A solid disk with a radius of $9 m$ $9 m$ and mass of $12 k g$ $12 kg$ is rotating on a frictionless surface. If $480 W$ $480 W$ of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at $15 H z$ $15 Hz$ ?

Question

1400 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-30T10:04:43

The torque is $= 5.09 N m$ $=5.09Nm$

mass of disc $= 12 k g$ $=12kg$

radius of disc $= 9 m$ $=9m$

The power is related to the torque by the equation

$P = τ \cdot ω$ $P=tau *omega$

$P = 480 W$ $P=480W$

Frequency of rotation is $f = 15 H z$ $f=15Hz$

$ω = f \cdot 2 π$ $omega=f*2pi$

angular velocity, $ω = 15 \cdot 2 π r a d s^{- 1}$ $omega=15*2pi rads^-1$

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

$= \frac{480}{30 π} = 5.09 N m$ $=480/(30pi)=5.09Nm$

A solid disk with a radius of 9m9 m and mass of 12kg12 kg is rotating on a frictionless surface. If 480W480 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 15Hz15 Hz?