A solid disk with a radius of $2 m$ $2 m$ and mass of $12 k g$ $12 kg$ is rotating on a frictionless surface. If $15 W$ $15 W$ of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at $1 H z$ $1 Hz$ ?

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Answer 1 · 2018-01-06T14:56:27

The torque is $= 2.39 N m$ $=2.39Nm$

Apply the equation

$Power (W) = torque(Nm) \times angular velocity (r a d s^{- 1})$ $"Power (W)" ="torque(Nm)"xx"angular velocity" (rad s^-1)$

The power is $P = 15 W$ $P=15W$

The frequency is $f = 1 H z$ $f=1Hz$

The angular velocity is

$ω = 2 π f = 2 \times π \times 1 = (2 π) r a d s^{- 1}$ $omega=2pif=2xxpixx1=(2pi)rads^-1$

Therefore,

The torque is

$τ = \frac{P}{ω} = \frac{15}{2 π} = 2.39 N m$ $tau=P/omega=15/(2pi)=2.39Nm$

A solid disk with a radius of 2m2 m and mass of 12kg12 kg is rotating on a frictionless surface. If 15W15 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 1Hz1 Hz?