A solid disk with a radius of #3 m# and mass of #2 kg# is rotating on a frictionless surface. If #480 W# of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at #5 Hz#?

1 Answer
Feb 9, 2017

Answer:

#tau=(480)/(10pi)~~15.28N*m#

Explanation:

We are told we have a rotating spinning coin.

http://www.mariowiki.com/coin

Step 1. Gather the information that you know and need

  • Mass of disk: #"mass"=m=2" kg"#
  • Radius of disk: #"radius"=r=3" m"#
  • Power applied to increase rotation: #P=480" W"#
  • Frequency: #f=5" Hz"#
  • Torque required: ??

Step 2. Determine the formula using the above givens

  • #P# is the power applied, #P=tau omega#
  • Therefore, the formula for torque is #tau=P"/"omega#
  • Angular speed, #omega=f*2pi=5*2pi=10pi#

Step 3. Plug your answer into the formula for torque

#tau=(480)/(10pi)~~15.28N*m#