Torque, Rotation, Pulley problem

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  • #1
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1. A 2.5 kg block and a 1.5 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 block.

Ok, the book says 2.2m/s^2, but i cannot come to that, and the book has been wrong in the past. I'm pretty sure the acceleration on the block will be the same as the tangential acceleration of the rope over the pulley, which is angular acceleration*r



2. a=angular acceleration/r
angular accereration=Net Torque/Moment of Inertia(I)
Net Torque=mass of b1*gr-mass of b2*gr
I=1/2(mass of pulley)(r)^2






3. Ok here we go.

I=1/2(.75kg)(.15)^2
=0084375kg*m^2

Net Torque=2.5kg(9.8m/s^2)(.15m)-1.5kg(9.8m/s^2)(.15m)
=3.675N*m-2.205N*m
=1.47N*m

angular acceleration= 1.47/.0084375
=174.22rad/s^2

a=174.22rad/s^2(.15m)
=26.133m/s^2

Who's right and who's wrong, or are both of us wrong?
 

Answers and Replies

  • #2
Doc Al
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Net Torque=mass of b1*gr-mass of b2*gr
This is incorrect. The net torque on the pulley would be due to the tensions in the rope, not the weight of the blocks. (The tension in the rope does not equal the weight of the blocks.)
 
  • #3
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But assumimg the rope is massless and the pulley is frictionless, wouldn't the tension in the rope be equal to the weight of the blocks?
 
  • #4
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But assumimg the rope is massless and the pulley is frictionless, wouldn't the tension in the rope be equal to the weight of the blocks?

that would be assuming it is in equilibrium. But there is acceleration.
 
  • #5
Doc Al
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But assumimg the rope is massless and the pulley is frictionless, wouldn't the tension in the rope be equal to the weight of the blocks?
No. As azicker93 points out, if that were true the blocks would be in equilibrium.

Perhaps you're thinking of when you have a massless rope and a frictionless and massless pulley. In that case, the tension would be uniform throughout the rope--the same on both sides of the pulley. But that's not relevant here.
 
  • #6
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Ok so the tensions in each side are different. But are the accelerations of both blocks the same?
 
  • #7
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Ok here is my new approach.

alpha=angular acceleration
r-radius
T=Tension
M=mass of pulley
m1=1.5 kg block
m2=2.5 kg block
a=acceleration of blocks
alpha=Net Torque/I
=T1r-T2r/.5Mr^2
=r(T1-T2)/(1/2)Mr^2
alpha =2(T1-T2)/Mr

acceleration is in the negative direction since the heavier block is moving down
a= -alpha*r
= -r*[2(T1-T2)/Mr]
a= -2(T1-T2)/M

I rearrange the above equation to get:
T1-T2= -(1/2)Ma
Phew, ok, I do Newton's second law for
each block and get:
m1a=T1-m1g
m1[-2(T1-T2)/M]=T1-m1g
and
m2a=T2-m2g
m2[-2(T1-T2)/M]=T2-m2g

Now i use substitution with these 3 equations to solve for T1, T2, and a.

T1-T2= -(1/2)Ma
m1[-2(T1-T2)/M]=T1-m1g
m2[-2(T1-T2)/M]=T2-m2g

I come out with a= -15.68, T1= -8.82, and T2= -14.7

And -15.68 m/s^2 doesn't agree with the books answer of 2.2 m/s^2

Where did i go wrong this time? I have a feeling it may have something to do with both blocks are accelerating in different directions.
 
  • #8
Doc Al
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I think you're messing up with signs.
Ok here is my new approach.

alpha=angular acceleration
r-radius
T=Tension
M=mass of pulley
m1=1.5 kg block
m2=2.5 kg block
a=acceleration of blocks
alpha=Net Torque/I
=T1r-T2r/.5Mr^2
=r(T1-T2)/(1/2)Mr^2
alpha =2(T1-T2)/Mr
Note that since T2 > T1, this makes alpha negative. Not a good idea. Let alpha be the magnitude of the acceleration, and write it with T2 - T1, instead of T1 - T2.

acceleration is in the negative direction since the heavier block is moving down
a= -alpha*r
= -r*[2(T1-T2)/Mr]
a= -2(T1-T2)/M
Let 'a' stand for the magnitude of the acceleration. Then the acceleration of m_2 will be -a (using the standard convention of up = positive) and the acceleration of m_1 will be +a.

Redo your equations with that in mind.
Where did i go wrong this time? I have a feeling it may have something to do with both blocks are accelerating in different directions.
As I point out above, I believe you made mistakes with the sign of the acceleration.

(Note: If you ever get a negative value for tension, you know you made an error somewhere.)
 
  • #9
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Ok that worked. Now, is there an easier way to solve this?
 

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