A solid disk with a radius of $8 m$ $8 m$ and mass of $9 k g$ $9 kg$ is rotating on a frictionless surface. If $960 W$ $960 W$ of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at $28 H z$ $28 Hz$ ?

Question

1411 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-03T06:23:32

The torque is $= 5.46 N m$ $=5.46Nm$

Apply the equation

$Power (W) = torque(Nm) \times angular velocity (r a d s^{- 1})$ $"Power (W)" ="torque(Nm)"xx"angular velocity" (rad s^-1)$

The power is $P = 960 W$ $P=960W$

The frequency is $f = 28 H z$ $f=28Hz$

The angular velocity is

$ω = 2 π f = 2 \times π \times 28 = (56 π) r a d s^{- 1}$ $omega=2pif=2xxpixx28=(56pi)rads^-1$

Therefore,

The torque is

$τ = \frac{P}{ω} = \frac{960}{56 π} = 5.46 N m$ $tau=P/omega=960/(56pi)=5.46Nm$

A solid disk with a radius of 8m8 m and mass of 9kg9 kg is rotating on a frictionless surface. If 960W960 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 28Hz28 Hz?