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 $5 H z$ $5 Hz$ ?

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Answer 1 · 2017-08-28T12:57:41

The torque is $= 1.91 N m$ $=1.91Nm$

The mass of the disc is $m = 6 k g$ $m=6kg$

The radius of the disc is $r = 8 m$ $r=8m$

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

The frequency is $f = 5 H z$ $f=5Hz$

The angular velocity is $ω = 2 π f = 2 \cdot π \cdot 5 = 10 π r a d s^{- 1}$ $omega=2pif=2*pi*5=10pirads^-1$

Therefore,

the torque is $τ = \frac{P}{ω} = \frac{60}{10 π} = 1.91 N m$ $tau=P/omega=60/(10pi)=1.91Nm$

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 5Hz5 Hz?