A solid disk with a radius of $2 m$ $2 m$ and mass of $3 k g$ $3 kg$ is rotating on a frictionless surface. If $36 W$ $36 W$ of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at $3 H z$ $3 Hz$ ?

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Answer 1 · 2017-06-24T06:30:38

The torque is $= 1.91 N m$ $=1.91Nm$

The mass of the disc is $m = 3 k g$ $m=3 kg$

The radius of the disc is $r = 2 m$ $r=2m$

The power and the torque are related by the formula

$P = τ ω$ $P=tau omega$

The angular velocity is $ω = 2 π f = 2 π \cdot 3 = 6 π r a d s^{- 1}$ $omega=2pif=2pi*3=6pirads^-1$

The power is $P = 36 W$ $P=36W$

Therefore,

The torque is

$τ = \frac{P}{ω} = \frac{36}{6 π} = 1.91 N m$ $tau=P/omega=36/(6pi)=1.91Nm$

A solid disk with a radius of 2m2 m and mass of 3kg3 kg is rotating on a frictionless surface. If 36W36 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 3Hz3 Hz?