Chapter 8, Problem 92A

(a)

To determine

The distance of the center of mass of the cylinder moved.

Answer to Problem 92A

The distance of the center of mass of the cylinder moved is $2.50\text{ m}$ .

Explanation of Solution

Given:

The diameter of the cylinder is $D=50\text{ cm}$ .

The distance traveled by the cylinder is $d=2.50\text{ m}$ .

Formula used:

The expression for the distance traveled by a body at a distance from the center and rotating at an angle $\left(\theta \right)$ is,

  $d=r\theta$

Here, d is the distance traveled by the cylinder and r is the radius of the cylinder.

Calculation:

Consider the initial velocity of the uniform cylinder to be zero over the point of contact with the surface.

The center of mass of the uniform cylinder is always over the point of contact with the surface.

If the rope is pulled a distance of $2.50\text{ m}$ at a constant speed, then the center of mass is moved at a distance of $2.50\text{m}$

Conclusion:

Therefore, the distance of the center of mass of the cylinder moved is $2.50\text{ m}$ .

(b)

To determine

The speed of the center of mass of cylinder.

Answer to Problem 92A

The speed at which the center of mass of the cylinder moving is $v=2.0\text{m}/\text{s}$ .

Explanation of Solution

Given:

The diameter of the cylinder is $D=50\text{ cm}$ .

The distance traveled by the cylinder is $d=2.50\text{ m}$ .

The time taken by the cylinder to reach the distance of $2.50\text{ m}$ is $t=1.25\text{ s}$ .

Formula used:

The expression for the linear velocity is,

  $v=\frac{d}{t}$

Here, $d$ is the distance traveled by the cylinder and $t$ is the time.

Calculation:

The speed at which the center of mass of the cylinder moving is,

  $\begin{array}{l}v=\frac{d}{t}\\ v=\frac{2.50\text{ m}}{1.25\text{ s}}\\ v=2.0\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the speed at which the center of mass of the cylinder moving is $v=2.0\text{m}/\text{s}$ .

(c)

To determine

The angular velocity of the cylinder.

Answer to Problem 92A

The angular velocity of the cylinder is $\omega =0.08\text{rad}/\text{s}$ .

Explanation of Solution

Given:

The diameter of the cylinder is $D=50\text{ cm}$ .

The distance traveled by the cylinder is $d=2.50\text{ m}$ .

Formula used:

The expression for the angular velocity $\left(\omega \right)$ is,

  $\omega =\frac{v}{r}$

Here, $v$ is the linear velocity of the cylinder and r is the radius of the cylinder.

Calculation:

Refer Part (b).

The linear velocity of the cylinder is $v=2.0\text{m}/\text{s}$ .

The radius $\left(r\right)$ of cylinder is,

  $\begin{array}{l}r=\frac{D}{2}\\ r=\frac{50\text{ cm}\times \frac{1\text{ m}}{100\text{ cm}}}{2}\\ r=0.25\text{m}\end{array}$

The angular velocity of the cylinder is,

  $\begin{array}{l}\omega =\frac{v}{r}\\ \omega =\frac{2.0\text{m}/\text{s}}{0.25\text{ m}}\\ \omega =8\text{rad}/\text{s}\end{array}$

Conclusion:

Therefore, the angular velocity of the cylinder is $\omega =8.0\text{rad}/\text{s}$ .