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Work and rotational Kinetic Energy

  • Thread starter frig0018
  • Start date
  • #1
5
0

Homework Statement


A 32.0 kg wheel, essentially a thin hoop with r=1.20 m is rotating at 280 rev/min. It must be brought to stop in 15.0 s. How much work must be done to stop it?


Homework Equations

/

The Attempt at a Solution


I=mr^2=46.08 kg*m^2
W=(1/2)I[tex]\omega[/tex]f^2 - (1/2)I[tex]\omega[/tex]i^2
W=[tex]\tau[/tex]([tex]\Delta[/tex][tex]\theta[/tex])
[tex]\tau[/tex]=I[tex]\alpha[/tex]
Rev in 15 sec = 70 rev => 140[tex]\Pi[/tex] radians in 15 sec.

I'm not sure which equation(s) to use. I've tried plugging numbers into all of them and getting stuck or wrong answers. Thanks!!
 

Answers and Replies

  • #2
Nabeshin
Science Advisor
2,205
16
Remember, work is a change in kinetic energy. Do you know the formula for kinetic energy of a rotating object?
 
  • #3
5
0
i thought it was
W = delta K = 1/2(I)(omega final)^2 - 1/2(I)(omega initial)^2...?
or do i use the formula
W= (integral from theta initial to theta final) torque d-theta?
 
  • #4
tiny-tim
Science Advisor
Homework Helper
25,832
249
… one step at a time …

A 32.0 kg wheel, essentially a thin hoop with r=1.20 m is rotating at 280 rev/min. It must be brought to stop in 15.0 s. How much work must be done to stop it?
Hi frig0018! :smile:

First step: what angular acceleration is needed to stop it in 15 s? :smile:
 

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