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Moment of inertia, tension, acceleration

  • Thread starter dinonichas
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  • #1
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A 10kg block is attached to a massless cord that doesn't stretch or sag which is rapped around a disc like wheel, m=5Kg, I= (1/2)mR^2. If the block is allowed to drop straight down held back only by the rotation of the wheel:

1. What is the tension of the rope?
2. What is the acceleration of the block?

I think that "I" is the moment of inertia of the disc, but I can't figure out how I can connect that with the fact that the falling block has to be decelerated by it and not reach the gravitational acceleration
 
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  • #2
Hootenanny
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Hi dinonichas and welcome to PF,

What have you attempted thus far? What are your thoughts?
 
  • #3
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Should i link the kinetic energy of the disc K.E=(1/2)Iw^2 with that of the falling block K.E=(1/2)mg^2 ? how can i do that?
 
  • #4
Hootenanny
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Should i link the kinetic energy of the disc K.E=(1/2)Iw^2 with that of the falling block K.E=(1/2)mg^2 ? how can i do that?
Since the question is asking for forces and accelerations, I think that it would be much better to consider the forces acting on the block and the wheel and link them to the accelerations (angular and linear).
 
  • #5
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the rate of change of angular momentum of disc = rate of change of linear momentum of block
d(Iw)/dt=dmv <=> (MR^2)/2 dw/dt = mg <=> dw/dt=2mg/MR^2
So I have the angular acceleration of the disc.
is the tension of the rope = mg ?
 
  • #6
Hootenanny
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the rate of change of angular momentum of disc = rate of change of linear momentum of block
d(Iw)/dt=dmv <=> (MR^2)/2 dw/dt = mg <=> dw/dt=2mg/MR^2
This isn't true in general.
is the tension of the rope = mg ?
No.

I suggest you start by considering the forces acting on the block and then use Newton's second law to determine the linear acceleration of the block.
 
  • #7
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Well, the block has gravitational force, but the tension of the rope is on the other direction, so the force acted on the wheel is mg - tension of rope.
so that force causes acceleration on the wheel. centripetal acceleration= (w^2).r
i am stuck. how will I use moment of inertia here, it is not connected to acceleration..
 
  • #8
Hootenanny
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Well, the block has gravitational force, but the tension of the rope is on the other direction, so the force acted on the wheel is mg - tension of rope.
Not quite, this is the net force acting on the block. The tension is the only force acting on the wheel. So, can you know use Newton's second law to write an expression for the acceleration of the block?
 
  • #9
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acceleration of the block= (mg-tension)/m
 
  • #10
Hootenanny
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acceleration of the block= (mg-tension)/m
Good. :approve:

Now, how does the torque exerted on a body relate to the angular acceleration produced?
 
  • #11
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torque = inertia * angular acceleration = 1/2 * m * R^2 * angular acceleration
 
  • #12
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Also torque=tesnsion*R
 
  • #13
Hootenanny
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torque = inertia * angular acceleration = 1/2 * m * R^2 * angular acceleration
Correct.

Next you need to determine the torque in terms of the tension and relate the angular acceleration of the wheel to the linear acceleration of the block. Note that the linear acceleration of the block is the same as the tangential acceleration of the wheel.

Then you should be done :biggrin:
 
  • #14
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and linear acceleration * R = angular acceleration

So linear acceleration= 2*tension/(a*r^3) ?
 
  • #15
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but what about the tension?
isn't it equal to m.g= 10*9.8=98N ??
 
  • #16
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is torque=tesnsion*R ???
 
  • #17
Hootenanny
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and linear acceleration * R = angular acceleration
Correct :approve:
So linear acceleration= 2*tension/(a*r^3) ?
Not quite sure where your getting this from.
but what about the tension?
isn't it equal to m.g= 10*9.8=98N ??
No, the tension isn't equal to the weight of the block. The tension is unknown as is the acceleration. You should be able to write down two equations involving the tension and acceleration and then solve for both.
 

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