# Total Torque

#### craig22

1. Homework Statement

For the picture, go to http://s16.photobucket.com/albums/b21/arbilvoldy/pulley.jpg" [Broken].
T1 is for the hanging string with mass m1, and T2 is for the string over the table with mass m2.

Two 6.20 kg blocks are connected by a massless string over a pulley of radius 2.40 cm and rotational inertia 7.40x10^-4 kgm^2. The string does not slip on the pulley; it is not known whether there is friction between the table and the sliding block; the pulley's axis is frictionless. When this system is released from rest, the pulley turns through 1.30 rad in 91.0 ms and the acceleration of the blocks is constant. What are (a) the magnitude of the pulley's angular acceleration, (b) the magnitude of either block's acceleration, (c) string tension T1, and (d) string tension T1?

2. Homework Equations

The one with which I'm concerned: Torque = FR sin theta

3. The Attempt at a Solution

From drawing free-body diagrams, you get
m1g-T1=m1a1 and T2-f2=m2a2

The part I don't get is with getting the net torque on the pulley. The torque is 0 for both the normal force and the weight, that I do know. However, in the student solutions manual, they have the net torque as T1R - T2R = Torque. Why is it that way? Wouldn't the torque from T2 be 0 because the sin of the angle b/w the center of the pulley and the radius is sin0=0?

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#### alphysicist

Homework Helper
Hi craig22,

1. Homework Statement

For the picture, go to http://s16.photobucket.com/albums/b21/arbilvoldy/pulley.jpg" [Broken].
T1 is for the hanging string with mass m1, and T2 is for the string over the table with mass m2.

Two 6.20 kg blocks are connected by a massless string over a pulley of radius 2.40 cm and rotational inertia 7.40x10^-4 kgm^2. The string does not slip on the pulley; it is not known whether there is friction between the table and the sliding block; the pulley's axis is frictionless. When this system is released from rest, the pulley turns through 1.30 rad in 91.0 ms and the acceleration of the blocks is constant. What are (a) the magnitude of the pulley's angular acceleration, (b) the magnitude of either block's acceleration, (c) string tension T1, and (d) string tension T1?

2. Homework Equations

The one with which I'm concerned: Torque = FR sin theta

3. The Attempt at a Solution

From drawing free-body diagrams, you get
m1g-T1=m1a1 and T2-f2=m2a2

The part I don't get is with getting the net torque on the pulley. The torque is 0 for both the normal force and the weight, that I do know. However, in the student solutions manual, they have the net torque as T1R - T2R = Torque. Why is it that way? Wouldn't the torque from T2 be 0 because the sin of the angle b/w the center of the pulley and the radius is sin0=0?
Your drawing is not quite accurate and I think it is leading you in the wrong direction. The top part of the rope will leave the pulley at the highest point, like this:

http://img159.imageshack.us/img159/70/pulley.jpg [Broken]

(If that does not make sense, try it: wrap a string around something round and see how it behaves if one end come off horizontal and the other side of the string is vertical.) Do you see why the torque is not zero?

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#### jambaugh

Science Advisor
Gold Member
The pulley has an angular acceleration and rotational inertia. There must be a torque increasing the rate of rotation of the pulley.

Add this to alphysicist's improved diagram and comments.

#### craig22

I agree that when the system is released from rest, there's a torque on both T1 and T2. However, at rest, is that still true? If so, how? Sorry, this stuff is very new to me!

#### alphysicist

Homework Helper
I agree that when the system is released from rest, there's a torque on both T1 and T2.
Just to be careful, we would say there is a torque from T1 and T2 on the pulley.

However, at rest, is that still true? If so, how? Sorry, this stuff is very new to me!
The problem is asking about what happens after the system is released so it is not at rest, and the two tensions are applying a torque to the pulley.

(Before it is released, there is something keeping it from moving: there might be something holding the pulley in place, which would mean an extra torque involved, or something might be holding up the hanging block, so that the tension would be zero, etc. The details aren't given, but the point is you just need to figure out what happens after it is released.)

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