A solid disk with a radius of $6 m$ $6 m$ and mass of $5 k g$ $5 kg$ is rotating on a frictionless surface. If $240 W$ $240 W$ of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at $8 H z$ $8 Hz$ ?

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Answer 1 · 2017-08-31T08:38:19

The torque is $= 1.19 N m$ $=1.19Nm$

The mass of the disc is $m = 5 k g$ $m=5kg$

The radius of the disc is $r = 6 m$ $r=6m$

The power $= P$ $=P$ and the torque $= τ$ $=tau$ are related by the following equation

$P = τ ω$ $P=tau omega$

where $ω =$ $omega=$ the angular velocity

Here,

The power is $P = 240 W$ $P=240W$

The frequency is $f = 8 H z$ $f=8Hz$

The angular velocity is $ω = 2 π f = 2 \cdot π \cdot 8 = 16 π r a d s^{- 1}$ $omega=2pif=2*pi*8=16pirads^-1$

Therefore,

the torque is $τ = \frac{P}{ω} = \frac{60}{16 π} = 1.19 N m$ $tau=P/omega=60/(16pi)=1.19Nm$

A solid disk with a radius of 6m6 m and mass of 5kg5 kg is rotating on a frictionless surface. If 240W240 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 8Hz8 Hz?