# A solid disk with a radius of 2 m and mass of 2 kg is rotating on a frictionless surface. If 640 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 16 Hz?

Feb 9, 2017

$\tau = \frac{640 W}{32 \pi} \approx 6.37 N \cdot m$

#### Explanation:

We are told we have a rotating spinning coin.

Step 1. Gather the information that you know and need

• Mass of disk: $\text{mass"=m=2" kg}$
• Radius of disk: $\text{radius"=r=2" m}$
• Power applied to increase rotation: $P = 640 \text{ W}$
• Frequency: $f = 16 \text{ Hz}$
• Torque required: ??

Step 2. Determine the formula using the above givens

• $P$ is the power applied, $P = \tau \omega$
• Therefore, torque is $\tau = P \text{/} \omega$
• Angular speed, $\omega = f \cdot 2 \pi = 16 \cdot 2 \pi = 32 \pi$

Step 3. Plug your knows into the formula for torque

$\tau = \frac{640 W}{32 \pi} \approx 6.37 N \cdot m$