GREPhysics.NET: GR8677 #44

GR8677 #44

Problem

GREPhysics.NET Official Solution

Alternate Solutions

Mechanics $\Rightarrow$ }Conservation of Momentum

One could use energy, but then one would have to take into account the inertia. Momentum might be easier,

$(p_i=mv) = (p_f=MV)\Rightarrow V=\frac{m}{M}v,<br />$

where the final momentum takes into account the fact that the final velocity of the particle is at rest (0). And, so it is (A)!

Alternate Solutions

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Comments

eliasds
2008-11-02 14:36:44

I'm still confused. the equation $\(p_{i})$ =MV in no way shows the linear momentum of the particle contributing to the new angular momentum of the stick. Can someone clearly explain why angular momentum does not show up in the equation?

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phys2718
2008-10-16 13:54:11

I think the crucial point here is that the rod is a rigid body; thus all parts of the rod rotate about the center with the same frequency, and the linear momentum of one side of the rod due to the rotation is completely canceled by the momentum of the other side. So only the linear momentum of the center of mass needs to be taken into account.

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sabinscabin
2008-10-15 21:51:31

I kept seeing the word "stick" in the problem; it is mentioned 4 times. So I thought that the mass "sticks" to the rod, which made me pick B.

D'oh!!!!!

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Rune
2007-10-13 23:05:52

I'm a bit confused because it seems to me that if the surface is frictionless and you hit a rod at one end, some of the momentum would go to rotating the rod. Why doesn't this happen?

jsdillon
2008-04-11 19:48:56

It does rotate. However, since the problem is asking for the velocity of the center of mass of the stick, it doesn't matter. The solution of the problem doesn't require a consideration of the conservation of angular momentum (although it is, indeed, conserved)--conservation of linear momentum is plenty.

p3ace
2008-05-04 14:34:50

Only linear momentum for center-of-mass coordinates is needed.

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grscjo3
2006-11-03 21:01:37

When you say that by using energy to solve this, one would have to take into account the inertia, do you mean the rotational inertia of the rod?

yosun
2007-02-22 15:11:24

Yes, referring to this kind of inertia: $I=\int dm r^2$

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I kept seeing the word "stick" in the problem; it is mentioned 4 times. So I thought that the mass "sticks" to the rod, which made me pick B. D'oh!!!!!

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LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces $\alpha^2_0$ .
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$\langle my \rangle$	$\langle my \rangle$
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