A solid disk with a radius of $24 m$ $24 m$ and mass of $8 k g$ $8 kg$ is rotating on a frictionless surface. If $480 W$ $480 W$ 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 · 2017-06-09T05:45:07

The torque is $= 25.46 N m$ $=25.46Nm$

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

Radius of disc $= 24 m$ $=24m$

The power is related to the torque by the equation

$P = τ \cdot ω$ $P=tau *omega$

$P = 480 W$ $P=480W$

Frequency of rotation is $f = 3 H z$ $f=3Hz$

$ω = f \cdot 2 π$ $omega=f*2pi$

angular velocity, $ω = 3 \cdot 2 π r a d s^{- 1}$ $omega=3*2pi rads^-1$

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

$= \frac{480}{6 π} = 25.46 N m$ $=480/(6pi)=25.46Nm$

A solid disk with a radius of 24m24 m and mass of 8kg8 kg is rotating on a frictionless surface. If 480W480 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?