# Pulley Problem: Find acceleration of block

by sona1177
Tags: acceleration, block, pulley
 P: 173 1. The problem statement, all variables and given/known data A 1.5 kg block and a 2.5 kg block are attached to opposite ends of a light rope. The rope hangs over a solid, frictionless pulley that is 30 cm in diameter and has a mass of .75 kg. When the blocks are released, what is the acceleration of the lighter rock? 2. Relevant equations alpha= net torque/moment of intertia F=ma 3. The attempt at a solution I know Newton's law for the 1.5 kg block is T-W=ma and for the 2.5 kg block is T-W=-ma now the problem is the pulley. How do I calculate this torque with the signs correctly? Won't this vary depending on how I draw the diagram? Since the pulley has mass, the tensions on the both sides of the rope are not equal. Can someone please help me? I don't know how to go about finding the torque for the pulley. 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution
 PF Gold P: 712 Why don't you make the rope as your axis? That's how I always approached pulley problems. Linear acceleration to the right is positive, while to the left is negative. For torque, counterclockwise is positive alpha, clockwise is negative alpha.
P: 173
 Quote by thrill3rnit3 Why don't you make the rope as your axis? That's how I always approached pulley problems. Linear acceleration to the right is positive, while to the left is negative.
I am afraid I don't understand what you mean by "make the rope as your axis". I always thought a force that makes the pulley go counter clockwise is positive but clockwise is negative. Your help is greatly appreciated :)

PF Gold
P: 712
Pulley Problem: Find acceleration of block

Your linear acceleration eq is wrong

 1.5 kg block is T-W=ma and for the 2.5 kg block is T-W=-ma
so would you have ma=-ma?

Check your diagram. I was just suggesting another way of setting up the axes, but it's totally up to you.
P: 173
 Quote by thrill3rnit3 Your linear acceleration eq is wrong so would you have ma=-ma? Check your diagram. I was just suggesting another way of setting up the axes, but it's totally up to you.
Why is it wrong? The 2.5 kg block is accelerating down so therefore I thought its acceleration is negative.
 PF Gold P: 712 edit: nevermind everything I just said. I thought T stood for the same unknown.
 PF Gold P: 712 Anyways, just continue with what you were doing. Counter clockwise rotation produces positive torque, clockwise rotation produces negative torque. and a=alpha(radius), so that should give you the link between alpha and a
 P: 173 That is my question I don't know the signs of the torques. Are they both clockwise ?
 PF Gold P: 712 Make an assumption. Do you suppose that the rope will slide left or right? After you make the initial assumption, just make sure everything else remains consistent. You won't make a mistake if everything that follows is consistent. If you get a negative answer, then that means it's actually sliding the other way.
 P: 173 Ok I changed my mind so that the 2.5 kg block is on the left and the 1.5 kg block is on the right. This means the rope Is moving towards the right (counter clockwise). This means: Net torque= T1r + T2r where capital T1 stands for the tension of the rope attached to the 2.5 kg block and T2 for the tension of the rope attached to the 1.5 kg block. And I=.5MR^2 So alpha = 2 (T1 + T2)/MR where capital M stands for the mass of the pulley. This means linear acceleration is alpha * r therefore it is = 2(T1 + T2)/M and by Newton's second law since T1= -m1a + w1 then a = 2(-m1a + w1 + T2)/M so T2=(Ma/2) + m1a -w1. Plugging this into T2-w2=m2a for T2 and doing algebra gives 28.5 m/s. This is wrong :( can someone please help?
PF Gold
P: 712
 Quote by sona1177 Net torque= T1r + T2r
The right rope is torquing it clockwise [ negative ], the left rope is torquing it counterclockwise [ positive ], and you assumed that the system is moving right -> clockwise -> negative alpha.

-T1r+T2r= - Ialpha

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