Work done by wheel.

  • Thread starter bobby3280
  • Start date
  • #1
12
0
A flywheel of mass 183 kg has an effective radius of 0.63 m (assume the mass is concentrated along a circumference located at the effective radius of the flywheel).

(a) What torque is required to bring this wheel from rest to a speed of 120 rpm in a time interval of 34.3 s?

(b) How much work is done during the 34.3 s?

Alright I figured out (a) using the moment of Inertia and the angular acceleration to be 26.65 N * m

So for part (b) I tried using W = torque * theta
to find theta i used many ways all coming up with right around 432 rad
so W = 26.65 * 432
= 11.5 kj
But this answer isn't correct any suggestions as to where i went wrong??
 

Answers and Replies

  • #2
176
0
All the work done on the flywheel is converted to rotational energy. Do you know an equation that will give you the energy of a rotating mass?

Hope this helps,
Sam
 
  • #3
radou
Homework Helper
3,120
6
For (b), simply use the fact that the work done equals the change of kinetic energy.
 

Related Threads on Work done by wheel.

  • Last Post
Replies
7
Views
5K
Replies
7
Views
2K
  • Last Post
Replies
6
Views
775
Replies
4
Views
4K
  • Last Post
Replies
1
Views
954
Replies
5
Views
7K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
2K
Top