Chapter 8, Problem 93A

a.

To determine

The initial angular velocity of the basketball.

Answer to Problem 93A

  $20.75{\text{rad/s}}^{\text{2}}$

Explanation of Solution

Given:

The diameter of the basketball, $D=24.1\text{}\text{cm = 0}\text{.241}\text{m}$

Mass of the ball, $m=0.60\text{kg}$

Moment of Inertia of the basketball, $I=5.8\times {10}^{-3}\text{kg}\cdot {\text{m}}^{\text{2}}$

Initial velocity of the basketball, $v_i=2.5\text{m/s}$

Formula used:

For a rotating and rigid object, its linear velocity( v ) can be related to its angular velocity $\left(\omega \right)$ as,

  $v=r\omega$

Where, r is the radius of the object.

Calculation:

Radius of the basketball will be,

  $r=\frac{D}{2}=\frac{0.241}{2}=0.1205\text{m}$

Then, the initial angular velocity of the basketball will be,

  $\begin{array}{c}\omega =\frac{v}{r}\\ =\frac{2.5}{0.1205}\\ =20.75{\text{rad/s}}^{\text{2}}\end{array}$

Conclusion:

Thus, the initial angular velocity of the basketball is $20.75{\text{rad/s}}^{\text{2}}$ .

b.

To determine

The number of revolutions undertaken by the basketball if it rolls for $12\text{m}$ .

Answer to Problem 93A

  $16$

Explanation of Solution

Given:

The diameter of the basketball, $D=24.1\text{}\text{cm = 0}\text{.241}\text{m}$

Mass of the ball, $m=0.60\text{kg}$

Moment of Inertia of the basketball, $I=5.8\times {10}^{-3}\text{kg}\cdot {\text{m}}^{\text{2}}$

Initial velocity of the basketball, $v_i=2.5\text{m/s}$

Formula used:

For a rotating and rigid object, its distance( d ) covered can be related to its angular displacement $\left(\theta \right)$ as,

  $d=r\theta$

Where r is the radius of the object.

Calculation:

Radius of the basketball will be,

  $r=\frac{D}{2}=\frac{0.241}{2}=0.1205\text{m}$

Then, the number of revolution undertaken by the basketball will be,

  $\begin{array}{c}\theta =\frac{d}{r}\\ =\frac{12}{0.1205}\\ =99.58\text{rad}\times \frac{1\text{rev}}{2\pi \text{rad}}\\ \approx 16\end{array}$

Conclusion:

Thus, the number of revolutions undertaken by the basketball is $16$ .

c.

To determine

The total angular displacement of the basketball if it rolls for $12\text{m}$ .

Answer to Problem 93A

  $100\text{rad}$

Explanation of Solution

Given:

The diameter of the basketball, $D=24.1\text{}\text{cm = 0}\text{.241}\text{m}$

Mass of the ball, $m=0.60\text{kg}$

Moment of Inertia of the basketball, $I=5.8\times {10}^{-3}\text{kg}\cdot {\text{m}}^{\text{2}}$

Initial velocity of the basketball, $v_i=2.5\text{m/s}$

Formula used:

For a rotating and rigid object, its distance( d ) covered can be related to its angular displacement $\left(\theta \right)$ as,

  $d=r\theta$

Where r is the radius of the object.

Calculation:

Radius of the basketball will be,

  $r=\frac{D}{2}=\frac{0.241}{2}=0.1205\text{m}$

Then, the number of revolution undertaken by the basketball will be,

  $\begin{array}{c}\theta =\frac{d}{r}\\ =\frac{12}{0.1205}\\ \approx 100\text{rad}\end{array}$

Conclusion:

Thus, the total angular displacement of the basketball is $100\text{rad}$ .