Chapter 9, Problem 101A

To determine

To Find: $\alpha ,\omega$ at time 2.00 s and $I$ .

Answer to Problem 101A

The values are $\alpha =4.17{\text{ rad/s}}^2$ , $\omega =8.34\text{ rad/s}$ and $I=5.755\text{ kg}\cdot {\text{m}}^2$

Explanation of Solution

Given info:

Diameter of drum, $R=0.600\text{ m}$

A machine pulling the rope for 2.00 s with a constant force is 40.0 N in this time 5.00 m of rope unwinds.

Formula used:

The angular acceleration can be obtained by using equation

  $\begin{array}{l}\Delta \theta =\frac{1}{2}\alpha t^2\\ \Rightarrow \alpha =\frac{2\Delta \theta }{t^2}\end{array}$

Here $\alpha$ is angular acceleration and t is time.

And $\Delta S=\Delta \theta R$

Here $\Delta \theta$ is angular displacement.

Calculation:

  $\begin{array}{c}\alpha =\frac{2\Delta \theta }{t^2}\\ \alpha =\frac{2\Delta S}{R{t}^2}\\ \alpha =\frac{2\times 5.00}{0.600\times {2.00}^2}\\ \alpha =4.17{\text{ rad/s}}^2\end{array}$

The angular speed

  $\begin{array}{c}\omega =\alpha t\\ \omega =4.17{\text{ rad/s}}^2\times 2\text{ s}\\ \omega =8.34\text{ rad/s}\end{array}$

The moment of inertia

  $\begin{array}{c}I=\frac{\tau }{\alpha }=\frac{FR}{\alpha }\\ I=\frac{\left(40\text{ N}\right)\left(0.600\text{ m}\right)}{4.17}\\ I=5.755\text{ kg}\cdot {\text{m}}^2\end{array}$

Conclusion:

Hence, the values are $\alpha =4.17{\text{ rad/s}}^2$ , $\omega =8.34\text{ rad/s}$ and $I=5.755\text{ kg}\cdot {\text{m}}^2$