A solid disk with a radius of $4 m$ $4 m$ and mass of $6 k g$ $6 kg$ is rotating on a frictionless surface. If $960 W$ $960 W$ of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at $12 H z$ $12 Hz$ ?

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Answer 1 · 2017-02-09T19:36:19

$τ = \frac{960}{24 π} \approx 12.73 N \cdot m$ $tau=(960)/(24pi)~~12.73N*m$

We are told we have a rotating spinning coin.

![http://www.mariowiki.com/coin]( useruploads.socratic.org )

Step 1. Gather the information that you know and need

Step 2. Determine the formula using the above givens

$P$ $P$ is the power applied, $P = τ ω$ $P=tau omega$
Therefore, torque is $τ = P / ω$ $tau=P"/"omega$
Angular speed, $ω = f \cdot 2 π = 12 \cdot 2 π = 24 π$ $omega=f*2pi=12*2pi=24pi$

Step 3. Plug your knows into the formula for torque

$τ = \frac{960}{24 π} \approx 12.73 N \cdot m$ $tau=(960)/(24pi)~~12.73N*m$

A solid disk with a radius of 4m4 m and mass of 6kg6 kg is rotating on a frictionless surface. If 960W960 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 12Hz12 Hz?