A solid disk with a radius of $6 m$ $6 m$ and mass of $8 k g$ $8 kg$ is rotating on a frictionless surface. If $18 W$ $18 W$ of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at $5 H z$ $5 Hz$ ?

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Answer 1 · 2017-03-28T09:30:13

The torque is $= 0.57 N m$ $=0.57Nm$

mass of disc $= 8 k g$ $=8kg$

radius of disc $= 6 m$ $=6m$

The power is related to the torque by the equation

$P = τ \cdot ω$ $P=tau *omega$

$P = 18 W$ $P=18W$

Frequency of rotation is $f = 5 H z$ $f=5Hz$

$ω = f \cdot 2 π$ $omega=f*2pi$

angular velocity, $ω = 5 \cdot 2 π r a d s^{- 1}$ $omega=5*2pi rads^-1$

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

$= \frac{18}{10 π} = 0.57 N m$ $=18/(10pi)=0.57Nm$

A solid disk with a radius of 6m6 m and mass of 8kg8 kg is rotating on a frictionless surface. If 18W18 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 5Hz5 Hz?