Question

A $6.0\mathrm{cm}$-diameter gear rotates with angular velocity role="math" localid="1647955750668" $\omega =(2.0+\frac{1}{2}{\mathrm{t}}^2)\mathrm{rad}/\mathrm{s}$, where t is in seconds. At role="math" localid="1647955660518" $\mathrm{t}=4.0\mathrm{s}$, what are:

a. The gear’s angular acceleration?

b. The tangential acceleration of a tooth on the gear?

Short answer

a. The gear’s angular acceleration is$\mathrm{\alpha }=4\mathrm{rad}/{\mathrm{s}}^2$.

b. The tangential acceleration of a tooth on the gear islocalid="1647956044997" ${\mathrm{a}}_{\mathrm{t}}=0.12\mathrm{m}/{\mathrm{s}}^2$.

Step-by-step solution

Given information

We are given with:

$$\begin{gathered} \mathrm{diameter}=\mathrm{d}=6\mathrm{cm} \\ \mathrm{radius}=\mathrm{r}=3\mathrm{cm}=0.03\mathrm{m} \\ \mathrm{angular}\mathrm{velocity}=\mathrm{\omega }=(2.0+\frac{1}{2}{\mathrm{t}}^2)\mathrm{rad}/\mathrm{s} \end{gathered}$$

Part a) Step 1: Calculation

The gear’s angular acceleration is:

$$\begin{gathered} \mathrm{\alpha }=\frac{\mathrm{d\omega }}{\mathrm{dt}} \\ \mathrm{\alpha }=\frac{\mathrm{d}}{\mathrm{dt}}\left[(2+\frac{1}{2}{\mathrm{t}}^2)\mathrm{rad}/\mathrm{s}\right] \\ \mathrm{\alpha }=(0+\frac{2\mathrm{t}}{2})\mathrm{rad}/{\mathrm{s}}^2 \\ \mathrm{\alpha }=\mathrm{t}\mathrm{rad}/{\mathrm{s}}^2 \\ \mathrm{\alpha }=4\mathrm{rad}/{\mathrm{s}}^2 \end{gathered}$$

Part b) Step 1:Calculation

The tangential acceleration of a tooth on the gear is:

$$\begin{gathered} {\mathrm{a}}_{\mathrm{t}}=\mathrm{r\alpha } \\ {\mathrm{a}}_{\mathrm{t}}=(0.03\mathrm{m})(4\mathrm{rad}/{\mathrm{s}}^2) \\ {\mathrm{a}}_{\mathrm{t}}=0.12\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$