Question

A computer hard disk $8.0\mathrm{cm}$in diameter is initially at rest. A small dot is painted on the edge of the disk. The disk accelerates at $600\mathrm{rad}/{\mathrm{s}}^2$for $\frac{1}{2}\mathrm{s}$, then coasts at a steady angular velocity for another $\frac{1}{2}\mathrm{s}$.

a. What is the speed of the dot at $\mathrm{t}=1.0\mathrm{s}$?

b. Through how many revolutions has the disk turned?

Short answer

a. The speed of the dot at $\mathrm{t}=1.0\mathrm{s}$is$12\mathrm{m}/\mathrm{s}$.

b. The revolutions the disk has turned is$36\mathrm{rev}$.

Step-by-step solution

Given information

We are given with:

$$\begin{gathered} \mathrm{diameter}=\mathrm{d}=8\mathrm{cm} \\ \mathrm{angular}\mathrm{acceleration}=\mathrm{\alpha }=600\mathrm{rad}/{\mathrm{s}}^2 \\ \mathrm{time}=\mathrm{t}=\frac{1}{2}\mathrm{s} \\ \mathrm{a}.\mathrm{t}=1\mathrm{s} \end{gathered}$$

Part a) Step 1: Calculation

$$\begin{gathered} {\mathrm{\omega }}_{\mathrm{f}}={\mathrm{\omega }}_{\mathrm{i}}+\mathrm{\alpha t} \\ {\mathrm{\omega }}_{\mathrm{f}}=0+600\mathrm{rad}/{\mathrm{s}}^2(\frac{1}{2}\mathrm{s}) \\ {\mathrm{\omega }}_{\mathrm{f}}=300\mathrm{rad}/\mathrm{s} \end{gathered}$$

For radius:

$$\begin{gathered} \mathrm{r}=\frac{\mathrm{d}}{2} \\ \mathrm{r}=\frac{8\mathrm{cm}}{2} \\ \mathrm{r}=4\mathrm{cm} \\ \mathrm{r}=\frac{4}{100}\mathrm{m} \\ \mathrm{r}=0.04\mathrm{m} \end{gathered}$$

To convert Angular velocity in linear velocity:

$$\begin{gathered} \mathrm{v}={\mathrm{\omega }}_{\mathrm{f}}\mathrm{r} \\ \mathrm{v}=(300\mathrm{rad}/\mathrm{s})(0.04\mathrm{s}) \\ \mathrm{v}=12\mathrm{m}/\mathrm{s} \end{gathered}$$

As the speed doesn't change so the velocity is same for$\mathrm{t}=\frac{1}{2}\mathrm{s}$and$\mathrm{t}=1\mathrm{s}$.

Part b) Step 1: Calculation

The revolutions for $\mathrm{t}=\frac{1}{2}\mathrm{s}$is:

localid="1649070914019" $$\begin{gathered} {\mathrm{\theta }}_1-{\mathrm{\theta }}_0={\mathrm{\omega }}_{\mathrm{i}}\mathrm{t}+\frac{1}{2}{\mathrm{\alpha t}}^2 \\ {\mathrm{\theta }}_1=0+\left(\frac{1}{2}\right)(600\mathrm{rad}/{\mathrm{s}}^{\mathrm{s}}){(0.5\mathrm{s})}^2 \\ {\mathrm{\theta }}_1=75\mathrm{rad} \end{gathered}$$

Now number of revolutions for $\mathrm{t}=1\mathrm{s}$

localid="1648909265788" $$\begin{gathered} {\mathrm{\theta }}_2-{\mathrm{\theta }}_1={\mathrm{\omega }}_{\mathrm{f}}\mathrm{t}+\frac{1}{2}{\mathrm{\alpha t}}^2 \\ {\mathrm{\theta }}_2={\mathrm{\theta }}_1+{\mathrm{\omega }}_{\mathrm{f}}\mathrm{t}+\frac{1}{2}{\mathrm{\alpha t}}^2 \\ {\mathrm{\theta }}_2=75\mathrm{rad}+(300\mathrm{rad}/\mathrm{s})(0.5\mathrm{s})+0 \\ {\mathrm{\theta }}_2=225\mathrm{rad} \end{gathered}$$

Converting radian in revolution

localid="1649070935550" $$\begin{gathered} {\mathrm{\theta }}_2=225\mathrm{rad}\times \frac{1}{2\mathrm{\pi }\mathrm{rad}} \\ {\mathrm{\theta }}_2=35.8\mathrm{rev} \\ {\mathrm{\theta }}_2=36\mathrm{rev} \end{gathered}$$