Atwoods: Deriving Torque Relation with R

[breez]

A pulley with radius R is free to rotate on a horizontal fixed axis through its center. A string passes over the pulley. Mass m1 is attached to one end and mass m2 is attached to the other. The portion of the string attached to m1 has tension T1 and the portion attached to m2 has tension T2. The magnitude of the total external torque, about the pulley center, acting on the masses and pulley, considered as a system, is given by:

(m1 – m2)gR

Can someone explain how to derive this relation? Why is the distance just R?

 

[Doc Al]

The force (mg) acts vertically and R is the perpendicular distance to the axis. So each torque is just:

\tau = \vec{r} \times \vec{F} = RF = mgR

 

[breez]

I thought the force a acting perpendicular to the center of pulley is tension, not gravity.

 

[Doc Al]

I thought the force a acting perpendicular to the center of pulley is tension, not gravity.

The force on the pulley is the tension, not the weight. But I thought you were asking about the net torque on the entire system, not just on the pulley. (A perfectly valid way to analyze the problem, but I would not recommend it.)

 

[breez]

Hmm,

The problem asks for the net torque on the entire system about the center of the pulley. I believe the gravitational forces, m1g and m2g are acting at the center of masses of the two masses, which are some distance from the center. I am confused how to calculate the net torque because I thought you had to add all the individual torques, which would mean I would need to know the distance at any time between the center of mass of each mass and the center of the pulley.

I'm sorry, but I do not see the reasoning behind just adding up all the forces on the system and multiplying by R.

 

[Doc Al]

The problem asks for the net torque on the entire system about the center of the pulley. I believe the gravitational forces, m1g and m2g are acting at the center of masses of the two masses, which are some distance from the center.

You can certainly do it that way also. You'll get the same answer.

I am confused how to calculate the net torque because I thought you had to add all the individual torques, which would mean I would need to know the distance at any time between the center of mass of each mass and the center of the pulley.

All you need is the perpendicular distance. Consider this:
\tau = \vec{r} \times \vec{F} = rF\sin\theta

While the distance r is constantly changing, what counts is r\sin\theta--which is constant and equal to R.

I'm sorry, but I do not see the reasoning behind just adding up all the forces on the system and multiplying by R.

You need to firm up your understanding of torque and how to calculate it.

 

[breez]

Yeah, you're right. I forgot it was the moment arm, since it's the cross product. Wow.