# Non-massless pulley

1. Mar 7, 2006

### nahya

A block of mass m1 = 1 kg rests on a table with which it has a coefficient of friction µ = 0.55. A string attached to the block passes over a pulley to a block of mass m3 = 3 kg. The pulley is a uniform disk of mass m2 = 0.7 kg and radius 15 cm. As the mass m3 falls, the string does not slip on the pulley.

With what acceleration does the mass m3 fall?
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because of the constraint that the string does not slip on the pulley, the angular acceleration of the pulley is equal to the acceleration at which mass m3 falls, right?
alpha = torque / I
by definition, I in this case is 1/2 * m2 * R^2,
torque = I * alpha = I * a, and
T + µ*m1 - m3g = m3a, right?
but how do i find out what the tension would be?
wouldn't i need the acceleration to find the tension?
but what i'm trying to find out is the acceleration... bleh.
the next two questions ask about the horizontal tension and the vertical tension (of the rope). so i'm guessing the two tensions are different...
now the problem just got more confusing.

am i even heading to the right direction?
thanks.

2. Mar 8, 2006

### Andrew Mason

This net force is supplied by the difference in tension in the rope across the pulley. The tension in the rope between the pulley and m1 supplies the acceleration to m1 and overcomes the force of friction as well. The additional tension in the rope between m3 and the pulley provides the torque to the pulley.

So express the acceleration of each part in terms of the tensions in the rope. Work that out to find a.

AM