Conservation of Angular momentum problem

  • #26
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Because we're discussing the angular momentum about point O, i.e. the top of the rod. How the particle came to be travelling horizontally at this speed is no longer of interest, so h plays no further part (beyond determining the speed).
shall i replace h with d?
 
  • #28
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Yes.
okay, then I have w=sqrt(mgh)/d
could u suggest a way to find the angle ø?(I haven't learnt this chapter of physics course yet, limited knowlege)
 
  • #29
haruspex
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then I have w=sqrt(mgh)/d
No. Reread your own post #24.
could u suggest a way to find the angle ø?
Reread the last lines of posts #4 and #17.
 
  • #30
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No. Reread your own post #24.

Reread the last lines of posts #4 and #17.
w=sqrt(2gh)/d

use energy conservation, so delta(GPE)=1/2Iw^2?
 
  • #31
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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
 
  • #34
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l is not equal to h
 
  • #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?
 
  • #37
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How did you get l=h?
 
  • #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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