A solid disk with a radius of $3 m$ $3 m$ and mass of $2 k g$ $2 kg$ is rotating on a frictionless surface. If $240 W$ $240 W$ of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at $9 H z$ $9 Hz$ ?

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Answer 1 · 2017-11-08T07:02:24

The torque is $= 4.24 N m$ $=4.24Nm$

Apply the equation

$P = τ ω$ $P=tau omega$

The power is $P = 240 W$ $P=240W$

The frequency is $f = 9 H z$ $f=9Hz$

The angular velocity is

$ω = 2 π f = 2 \times π \times 9 = 18 π r a d s^{- 1}$ $omega=2pif=2xxpixx9=18pirads^-1$

Therefore,

The torque is

$τ = \frac{P}{ω} = \frac{240}{18 π} = 4.24 N m$ $tau=P/omega=240/(18pi)=4.24Nm$

A solid disk with a radius of 3m3 m and mass of 2kg2 kg is rotating on a frictionless surface. If 240W240 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 9Hz9 Hz?