A solid disk with a radius of $8 m$ $8 m$ and mass of $6 k g$ $6 kg$ is rotating on a frictionless surface. If $60 W$ $60 W$ of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at $3 H z$ $3 Hz$ ?

Question

1441 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-09T09:17:54

The torque is $= 3.18 N m$ $=3.18Nm$

mass of disc $= 2 k g$ $=2kg$

radius of disc $= 2 m$ $=2m$

The power is related to the torque by the equation

$P = τ \cdot ω$ $P=tau *omega$

$P = 60 W$ $P=60W$

Frequency of rotation is $f = 3 H z$ $f=3Hz$

$ω = f \cdot 2 π$ $omega=f*2pi$

angular velocity, $ω = 3 \cdot 2 π r a d s^{- 1}$ $omega=3*2pi rads^-1$

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

$= \frac{60}{6 π} = 3.18 N m$ $=60/(6pi)=3.18Nm$

A solid disk with a radius of 8m8 m and mass of 6kg6 kg is rotating on a frictionless surface. If 60W60 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 3Hz3 Hz?