Question

A painted tooth on a spinning gear has angular acceleration $\mathrm{\alpha }=(20-\mathrm{t})\mathrm{rad}/{\mathrm{s}}^2$ , where t is in s. Its initial angular velocity, at $\mathrm{t}=0\mathrm{s}$, is $300\mathrm{rpm}$. What is the tooth’s angular velocity in rpm at $\mathrm{t}=20\mathrm{s}$?

Short answer

The tooth’s angular velocity is $2210\mathrm{rpm}$.

Step-by-step solution

Given information

$$\begin{gathered} \mathrm{angular}\mathrm{acceleration}=\mathrm{\alpha }=(20-\mathrm{t})\mathrm{rad}/{\mathrm{s}}^2 \\ \mathrm{angular}\mathrm{velocity}={\mathrm{\omega }}_1=300\mathrm{rpm} \end{gathered}$$

Calculation

The angular acceleration is:

$$\begin{gathered} \mathrm{\alpha }=\frac{\mathrm{d\omega }}{\mathrm{dt}} \\ \mathrm{d\omega }=\mathrm{\alpha }.\mathrm{dt} \\ \mathrm{d\omega }=\left[(20-\mathrm{t})\mathrm{rad}/{\mathrm{s}}^2\right]\mathrm{dt} \\ {\int }_{{\mathrm{\omega }}_1}^{{\mathrm{\omega }}_2}\mathrm{d\omega }={\int }_0^{20}\left[(20-\mathrm{t})\mathrm{rad}/{\mathrm{s}}^2\right]\mathrm{dt} \\ {\mathrm{\omega }}_2-{\mathrm{\omega }}_1=\left[\left(20\right)\left(20\right)-\left(\frac{1}{2}\right){\left(20\right)}^2\right]\mathrm{rad}/\mathrm{s} \\ {\mathrm{\omega }}_2-{\mathrm{\omega }}_1=200\mathrm{rad}/\mathrm{s} \end{gathered}$$

Now converting angular velocity from radian per second to revolution per minute:

$$\begin{gathered} {\mathrm{\omega }}_2-{\mathrm{\omega }}_1=200\frac{\mathrm{rad}}{\mathrm{s}}\times \frac{60\mathrm{s}}{1\min }\times \frac{1\mathrm{rev}}{2\mathrm{\pi }\mathrm{rad}} \\ {\mathrm{\omega }}_2-{\mathrm{\omega }}_1=1910\mathrm{rpm} \end{gathered}$$

The tooth’s angular velocity

$$\begin{gathered} {\mathrm{\omega }}_2=1910\mathrm{rpm}+{\mathrm{\omega }}_1 \\ {\mathrm{\omega }}_2=1910\mathrm{rpm}+300\mathrm{rpm} \\ {\mathrm{\omega }}_2=2210\mathrm{rpm} \end{gathered}$$