Chapter 8, Problem 94A

(a)

To determine

The angular acceleration of the basket ball.

Answer to Problem 94A

The angular acceleration of the basketball is $2.16\text{rad}/{\text{s}}^2$

Explanation of Solution

Given Info:

The initial velocity of the basketball is, $u=2.5\text{m}/\text{s}$ .

The diameter of the basketball is: $d=24.1\text{cm}$ .

The distance covered by the basketball is, $x=12\text{m}$ .

Formula Used:

The expression to calculate the angular acceleration of the basketball is,

  $\alpha =\frac{u^2}{dx}$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}\alpha =\frac{{\left(2.5\text{m}/\text{s}\right)}^2}{\left(24.1\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\left(12\text{m}\right)}\\ =2.16\text{rad}/{\text{s}}^2\end{array}$

Thus, the angular acceleration of the basketball is $2.16\text{rad}/{\text{s}}^2$ .

To determine

The torque acts on the basketball to slow it.

Answer to Problem 94A

The torque acts on the basketball to slow the basketball is $12.53\times {10}^{-3}\text{N}\cdot \text{m}$ .

Explanation of Solution

Given Info:

The moment of inertia of the basketball is, $I=5.8\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ .

Formula Used:

The expression to calculate the torque acts on the basketball to slow the basketball is,

  $\tau =I\alpha$

Calculation:

Substitute all the values in the above expression

  $\begin{array}{c}\tau =\left(5.8\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\right)\left(2.16\text{rad}/{\text{s}}^2\right)\\ =12.53\times {10}^{-3}\text{N}\cdot \text{m}\end{array}$

Thus, the torque acts on the basketball to slow the basketball is $12.53\times {10}^{-3}\text{N}\cdot \text{m}$ .