A solid disk with a radius of $3 m$ $3 m$ 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 $15 H z$ $15 Hz$ ?

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Answer 1 · 2018-02-05T10:45:05

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

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

The frequency is $f = 15 H z$ $f=15Hz$

The angular velocity is

$ω = 2 π f = 2 \times π \times 15 = (30 π) r a d s^{- 1}$ $omega=2pif=2xxpixx15=(30pi)rads^-1$

Therefore,

The torque is

$τ = \frac{P}{ω} = \frac{480}{30 π} = 5.09 N m$ $tau=P/omega=480/(30pi)=5.09Nm$

A solid disk with a radius of 3 m3m3 m and mass of 12 kg12kg12 kg is rotating on a frictionless surface. If 480 W480W480 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 15 Hz15Hz15 Hz?