How Do You Calculate Torque in a Pulley System?

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To calculate torque in a pulley system, the net torque equation is expressed as T2R - T1R = I alpha, where T2 and T1 are the tensions in the string on either side of the pulley. The direction of rotation affects the signs of the torques, with counterclockwise torque considered positive and clockwise negative. The angle theta is typically 90 degrees in this context, simplifying the torque equation. The discussion also clarifies that the heavier block influences the acceleration direction but does not change the fundamental torque equation. Understanding these principles is crucial for accurately solving problems involving pulleys and torque.
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Homework Statement


A block of mass m1 = 2kg and a block of mass m2 = 6kg are connected by a massless string over a pulley in the shape of a solid disk having radius R = 0.250m and M = 10kg. The fixed, wedge-shaped ramp makes an angle of 30 degrees. Coefficient of kinetic friction is .360 for both blocks. Find acceleration of the two blocks.

http://img685.imageshack.us/i/img0230oo.jpg/

Homework Equations


Net torque = I alpha
F = ma
I= 1/2 mr^2
a =r alpha

The Attempt at a Solution


I isolated the objects (3)

equation for object 1
T1-ukm1g= m1a
m1g-FN= 0

equation for object 2
m2gsintheta - T2ukm2gcostheta = m2a
FN-m2gcostheta = 0

TORQUE equation for disk
T2 R sintheta - T1 R sintheta = I alpha
I of a disk is 1/2 Mr^2 and i can also rewrite alpha as a/r so
T2 R sintheta - T1R sintheta = 1/2 Mr^2 a/r
all the R's cancel each other and theta is 90 so.
T2 - T1 = 1/2MaWhen i set all the three equation together i got the right answer of .31 m/s^2. My question is though, for the TORQUE equation of the disk why is it
T2R sin theta - T1R sin Theta = I alpha instead of
T1R sin theta - T2R sin theta - I alpha?
I don't believe it is because the system is accelerating down towards the second object since the 2nd object is heavier than the 1st(i know this is true for my force equation) Doesn't force have nothing to do with direction of torque? If you have a pulley system with two objects hanging on the sides for example.
If I call the mass on the left m1 and right m2. Is the net torque always going to be TR of right object - TR of left object? does it matter which object is heavier? this doesn't seem right.

The left object is making the disc rotate counter clock wise while the right object rotates the disk clockwise so why is it NOT TR of left object - TR of right object? isn't clockwise negative and counterclockwise positive?

also is theta always going to be 90 in case of a pulley system? since the disk has to spin so it has to be 90?

THANK YOU.
 
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RU2490 said:
My question is though, for the TORQUE equation of the disk why is it
T2R sin theta - T1R sin Theta = I alpha instead of
T1R sin theta - T2R sin theta - I alpha?

I would say that this is the right equation:T2R sin - T1R sin = I alpha .

The disk is rotating clockwise, so the torque "I alpha" is directed counterclockwise. It's similar to the reactional force m*a which is directed opposite from the acceleration.
 
method_man said:
I would say that this is the right equation:T2R sin - T1R sin = I alpha .

The disk is rotating clockwise, so the torque "I alpha" is directed counterclockwise. It's similar to the reactional force m*a which is directed opposite from the acceleration.

What is sin?

it will only be

T2R - T1R = Iα
 
cupid.callin said:
What is sin?

it will only be

T2R - T1R = Iα

Formula for torque is f r sin theta. In this case Theta is 90 so it becomes t2-t1= I alpha
 
cupid.callin said:
What is sin?

it will only be

T2R - T1R = Iα

I forgot to erase that when I quoted. My bad.
 
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