Find the net torque acting on the system

1. Sep 8, 2008

kehler

1. The problem statement, all variables and given/known data
In the figure below, the incline is frictionless and the string passes through the centre of mass of each block. The pulley has moment of inertia I and radius R.
a) Find the net torque acting on the system (the two masses, string and pulley) about the centre of the pulley
b) Write an expression for the total angular momentum of the system about the center of the pulley when the masses are moving with a speed v
c) Find the acceleration of the masses from your results for a and b by setting the net torque equal to the rate of change of the angular momentum of the system

2. Relevant equations
torque = r x F
angular momentum = r x p or Iw

3. The attempt at a solution
a) I got an answer for this but I dunno if its right
F on m2 in direction parallel to incline = m2 x g sin theta
F on m1 = m1 x g
So torque about centre of pulley due to force on m2 is R x m2 x g sin theta, out of the page
and torque about centre of pulley due to force on m1 is R x m x g, into the page
Net torque is (R x m2 x g sin theta) - (R x m x g), out of the page

b) I'm not sure how to do this. Do I calculate the angular momentum of each of the masses about the centre of the pulley and add it with the angular momentum of the pulley? But if this is the case, I don't know what the angular velocity of the pulley is :S
Angular momentum of m1 about centre = R x m1 v, into the page
Angular momentum of m2 about centre = R x m2 v, into the page
Angular momentum of pulley about it's centre = Iw = 0.5MR^2 w but we don't know w

c)No idea. Do I just set the answer for a to be equal to the differential of the answer to b???

Once again, any help would be much appreciated :)

Last edited: Sep 8, 2008
2. Sep 8, 2008

Staff: Mentor

This is good, except that you seem to have the directions reversed. Using the right hand rule for cross products, r X F for the force on m2 should be into the page.

Yes.
There's a simple relationship between angular speed of the pulley and the linear speed of the masses.
OK, but again, the directions are off. Let's assume that m2 moves up, m1 down. Thus the angular momentum of m2, r X m2v, should be out of the page.
Express w in terms of v.

Yes, "rate of change" means take the derivative. What's the derivative of v?

3. Sep 8, 2008

kehler

Thanks Doc Al.
Oops, I don't know what I was thinking while typing out my solution to part (a)! I got the directions right on the piece of paper I was writing on.
For part (b), I intended for m2 to move down and m1 to move up. I'm pretty sure the angular momentum for both masses should be into the page if I do it like that.
Is the total angular momentum (R x m2 v) + (R x m1 v) + I(v/R), into the page (based on the assumption that m2 moves down and m1 up) ?
So for (c), is the equation I'm meant to get
(R x m2 a) + (R x m2 a) + (I/R)a = (R x m2 x g sin theta) - (R x m x g)?
Which would mean a = (M2 x gsintheta - M1g)/(M1 + M2 + (I/R^2))
Did I get that right? :)

4. Sep 8, 2008

Staff: Mentor

Absolutely. You were right the first time.
Looks good.
Yes, except for a few typos. (Make sure your m's are properly labeled.)
Looks good to me.

5. Sep 8, 2008

kehler

Awesome :). Thanks again.

6. Sep 8, 2008

Staff: Mentor

You are most welcome.

Just for fun, I recommend that you solve for the acceleration the "usual" way (by applying Newton's 2nd law to each block and the pulley and then combining the resulting equations) and confirm that you get the same answer.