Chapter 8, Problem 34P

(a)

To determine

To find: The moment of inertia.

Answer to Problem 34P

The value is $I=2.0953\times {10}^{-3}{\text{kg·m}}^{\text{2}}$ .

Explanation of Solution

Given information:

Mass of cylinder $M=0.58\text{kg}$

Radius of the cylinder $r=8.5\text{cm}=\text{0}\text{.085}\text{m}$

Formula used:

  $I=\frac{1}{2}Mr$

Here,

  $I$ is the moment of inertia,

  $M$ is the mass of the cylinder,

  $r$ is the radius of cylinder.

Calculation:

The moment of Inertial about its center is

  $I=\frac{1}{2}Mr$

  $\begin{array}{l}I=\frac{1}{2}\times 0.58\times {0.085}^2\text{kg}\cdot {\text{m}}^{\text{2}}\\ I=2.0953\times {10}^{-3}\text{kg}\cdot {\text{m}}^{\text{2}}\end{array}$

Conclusion:

The value is $I=2.0953\times {10}^{-3}\text{kg}\cdot {\text{m}}^{\text{2}}$ .

(b)

To determine

To find: The value of the applied torque.

Answer to Problem 34P

The value is $7.18\times {10}^{-2}\text{N}\cdot \text{m}$ .

Explanation of Solution

Given information:

A grinding wheel is a uniform cylinder with a radius or $3.50\text{cm}$ and a mass or $0.580\text{}\text{kg}$ .

Formula used:

  ${\omega }_0=2\pi f$

Here,

  ${\omega }_0$ is the angular velocity,

  $f$ is the rotational frequency.

Calculation:

As the cylinder slows down, let us calculate the negative angular acceleration.

Initial rotational frequency $f=1500\text{rpm}$

  $=25\text{rev/s}$

Hence initial angular velocity ${\omega }_0=2\pi f$

  $=50\pi \text{rad/s}$

Time taken for wheel to come to rest, $t=55\mathrm{s}$

Hence if ${\alpha }^1$ be the negative angular acceleration then

  $\begin{array}{l}\omega ={\omega }_0-{\alpha }^{1}t\\ \text{or}0={\omega }_0-{\alpha }^{1}t\\ \text{or}{\alpha }^1=\frac{{\omega }_0}{t}\\ {\alpha }^1=\frac{50\pi }{55}{\text{rad/s}}^{\text{2}}\\ {\alpha }^1=2.86{\text{rad/s}}^2\end{array}$

If the cylinder has to accelerate from rest to $1500\text{rpm}$ in time $t^1=5\mathrm{s}$ then applied angular acceleration a is given as

  $\begin{array}{l}\omega ={\omega }_0+\alpha t\\ \text{or}\omega =0+\left(\alpha -{\alpha }^1\right)t^1\\ \text{or}\alpha =\frac{\omega }{t}+{\alpha }^1\\ \alpha =\frac{50\pi }{5}+2.86{\text{rad/s}}^{\text{2}}\\ \alpha =34.28{\text{rad/s}}^{\text{2}}\end{array}$

Hence torque that needs to be applied is $\tau =I\alpha$

  $\begin{array}{l}=2.0953\times {10}^{-3}\times 34.28\text{N}\cdot \text{m}\\ =7.18\times {10}^{-2}\text{N}\cdot \text{m}\end{array}$

Conclusion:

The value is $7.18\times {10}^{-2}\text{N}\cdot \text{m}$