GREPhysics.NET: GR8677 #97

GR8677 #97

Problem

GREPhysics.NET Official Solution

Mechanics $\Rightarrow$ }Conservation of Momentum

The rotational part of the angular momentum is $L=I\vec{\omega_0}$ . The translational part of the angular momentum is $L_{v_0}=m \vec{r} \times \vec{v_0}$ . (Note that, according to the diagram, this cross-product points in the other direction to the angular velocity.) Initially, the angular momentum about the point P is $L=\frac{1}{2}MR^2 \omega_0 - L_{v_0} = 0$ , since $v_0=\frac{1}{2}R^2 \omega_0^2$ and $\vec{L_{v_0}}=\vec{r}\times\vec{p}=-RMv_0=-\frac{1}{2}M\omega_0 R^2$ . QED.

Alternate Solutions

fred
2008-10-15 21:29:47

I think the answer sheet is right. At the instant the disks "collide" the moving disk has an angular momentum given by L = r x M $_{cm}$ v $_{0}$ , which gives L $_{v}$ = RMv $_{0}$ = 1/2 MR $^{2}$ $\omega\$ $_{o}$ .

But the disk also has an angular momentum due to it's rotation, which is L $\omega\$ = -1/2MR $^{2}$ .

Adding L $_{v}$ + L $\omega\$ = 0. Since angular moment is conserved, L= 0 before and after the collision at point P. So I think the answer is (A), which explains why (B) = (D).

Reply to this comment

Comments

fred
2008-10-15 21:29:47

Reply to this comment

Poop Loops
2008-09-23 20:02:33

My answer sheet says that the answer is A, which makes sense because if one is moving, the other stationary, and they are the same mass, they would each move at half of the speed that Disk I was moving at before the collision.

However, half the angular momentum of Disk 1 gets transfered to Disk 2, but in this case it is spinning in the opposite direction, giving the system zero total angular momentum. Since point P is right in between both disks, it also sees zero angular momentum.

rorytheherb
2008-10-11 13:41:17

yeah so what is the correct answer? my answer sheet says (A) zero. It does look like (B) = (D) so how is this a valid question... everyone seems to showing solutions that give (B) = (D) ....

Reply to this comment

neutrinosrule
2008-09-21 11:10:48

I am very confused... I had chosen D for my answer because that's the equation I had gotten to and I saw it in the answer choices... Now that I see that B is the right answer, I am having trouble seeing how B and D are not EXACTLY the same thing... the question DEFINES $v_{0}=\frac{1}{2}\omega_{0}R$ ...and if you just plug in this $\omega_{0}$ to choice B, you get choice D! What is the flaw in this logic? aren't $\omega_{0}, v_{0}$ , and R all constants? You yourself had $RMv_{0}$ before plugging in... did you just plug in for $v_{0}$ because you knew that the right answer was B or is there some reason why it is wrong to leave the answer like this? Is this possibly a mistake on the part of ETS?

Reply to this comment

u0455225
2008-06-14 16:10:17

It seems to me that answers (B) and (D) are the same thing, by plugging in $\left( v \right_{0}$ = $\frac{1}{2}$ $\left( w \right)_{0}$ R into (D). Doesn't this rule out both (B) and (D)?

Reply to this comment

grayza
2007-09-25 02:27:13

actually the parrelel axis theorem is needed. It's a trick question look at the arrows in the diagram. The translational angular momentum is in the opposite direction to the rotational angular momentum. The translational moment is $L_{com}= \frac{1}{2}Mr^2\omega \mapsto L_P = \frac{1}{2}Mr^2\omega+Mr^2\omega$ . The translational angular momentum is simply $MRv=MR^2\omega$ Therefore the total angular momentum is

$\frac{1}{2}Mr^2+ Mr^2-Mr^2=\frac{1}{2}MR^2$

Reply to this comment

comorado
2006-10-25 11:40:06

I think that there is an misprint, you wrote: $v_0 = 1/2 \Omega^2 R^2$ , must be $v_0 = 1/2 \Omega R$

Reply to this comment

Jung
2006-10-20 04:02:13

I have a trouble understanding this solution. From the text, we need to calculate the total angular mementum ABOUT THE POINT P, but I think the angular mementum of rataitional part, IWo, from your solution is not the POINT P, but about its certer. Is it because the moment of inertia I in the text is not about disk 1's center, but about the POINT P? But ETS does not mention about it. How do you think? I love this site! Thank you~

Jung
2006-10-20 04:29:53

Is the parallel-axis theorem not needed?

sharpstones
2007-03-20 15:01:48

I think the problem here is that the terms "rotational/translational angular momentum" are confusing and don't really explain what's going on. In general the angular momentum of a body is $\ L = L_{of center of mass} + L_{about center of mass}$ . The angular momentum about the center of mass is Iw, the angular momentum of the center of mass about P is MRv. That's how you get the answer.

Richard
2007-11-01 13:52:28

I want to second this comment.
It is absolutely wrong to concern yourself with the "rotational" angular momentum given the wording of the problem. Only consider the MOMENTUM ABOUT P:
$L=r\times p$
Just before the collision the linear momentum and the radius vector to $P$ are perpendicular:

$L=rp=rMv=\frac{1}{2}MR^2 \omega_0$

Choice (B).

Reply to this comment