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 $750 W$ $750 W$ of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at $4 H z$ $4 Hz$ ?

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Answer 1 · 2017-02-24T19:02:54

The torque is $= 119.4 N m$ $=119.4Nm$

The power is related to the torque, by the equation

$P = τ ω$ $P=tau omega$

$P =$ $P=$ power

$τ =$ $tau=$ torque

$ω =$ $omega=$ angular velocity

$P = 750 W$ $P=750W$

$ω = 4 \cdot 2 π = (8 π) r a d s^{- 1}$ $omega=4*2pi=(8pi)rads^-1$

$τ = \frac{750}{2 π} = 119.4 N m$ $tau = 750/(2pi) =119.4Nm$

A solid disk with a radius of 6m6 m and mass of 9kg9 kg is rotating on a frictionless surface. If 750W750 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 4Hz4 Hz?