How do I calculate the total torque exerted on a rotating disk?

  • Thread starter Thread starter Sheneron
  • Start date Start date
  • Tags Tags
    Rotational
Sheneron
Messages
360
Reaction score
0
[SOLVED] Easy rotational problem

Homework Statement


A string is wound around a uniform disk of radius 0.25m and mass 8.00kg. The end of the string is pulled with a constant force of 50.0 N, and there is a frictional torque acting at the axle of 3.00 Nm.
A) Find the total torque exerted on the disk

The Attempt at a Solution


I think I have it solved... just wanted to make sure and I have one question. It says that the frictional torque is acting on the axle, do I still use the radius r=0.25 to find the torque for that part? Here is what I did:

[tex]\Sigma \tau = \Sigma Fd[/tex]
[tex]\Sigma \tau = (50)(0.25) + (-3)(0.25)[/tex]
[tex]\Sigma \tau = 11.75 Nm[/tex]

So is that how I would find the net torque?
 
on Phys.org
Sheneron said:
It says that the frictional torque is acting on the axle, do I still use the radius r=0.25 to find the torque for that part?
Careful. For friction they gave you the torque, not the force. So you don't have to use any radius to find the frictional torque.
 
Oh yeah I just realized that. so I guess what i meant to say would the net force end up being 9.5 Nm?
 
Sheneron said:
so I guess what i meant to say would the net force end up being 9.5 Nm?
The net torque is 9.5 Nm.
 
oops again...
Thanks.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
Replies
26
Views
7K
Replies
5
Views
1K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 25 ·
Replies
25
Views
12K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
19
Views
4K
Replies
28
Views
5K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K