Question

12) When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius of 6.0 cm and a mass of 17 g, what is the torque exerted on it?

Answer 1

0.03 N/m

Explanation:

The first thing is to convert from rev/min to rad/s and rev a rad, like this:

`\begin{gathered} 450(rev)/(\min)\cdot\frac{2\pi\text{ rad}}{rev}\cdot\frac{1\min}{60\text{ sec}}=47.12\text{ rad/s} \\ 3\text{ rev}\cdot\frac{2\pi\text{ rad}}{rev}=18.85\text{ rad} \end{gathered}`

Now, we apply the following formula:

`\begin{gathered} w^2=w^2_0+2a\mleft(g_f-g_i\mright) \\ \text{ replacing:} \\ (47.12)^2=0+2\cdot a\cdot(18.85-0) \\ a=((47.12)^2)/(2\cdot18.85) \\ a=58.9rad/s^2 \end{gathered}`

We use the torque formula, like this:

`\begin{gathered} \tau=a\cdot I \\ I=(1)/(2)\cdot m\cdot r^2=(1)/(2)\cdot(0.017)\cdot(0.06)=0.00051 \\ \tau=58.9\cdot0.00051 \\ \tau=0.03\text{ N/m} \end{gathered}`

The torque exerted is 0.03 N/m