A solid disk with a radius of $3 m$ $3 m$ and mass of $4 k g$ $4 kg$ is rotating on a frictionless surface. If $120 W$ $120 W$ of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at $12 H z$ $12 Hz$ ?

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Answer 1 · 2017-07-02T06:35:41

The torque is $= 1.59 N m$ $=1.59Nm$

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

The radius of the disc is $r = 3 m$ $r=3m$

The power and the torque are related by the formula

$P = τ ω$ $P=tau omega$

The angular velocity is $ω = 2 π f = 2 π \cdot 12 = 24 π r a d s^{- 1}$ $omega=2pif=2pi*12=24pirads^-1$

The power is $P = 120 W$ $P=120W$

Therefore,

The torque is

$τ = \frac{P}{ω} = \frac{120}{24 π} = 1.59 N m$ $tau=P/omega=120/(24pi)=1.59Nm$

A solid disk with a radius of 3m3 m and mass of 4kg4 kg is rotating on a frictionless surface. If 120W120 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 12Hz12 Hz?