Chapter 8, Problem 52P

a.

To determine

To Find: The Angular momentum of wheel.

Answer to Problem 52P

The angular momentum of wheel is $7.128\text{kg}\cdot {\text{m}}^{\text{2}}\text{/s}$ .

Explanation of Solution

Given Information:

Mass of Cylindrical wheel, $M=2.8\text{kg}$

Radius of wheel, $R=18\text{cm}\text{=}\text{0}\text{.18}\text{m}$

Angular velocity, $\omega =1500\text{rpm}\text{= 1500}\times \frac{2\pi }{60}\text{rad/}s$

Concept used:

  $L=I\omega$

And $I=\frac{1}{2}{\text{mr}}^{\text{2}}$, so

  $L=\frac{1}{2}{\text{mr}}^{\text{2}}\text{ω}$

Here,

  $L$ is the angular momentum of wheel,

  $I$ is the moment of inertia,

  $m$ is the mass of the wheel,

  $r$ is the radius of wheel,

  $\omega$ is the angular velocity.

Calculation:

  $L=\frac{1}{2}{\text{mr}}^{\text{2}}\text{ω}$

By putting the values

  $\begin{array}{l}L=\frac{1}{2}(2.8){(}^{0.18}\times 1500\times \frac{2\pi }{60}\\ =7.128\text{kg}\cdot {\text{m}}^{\text{2}}\text{/s}\end{array}$

Conclusion:

The angular momentum of the wheel is

  $L=7.128\text{kg}\cdot {\text{m}}^{\text{2}}\text{/s}$ .

b.

To determine

To find: The change in Torque

Answer to Problem 52P

Change in Torque is $-1.188\text{N}\text{.m}$ (negative sign indicates that momentum act opposite of motion)

Explanation of Solution

Given Information:

Change in time $\Delta t=6.0\text{s}$

Initial Angular Momentum = $L_{I=}7.128\text{kg}\cdot {\text{m}}^{\text{2}}\text{/s}$

Concept used:

Change in Torque is $\Delta \tau =\frac{L_{f-}{L}_I}{\Delta t}$

Here,

  $\Delta \tau$ is the change in torque,

  $L_F$ is the final angular momentum,

  $L_I$ is the initial angular momentum,

  $\Delta t$ is the change in time.

Calculation:

As things is at rest so final angular momentum is zero so,

  $\Delta \tau =\frac{0-7.128}{6}$

  $\Delta \tau =-1.188\text{N}\cdot \text{m}$

Conclusion: Change in torque is $-1.188\text{N}\cdot \text{m}$ .