- #31

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Right. The final challenges are to calculate I correctly for the rod plus block, and express the ΔGPE as a function of the angle.w=sqrt(2gh)/d

use energy conservation, so delta(GPE)=1/2Iw^2?

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- Thread starter i_hate_math
- Start date

- #31

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- 7,711

Right. The final challenges are to calculate I correctly for the rod plus block, and express the ΔGPE as a function of the angle.w=sqrt(2gh)/d

use energy conservation, so delta(GPE)=1/2Iw^2?

- #32

- 150

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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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soz w=sqrt(2gh)/d i meant

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

- #34

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l is not equal to h

- #35

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I mean after subing in all the values, i got l=h, or did i do the wrong calculations?l is not equal to h

- #36

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is h the original height and l the final height?

- #37

- 9

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How did you get l=h?

- #38

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L=(0.5I*w^2)/mgHow 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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After the collison, mechanical energy is conservedHmm... I don't think energy is conserved as it is an inelastic collision.

- #41

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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 Inow w=sqrt(2gh)/d

mgl=0.5Iw^2, where l is change in height

As I said, you next need to figure out the right value for I

- #42

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Since the block is treated as a particle, the Imom would be 1/3mw^2 right?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 I_{mom}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 I_{mom}. This is for the rod and block as a combined system. Think carefully.

- #43

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(1/3)(m+M)w^2 is what i meant.Since the block is treated as a particle, the Imom would be 1/3mw^2 right?

- #44

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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

- #45

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First, you don't mean w(1/3)(m+M)w^2 is what i meant.

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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Please do not spoon-feed @i_hate_math. He/she needs to get the hang of figuring these things out.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

- #47

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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.First, you don't mean w^{2}, 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?

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