Question

A high-speed drill rotating ccw at $2400\mathrm{rpm}$comes to a halt in $2.5\mathrm{s}$.

a. What is the magnitude of the drill’s angular acceleration?

b. How many revolutions does it make as it stops?

Short answer

a. The magnitude of the drill’s angular acceleration$-100\mathrm{rad}/{\mathrm{s}}^2$.

b. $50\mathrm{rev}$it make as it stops.

Step-by-step solution

Given information

${\mathrm{\omega }}_{\mathrm{i}}=2400\mathrm{rpm}$

${\mathrm{\omega }}_{\mathrm{f}}=0$

$\mathrm{t}=2.5\mathrm{s}$

Part a) Step 1: Calculation

To convert angular velocity from revolution per minutes to the revolution per second:

$$\begin{gathered} {\mathrm{\omega }}_{\mathrm{i}}=2400\frac{\mathrm{rev}}{\min }\times \frac{1\min }{60\mathrm{s}}\times \frac{2\mathrm{\pi }\mathrm{rad}}{1\mathrm{rev}} \\ {\mathrm{\omega }}_{\mathrm{i}}=251\mathrm{rad}/\mathrm{s} \end{gathered}$$

To calculate angular acceleration:

$$\begin{gathered} \mathrm{\alpha }=\frac{{\mathrm{\omega }}_{\mathrm{f}}-{\mathrm{\omega }}_{\mathrm{i}}}{\mathrm{t}} \\ \mathrm{\alpha }=\frac{0\mathrm{rad}/\mathrm{s}-251\mathrm{rad}/\mathrm{s}}{2.5\mathrm{s}} \\ \mathrm{\alpha }=-100\mathrm{rad}/{\mathrm{s}}^2 \end{gathered}$$

Part b) Step 1: Calculation

To calculate number of revolution

$$\begin{gathered} 2\mathrm{\theta }=({\mathrm{\omega }}_{\mathrm{f}}+{\mathrm{\omega }}_{\mathrm{i}})\mathrm{t} \\ \mathrm{\theta }=\frac{({\mathrm{\omega }}_{\mathrm{f}}+{\mathrm{\omega }}_{\mathrm{i}})\mathrm{t}}{2} \\ \mathrm{\theta }=\frac{(0\mathrm{rad}/\mathrm{s}+251\mathrm{rad}/\mathrm{s})(2.5\mathrm{s})}{2} \\ \mathrm{\theta }=\frac{(251\mathrm{rad}/\mathrm{s})(2.5\mathrm{s})}{2} \\ \mathrm{\theta }=313.75\mathrm{rad} \end{gathered}$$

To convert radian in revolutions

$$\begin{gathered} \mathrm{\theta }=2\mathrm{\pi N} \\ \mathrm{N}=\frac{\mathrm{\theta }}{2\mathrm{\pi }} \\ \mathrm{N}=\frac{313.75\mathrm{rad}}{\left(2\right)(3.14\mathrm{rad})} \\ \mathrm{N}=49.96\mathrm{rev} \\ \mathrm{N}=50\mathrm{rev} \end{gathered}$$