## Question 97, 8677

AriAstronomer
Posts: 76
Joined: Thu May 12, 2011 4:53 pm

### Question 97, 8677

Hey everyone,
So on the 8677 test, question 97, looking at the solutions posted on grephysics.net website there seems to be massive disagreement. Some (including yosun, the creator of the site) think it's B, while ETS and a few others apparently think its A. Can anyone settle this once and for all?

Ari

HappyQuark
Posts: 762
Joined: Thu Apr 16, 2009 2:08 am

### Re: Question 97, 8677

AriAstronomer wrote:Hey everyone,
So on the 8677 test, question 97, looking at the solutions posted on grephysics.net website there seems to be massive disagreement. Some (including yosun, the creator of the site) think it's B, while ETS and a few others apparently think its A. Can anyone settle this once and for all?

Ari

I'm pretty sure it's (A)

bfollinprm
Posts: 1203
Joined: Sat Nov 07, 2009 11:44 am

### Re: Question 97, 8677

It's zero. it's easy to see if you consider any point before the collision. Since momentum is conserved through any collision (as long as you include all objects in your system), the total momentum then is equal to the momentum after. Taking both objects as a single system, the angular momentum is given by the rotational component of disk 1 + translational component of disk 1 moving in reference to the point p (disk 2 does not move with respect to point p, so it doesn't matter). The first term is straightforwardly $1/2 m R^2 \omega$, and the second term is $m r \times v= m (R)(1/2 \omega R)$. These are equal, and considering the system shows they're opposite (I guess you could also keep track of signs from the dot and cross products, but i never do).

EDIT: should say the fastest way to do this question is to eliminate B, C, and E (b and c are equivalent, and E makes no sense). Then A is less than $I \omega$ and D is more. Since the disk is revolving opposite its motion we know it's less, so the answer must be A.

AriAstronomer
Posts: 76
Joined: Thu May 12, 2011 4:53 pm

### Re: Question 97, 8677

thanks very much guys!

aby
Posts: 9
Joined: Fri Sep 30, 2011 6:41 am

### Re: Question 97, 8677

there's a problem here. the moment of inertia I given for the first disk is relative to it's c.m , so relative to point p, using the parallel axis theorem its moment of inertia is: I=(3/2)MR^2. then when adding the transitional part we get a non-zero angular momentum...

aby
Posts: 9
Joined: Fri Sep 30, 2011 6:41 am

### Re: Question 97, 8677

pls, does anybody know what's up with this question. like i said, when we take the moment of inertia about point p , we get eventually a non zero angular momentum, so the answer shouldn't be (B).

bfollinprm
Posts: 1203
Joined: Sat Nov 07, 2009 11:44 am

### Re: Question 97, 8677

aby wrote:pls, does anybody know what's up with this question. like i said, when we take the moment of inertia about point p , we get eventually a non zero angular momentum, so the answer shouldn't be (B).

The answer is (A). Anything else is a misprint or someone overthinking the problem. It's conservation of (angular) momentum--no torques on the system; consider $\vec{L}$ of the initial state.

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