Conservation of Angular momentum problem

  • #31
haruspex
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w=sqrt(2gh)/d

use energy conservation, so delta(GPE)=1/2Iw^2?
Right. The final challenges are to calculate I correctly for the rod plus block, and express the ΔGPE as a function of the angle.
 
  • #32
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okay,
now w=sqrt(mgh)/d
mgl=0.5Iw^2, where l is change in height
and by subing in all the mess l=h
and then ø=arccos[(d-h)/d]=36.8
 
  • #33
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okay,
now w=sqrt(mgh)/d
mgl=0.5Iw^2, where l is change in height
and by subing in all the mess l=h
and then ø=arccos[(d-h)/d]=36.8
soz w=sqrt(2gh)/d i meant
 
  • #35
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l is not equal to h
I mean after subing in all the values, i got l=h, or did i do the wrong calculations?
 
  • #36
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is h the original height and l the final height?
 
  • #39
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Hmm... I don't think energy is conserved as it is an inelastic collision.
 
  • #40
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Hmm... I don't think energy is conserved as it is an inelastic collision.
After the collison, mechanical energy is conserved
 
  • #41
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now w=sqrt(2gh)/d
mgl=0.5Iw^2, where l is change in height
Right, though we will have to be careful not to confuse l (lower case L,)with I (uppercase I). To avoid confusion I'll write H for the change in height and Imom for moment of inertia. Also, the m in there should be for the mass of the rod plus block, not just the block, and H is the change in height of the mass centre of that combination. We will return to that later.

As I said, you next need to figure out the right value for Imom. This is for the rod and block as a combined system. Think carefully.
 
  • #42
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Right, though we will have to be careful not to confuse l (lower case L,)with I (uppercase I). To avoid confusion I'll write H for the change in height and Imom for moment of inertia. Also, the m in there should be for the mass of the rod plus block, not just the block, and H is the change in height of the mass centre of that combination. We will return to that later.

As I said, you next need to figure out the right value for Imom. This is for the rod and block as a combined system. Think carefully.
Since the block is treated as a particle, the Imom would be 1/3mw^2 right?
 
  • #43
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Since the block is treated as a particle, the Imom would be 1/3mw^2 right?
(1/3)(m+M)w^2 is what i meant.
 
  • #44
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Let ω be the instantaneous angular velocity of the rod as well as the stuck block about the pivot just after impact.
Te expression for angular momentum of rod + block just after impact {(Md²/3) + (md²)}ω. Equating this with initial angular momentum of the block just before impact gives you ω
To find θ, find the change in potential energy of the rod and the stuck block, when the rod reaches the angle θ and momentarily come to rest. This then is ro be equated with initial rotational KE of the of the block using formula 0.5Iω², where I = {(M/3)+m}d². { Note I(rod) = M(d²/3) and that of stuck small block = md²}
 
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  • #45
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(1/3)(m+M)w^2 is what i meant.
First, you don't mean w2, I hope. A moment of inertia is a mass multiplied by the square of a distance.
Secondly, you need to consider each body separately and add up their moments. What did you find for the moment of the block about O in post #26?
 
  • #46
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Let ω be the instantaneous angular velocity of the rod as well as the stuck block about the pivot just after impact.
Te expression for angular momentum of rod + block just after impact {(Md²/3) + (md²)}ω. Equating this with initial angular momentum of the block just before impact gives you ω
To find θ, find the change in potential energy of the rod and the stuck block, when the rod reaches the angle θ and momentarily come to rest. This then is ro be equated with initial rotational KE of the of the block using formula 0.5Iω², where I = {(M/3)+m}d²
Please do not spoon-feed @i_hate_math. He/she needs to get the hang of figuring these things out.
 
  • #47
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First, you don't mean w2, I hope. A moment of inertia is a mass multiplied by the square of a distance.
Secondly, you need to consider each body separately and add up their moments. What did you find for the moment of the block about O in post #26?
Yeah i didnt. I think i know where i went wrong. Its MoI again, nasty little bugger, the MoI of a point is simply md^2, and i forgot to add this into my calculation.
 

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