# A solid disk with a radius of 8 m and mass of 8 kg is rotating on a frictionless surface. If 480 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 6 Hz?

Jul 2, 2016

$= 12.73 N m$

#### Explanation:

Given

• $m \to \text{Mass of solid disk} = 8 k g$

• $r \to \text{Radius of solid disk} = 8 m$

• $P \to \text{Power used to solid disk} = 480 W$

• $n \to \text{Frequency of rotaion of solid disk} = 6 H z$

We are to find out

• $\tau \to \text{Torque applied to solid disk}$

Formula

$\text{Power"(P)="Torque"(tau)xx"Angular velocity} \left(w\right)$

$\implies \text{Power"(P)="Torque"(tau)xx2pixx"Frequency of rotation} \left(n\right)$

$\implies 480 = \text{Torque} \left(\tau\right) \times 2 \pi \times 6$

$\text{Torque} \left(\tau\right) = \frac{480}{2 \pi \times 6} = 12.73 N m$