Rotational Work and Energy

  • Thread starter Torater
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  • #1
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A 5.0kg mass is initally held at rest, attached to a disk shaped 8.0kg pulley of radius 12.0cm by a massless cord. Using Newton's Laws, find the tension in the cord and the acceleration of the mass falling after it is released.

Please help
 

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  • #2
tiny-tim
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Hi Torater! :smile:

(btw, pleeeeease don't give two threads the same title! :redface:)
A 5.0kg mass is initally held at rest, attached to a disk shaped 8.0kg pulley of radius 12.0cm by a massless cord. Using Newton's Laws, find the tension in the cord and the acceleration of the mass falling after it is released.

Please help
Instead of a rolling condition, you'll have an un-rolling condition (but it'll be the same equation :wink:)…

show us what you get. :smile:
 
  • #3
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wouldn't there NOT be a X direction for this problem?

Giving me:
∑τ=Iα = Fr
∑Fy=-m2a = T-m2g

I= 1/2mr²

OR if I had an X equation it would be:
∑Fx=m1a = T-F

Therefore 1/2m1r²(a/r)= (m2(g-a) - m1a) (r)
1/2(8)(.12)²(a/.12) = (5(9.81-a)) (.12)

2.48a=5.886
a= 2.34m/s²

T= (5(9.81-2.34))
T= 37.4N

Is this the right approach?
 
  • #4
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I'm beginning to think we are in the same school Torater.. (Macewan?)

BTW, yeah I don't believe there would be an x direction because that would include friction which arises from a rolling sphere and in this case it's an unrolling condition.

I'm happy we've made it through these problems together hahah
 
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  • #5
tiny-tim
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Hi Torater! :smile:
wouldn't there NOT be a X direction for this problem?
That's right … the pulley doesn't move, it only rotates, so you don't have an F = ma for the pulley.
Therefore 1/2m1r²(a/r)= (m2(g-a) - m1a) (r)
1/2(8)(.12)²(a/.12) = (5(9.81-a)) (.12)
First, you could save yourself a lot of trouble by noticing that every term in the first equation is a multiple of r, so you can just ignore r, instead of remembering which power of .12 to multiply by. :wink:

(this isn't a conicidence, it happens in every question of this type)

Second, I can't quite follow what you've done, but I think you didn't move the a (from g-a) over to the LHS before solving for a. :confused:
 
  • #6
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Ok lets try this again:



∑τ=Iα = Tr
∑Fy=-m2a = T-m2g

I= 1/2mr²


Therefore 1/2m1r²(a/r)= m2(g-a) (r)
1/2m1 a = m2(9.81-a)

1/2 (8) (a) = 5(9.81-a)

a= 5.45m/s²

T= (5(9.81-5.45))
T= 21.8N

???


I hope your not in my class CHOD!! lol
 
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  • #7
tiny-tim
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Hi Torater! :wink:

Yes that's fine. :wink:

(note that, using gneill's trick in the https://www.physicsforums.com/showthread.php?p=3033838&posted=1#post3033838", you can say that the effective mass of the pulley is its "rolling mass", 8/2, so the whole thing has mass 5 + 8/2 = 9, and force 5g, so using F = ma you get a = F/m = 5g/9 = 5.45 … which is how I checked it! … though perhaps the question doesn't allow that short-cut:smile:)
 
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