A solid disk with a radius of $2 m$ $2 m$ and mass of $3 k g$ $3 kg$ is rotating on a frictionless surface. If $36 W$ $36 W$ 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 · 2017-06-12T15:15:37

The torque is $= 0.95 N m$ $=0.95Nm$

Mass of disc $= 3 k g$ $=3kg$

Radius of disc $= 2 m$ $=2m$

The power is related to the torque by the equation

$P = τ \cdot ω$ $P=tau *omega$

$P = 36 W$ $P=36W$

Frequency of rotation is $f = 6 H z$ $f=6Hz$

$ω = f \cdot 2 π$ $omega=f*2pi$

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

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

$= \frac{36}{12 π} = 0.95 N m$ $=36/(12pi)=0.95Nm$

A solid disk with a radius of 2m2 m and mass of 3kg3 kg is rotating on a frictionless surface. If 36W36 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?