A solid disk with a radius of $9 m$ $9 m$ and mass of $8 k g$ $8 kg$ is rotating on a frictionless surface. If $360 W$ $360 W$ of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at $6 H z$ $6 Hz$ ?

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Answer 1 · 2017-10-15T05:25:01

The torque is $= 9.55 N m$ $=9.55Nm$

The mass of the disc is $m = 8 k g$ $m=8kg$

The radius of the disc is $r = 9 m$ $r=9m$

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 $P = 360 W$ $P=360W$

The frequency is $f = 6 H z$ $f=6Hz$

The angular velocity is $ω = 2 π f = 2 \cdot π \cdot 6 = (12 π) r a d s^{- 1}$ $omega=2pif=2*pi*6=(12pi)rads^-1$

Therefore,

the torque is $τ = \frac{P}{ω} = \frac{360}{12 π} = 9.55 N m$ $tau=P/omega=360/(12pi)=9.55Nm$

A solid disk with a radius of 9m9 m and mass of 8kg8 kg is rotating on a frictionless surface. If 360W360 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 6Hz6 Hz?