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 $360 W$ $360 W$ of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at $6 H z$ $6 Hz$ ?

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Answer 1 · 2017-03-25T06:54:21

The torque is $= 9.55 N m$ $=9.55Nm$

mass of disc $= 4 k g$ $=4kg$

radius of disc $= 9 m$ $=9m$

The power is related to the torque by the equation

$P = τ \cdot ω$ $P=tau *omega$

$P = 360 W$ $P=360W$

angular velocity, $ω = 6 \cdot 2 π r a d s^{- 1}$ $omega=6*2pi rads^-1$

torque, $τ = \frac{P}{ω}$ $tau=P/omega$

$= \frac{360}{12 π} = \frac{30}{π} = 9.55 N m$ $=360/(12pi)=30/pi=9.55Nm$

A solid disk with a radius of 9m9 m and mass of 4kg4 kg is rotating on a frictionless surface. If 360W360 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 6Hz6 Hz?