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Pulley Problem: Find acceleration of block 
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#1
Aug510, 10:51 PM

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 TW=ma and for the 2.5 kg block is TW=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 


#2
Aug510, 11:07 PM

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. 


#3
Aug510, 11:10 PM

P: 173




#4
Aug510, 11:17 PM

PF Gold
P: 712

Pulley Problem: Find acceleration of block
Your linear acceleration eq is wrong
Check your diagram. I was just suggesting another way of setting up the axes, but it's totally up to you. 


#5
Aug510, 11:23 PM

P: 173




#6
Aug510, 11:26 PM

PF Gold
P: 712

edit: nevermind everything I just said. I thought T stood for the same unknown.



#7
Aug510, 11:29 PM

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 


#8
Aug610, 12:10 AM

P: 173

That is my question I don't know the signs of the torques. Are they both clockwise ?



#9
Aug610, 12:14 AM

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. 


#10
Aug610, 12:57 AM

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 T2w2=m2a for T2 and doing algebra gives 28.5 m/s. This is wrong :( can someone please help? 


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