

Substitute $11\text{rad}/\text{s}$ for ${\omega }_f$, $0\text{rad}/\text{s}$ for ${\omega }_i$, and $0.20\text{s}$ for $\Delta t$ in equation (I).

$\begin{array}{c}\alpha =\frac{11\text{rad}/\text{s}-0\text{rad}/\text{s}}{0.20\text{s}}\\ =55\text{rad}/{\text{s}}^{\text{2}}\end{array}$

Therefore, the angular acceleration is $55\text{rad}/{\text{s}}^{\text{2}}$.

(b)

To determine

The net torque on disc during $0.20\text{s}$ time interval.

Answer to Problem 131P

The net torque is $83\text{N}\cdot \text{m}$.

Explanation of Solution

The final angular speed is $11\text{rad}/\text{s}$, rotational inertia of disc is $1.5\text{kg}\cdot {\text{m}}^{\text{2}}$, and the radius of disc is $11.5\text{cm}$.

Write the equation to find the net torque.

${\tau }_{net}=I\alpha$

Here, ${\tau }_{net}$ is the net torque and $I$ is the rotational inertia.

Conclusion:

Substitute $1.5\text{kg}\cdot {\text{m}}^{\text{2}}$ for $I$ and $55\text{rad}/{\text{s}}^{\text{2}}$ for $\alpha$ in the above equation.

$\begin{array}{c}{\tau }_{net}=\left(1.5\text{kg}\cdot {\text{m}}^{\text{2}}\right)\left(55\text{rad}/{\text{s}}^{\text{2}}\right)\\ =83\text{N}\cdot \text{m}\end{array}$

Therefore, the net torque is $83\text{N}\cdot \text{m}$.

(c)

To determine

The angle of rotation of disc before it comes to rest.

Answer to Problem 131P

The angle of rotation is $7.3\text{rad}$.

Explanation of Solution

The final angular speed is $11\text{rad}/\text{s}$, rotational inertia of disc is $1.5\text{kg}\cdot {\text{m}}^{\text{2}}$, radius of disc is $11.5\text{cm}$, and the angular acceleration due to the frictional torque is $9.8\text{rad}/{\text{s}}^{\text{2}}$.

Write equation 5.21 to find the angle during the spin-up.

${\omega }^2-{\omega }_i^2=2{\alpha }_1\Delta {\theta }_1$

Here,$\omega$ is the final angular velocity, ${\alpha }_1$ is the angular acceleration during the spin-up and $\Delta {\theta }_1$ is the angle during the spin-up.

The angular acceleration during the spin-up is the angular acceleration due to applied torque found in part (a).

Rewrite the above equation in terms of $\Delta {\theta }_1$ by substituting $0\text{rad}/\text{s}$ for ${\omega }_i$.

$\begin{array}{c}{\omega }^2-{\left(0\text{rad}/\text{s}\right)}^2=2{\alpha }_1\Delta {\theta }_1\\ \Delta {\theta }_1=\frac{{\omega }^2}{2{\alpha }_1}\end{array}$ (II)

Note that the final angular velocity of spin-up motion is the initial angular velocity of spin-down motion.

Write equation 5.21 to find the angle during the spin-down.

${\omega }_f^2-{\omega }^2=2{\alpha }_2\Delta {\theta }_2$

Here, ${\alpha }_2$ is the angular acceleration during the spin-down motion and $\Delta {\theta }_2$ is the angle during the spin-down.

Rewrite the above equation in terms of $\Delta {\theta }_2$ by substituting $0\text{rad}/\text{s}$ for ${\omega }_f$.

$\begin{array}{c}{\left(0\text{rad}/\text{s}\right)}^2-{\omega }^2=2{\alpha }_2\Delta {\theta }_2\\ \Delta {\theta }_2=\frac{-{\omega }^2}{2{\alpha }_2}\end{array}$ (III)

Add equation (I) with equation (II).

$\begin{array}{c}\Delta {\theta }_1+\Delta {\theta }_2=\frac{{\omega }^2}{2{\alpha }_1}+\frac{-{\omega }^2}{2{\alpha }_2}\\ =\frac{{\omega }^2}{2}\left(\frac{1}{{\alpha }_1}-\frac{1}{{\alpha }_2}\right)\end{array}$

Conclusion:

Substitute $55\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_1$, $-9.8\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_2$, and $11\text{rad}/\text{s}$ for $\omega$ in the above equation to find $\Delta {\theta }_1+\Delta {\theta }_2$.

$\begin{array}{c}\Delta {\theta }_1+\Delta {\theta }_2=\frac{{\left(11\text{rad}/\text{s}\right)}^2}{2}\left(\frac{1}{55\text{rad}/{\text{s}}^{\text{2}}}-\frac{1}{-9.8\text{rad}/{\text{s}}^{\text{2}}}\right)\\ =\left(5.5{\text{rad}}^2/{\text{s}}^2\right)\left(1.33{\text{s}}^2/\text{rad}\right)\\ =7.3\text{rad}\end{array}$

Therefore, the angle of rotation is $7.3\text{rad}$.

(d)

To determine

The speed of point lies at the midpoint of distance between the rim of disc and its rotation axis after $0.20\text{s}$ of removing the applied torque.

Answer to Problem 131P

The speed is $0.52\text{m}/\text{s}$.

Explanation of Solution

The final angular speed is $11\text{rad}/\text{s}$, rotational inertia of disc is $1.5\text{kg}\cdot {\text{m}}^{\text{2}}$, radius of disc is $11.5\text{cm}$, and the angular acceleration due to the frictional torque is $9.8\text{rad}/{\text{s}}^{\text{2}}$.

Write the relation between angular speed and the angular acceleration.

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

Write the relation between ${\omega }_f$ and the linear speed.

${\omega }_f=\frac{v}{r}$

Here, $v$ is the linear speed and $r$ is the distance between the midpoint and rotational axis.

Rewrite the previous equation by substituting the above relation for ${\omega }_f$.

$\begin{array}{c}\left(\frac{v}{r}\right)-\omega =\alpha \Delta t\\ v=r\left(\alpha \Delta t+\omega \right)\end{array}$

Conclusion:

Substitute $\frac{11.5\text{cm}}{2}$ for $r$, $-9.8\text{rad}/{\text{s}}^{\text{2}}$ for $\alpha$, $0.20\text{s}$ for $\Delta t$, and $11\text{rad}/\text{s}$ for $\omega$ in the above equation.

$\begin{array}{c}v=\left(\frac{11.5\text{cm}}{2}\left(\frac{{\text{10}}^{-\text{2}}\text{m}}{\text{1}\text{cm}}\right)\right)\left(\left(\left(-9.8\text{rad}/{\text{s}}^{\text{2}}\right)\left(0.20\text{s}\right)\right)+11\text{rad}/\text{s}\right)\\ =\left(0.0575\text{m}\right)\left(9.04\text{rad}/\text{s}\right)\\ =0.52\text{m}/\text{s}\end{array}$

Therefore, the speed is $0.52\text{m}/\text{s}$.