A solid disk with a radius of $12 m$ $12 m$ and mass of $8 k g$ $8 kg$ is rotating on a frictionless surface. If $640 W$ $640 W$ of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at $16 H z$ $16 Hz$ ?

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Answer 1 · 2017-05-09T09:45:52

The torque is $= 6.4 N m$ $=6.4Nm$

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

Radius of disc $= 12 m$ $=12m$

The power is related to the torque by the equation

$P = τ \cdot ω$ $P=tau *omega$

$P = 640 W$ $P=640W$

Frequency of rotation is $f = 16 H z$ $f=16Hz$

$ω = f \cdot 2 π$ $omega=f*2pi$

angular velocity, $ω = 16 \cdot 2 π r a d s^{- 1}$ $omega=16*2pi rads^-1$

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

$= \frac{640}{32 π} = 6.4 N m$ $=640/(32pi)=6.4Nm$

A solid disk with a radius of 12m12 m and mass of 8kg8 kg is rotating on a frictionless surface. If 640W640 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 16Hz16 Hz?