Chapter 8, Problem 115P

(a)

To determine

The angular velocity when the hoop arrives at the bottom.

Answer to Problem 115P

The angular velocity when the hoop arrives at the bottom is $6.28\text{rad/s}$.

Explanation of Solution

Write an expression to calculate the angular velocity when the hoop arrives at the bottom.

$\begin{array}{c}{\omega }_f=2{\omega }_{av}\\ =2\left(\frac{v_{av}}{r}\right)\\ =2\left(\frac{\left(\Delta x/\Delta t\right)}{\left(C/2\pi \right)}\right)\\ =\frac{4\pi \Delta x}{C\Delta t}\end{array}$ (I)

Here, ${\omega }_f$ is the angular velocity when the hoop arrives at the bottom, ${\omega }_{av}$ is the average velocity, $v_{av}$ is the average velocity, $r$ is the radius, $\Delta x$ is the displacement, $\Delta t$ is the time and $C$ is the circumference.

Conclusion:

Substitute $10.0\text{m}$ for $\Delta x$, $2.00\text{m}$ for $C$ and $10.0\text{s}$ for $\Delta t$ in equation (I) to find ${\omega }_f$.

$\begin{array}{c}{\omega }_f=\frac{4\pi \left(10.0\text{m}\right)}{\left(2.00\text{m}\right)\left(10.0\text{s}\right)}\\ =\frac{4\pi \left(10.0\text{m}\right)}{\left(20.0\text{m}\cdot \text{s}\right)}\\ =6.28\text{rad/s}\end{array}$

Thus, the angular velocity when the hoop arrives at the bottom is $6.28\text{rad/s}$.

(b)

To determine

The angular momentum of the hoop at the bottom.

Answer to Problem 115P

The angular momentum of the hoop at the bottom is $0.955\text{kg}\cdot {\text{m}}^2\text{/s}$.

Explanation of Solution

Refer figure 1.

Write an expression for the angular momentum of the hoop at the bottom.

$\begin{array}{c}L=I\omega \\ =\left(m{r}^2\right)\omega \\ =m{\left(\frac{C}{2\pi }\right)}^2\omega \end{array}$ (II)

Here, $L$ is the angular momentum of the hoop at the bottom, $I$ is the moment of inertia, $m$ is the mass and $\omega$ is the angular velocity.

Conclusion:

Substitute $1.50\text{kg}$ for $m$, $2.00\text{m}$ for $r$ and $2\pi \text{rad/s}$ for $\omega$ in equation (II) to find ${\tau }_{\text{motor}}$.

$\begin{array}{c}L=\left(1.50\text{kg}\right){\left(\frac{2.00\text{m}}{2\pi }\right)}^2\left(2\pi \text{rad/s}\right)\\ =\left(1.50\text{kg}\right)\left(\frac{4.00{\text{m}}^2}{4{\pi }^2}\right)\left(2\pi \text{rad/s}\right)\\ =0.955\text{kg}\cdot {\text{m}}^2\text{/s}\end{array}$

Thus, the angular momentum of the hoop at the bottom is $0.955\text{kg}\cdot {\text{m}}^2\text{/s}$.

(c)

To determine

The forces applied the net torque to change the hoop’s angular momentum.

Answer to Problem 115P

The forces applied the net torque to change the hoop’s angular momentum is the friction.

Explanation of Solution

The gravitational force is acting at the geometric centre of the hoop. To have a torque to create a change in momentum, the force should act at a distance from the axis of rotation through the geometric centre. Thus, the gravitational force cannot produce a change in momentum.

The force of static friction acts at the rim of the hoop. The force of friction acts perpendicularly to the line between the axis of the hoop and the point of contact between the rim of the hoop and the inclined plane. This can produce torque and hence change in angular momentum. Thus, force of friction causes the net torque about the hoop’s axis.

(d)

To determine

Magnitude of the force that creates the torque.

Answer to Problem 115P

Magnitude of the force that creates the torque is $0.300\text{N}$.

Explanation of Solution

Write an expression for the magnitude of the force that creates the torque.

$\begin{array}{c}f=\frac{\tau }{r}\\ =\frac{\left(\Delta L/\Delta t\right)}{\left(C/2\pi \right)}\\ =\frac{2\pi \Delta L}{C\Delta t}\end{array}$ (III)

Here, $f$ is the force, $\tau$ is the torque and $\Delta L$ is the change in angular momentum.

Conclusion:

Substitute $0.955\text{kg}\cdot {\text{m}}^2\text{/s}$ for $\Delta L$, $2.00\text{m}$ for $C$ and $10.0\text{s}$ for $\Delta t$ in equation (III) to find $f$.

$\begin{array}{c}f=\frac{2\pi \left(0.955\text{kg}\cdot {\text{m}}^2\text{/s}\right)}{\left(2.00\text{m}\right)\left(10.0\text{s}\right)}\\ =\frac{2\pi \left(0.955\text{kg}\cdot {\text{m}}^2\text{/s}\right)}{20.0\text{m}\cdot \text{s}}\\ =0.300\text{N}\end{array}$

Thus, the magnitude of the force that creates the torque is $0.300\text{N}$.