A solid disk with a radius of $8 m$ $8 m$ and mass of $9 k g$ $9 kg$ is rotating on a frictionless surface. If $960 W$ $960 W$ of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at $32 H z$ $32 Hz$ ?

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Answer 1 · 2017-02-19T06:00:31

The torque is $= 4.77 N m$ $=4.77Nm$

Power is related to torque, by the equation

$P = τ \cdot ω$ $P=tau* omega$

$P =$ $P=$ power (W)

$τ =$ $tau=$ torque (Nm)

$ω =$ $omega=$ angular velocity (rads^-1)

Here,

$P = 960 W$ $P=960W$

$ω = 32 \cdot 2 π = 64 π r a d s^{- 1}$ $omega=32*2pi=64pi rads^-1$

Therefore,

$τ = \frac{960}{64 π} = 4.77 N m$ $tau=960/(64pi)=4.77Nm$

A solid disk with a radius of 8m8 m and mass of 9kg9 kg is rotating on a frictionless surface. If 960W960 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 32Hz32 Hz?