A solid disk with a radius of $6 m$ $6 m$ and mass of $9 k g$ $9 kg$ is rotating on a frictionless surface. If $120 W$ $120 W$ of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at $8 H z$ $8 Hz$ ?

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Answer 1 · 2017-03-29T05:38:29

The torque is $= 2.39 N m$ $=2.39Nm$

mass of disc $= 9 k g$ $=9kg$

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

The power is related to the torque by the equation

$P = τ \cdot ω$ $P=tau *omega$

$P = 120 W$ $P=120W$

Frequency of rotation is $f = 8 H z$ $f=8Hz$

$ω = f \cdot 2 π$ $omega=f*2pi$

angular velocity, $ω = 8 \cdot 2 π r a d s^{- 1}$ $omega=8*2pi rads^-1$

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

$= \frac{120}{16 π} = 2.39 N m$ $=120/(16pi)=2.39Nm$

A solid disk with a radius of 6 m6m6 m and mass of 9 kg9kg9 kg is rotating on a frictionless surface. If 120 W120W120 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 8 Hz8Hz8 Hz?