A solid disk with a radius of $9 m$ $9 m$ and mass of $4 k g$ $4 kg$ is rotating on a frictionless surface. If $750 W$ $750 W$ of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at $4 H z$ $4 Hz$ ?

Question

1448 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-23T05:55:23

The torque is $= 29.8 N m$ $=29.8Nm$

The mass of the disc is $m = 4 k g$ $m=4 kg$

The radius of the disc is $r = 9 m$ $r=9m$

The power and the torque are related by the formula

$P = τ ω$ $P=tau omega$

The angular velocity is $ω = 2 π f = 2 π \cdot 2 = 8 π r a d s^{- 1}$ $omega=2pif=2pi*2=8pirads^-1$

The power is $P = 750 W$ $P=750W$

Therefore,

The torque is

$τ = \frac{P}{ω} = \frac{750}{8 π} = 29.8 N m$ $tau=P/omega=750/(8pi)=29.8Nm$

A solid disk with a radius of 9 m9m9 m and mass of 4 kg4kg4 kg is rotating on a frictionless surface. If 750 W750W750 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 4 Hz4Hz4 Hz?