GR9277 #78


Problem


\prob{78}
One ice skater of mass m moves with speed 2v to the right, while another of the same mass m moves with speed v toward the left, as shown in Figure I. Their paths are separated by a distance b. At t=0, when they are both at x=0, they grasp a pole of length b and negligible mass. For t>0, consider the system as a rigid body of two masses m separated by distance b, as shown in Figure II. Which of the following is the correct formula for the motion after t=0 of the skator initially at y=b/2?






Mechanics}Multiple Particles
The angular momentum equation gives the angular frequency. , which relates the angular momentum to the moment of inertia, the angular velocity, the radius of gyration and the linear velocity.
The system spins about its center of mass, which is conserved. Since the pole is massless and the skaters are off the same mass, . The moment of inertia of the system is just .
Thus, the angular momentum equation gives , since the crossproducts point in the same direction. Now that one has the angular velocity, one eliminates all but choices (B) and (C).
Now, take the timederivative of x for choices (B) and (C), then evaluate it at .
For B, one has for .
For C, one has for .
Since the top skater is initially at , only choice (C) has the right initial condition. Choose choice (C).
(FYI: The center of mass velocity is given by . One can also arrive at (C) by noting conservation of center of mass velocity, since there is no net force.)


Alternate Solutions 
Niko 20101111 21:47:06  Throw out E because you know the thing is going to rotate.
One should notice that at t=0 the xcomponent of the velocity should be 2v.
Take the time derivative of all of the answers (shouldn't take more than 30s) and realize that only c satisfies.
Good rule of thumb:
When you have to choose the correct relation from a set of equations always look at the limits and initial conditions. Applying this and dimension analysis usually eliminates 2 or more of the choices.   flyboy621 20101022 20:58:00  There are 2 conserved quantities here: linear momentum and angular momentum.
Conservation of linear momentum tells us that the CM moves at the same velocity after the "collision" as before, namely
so it must be (C) or (E). To narrow it down, we need to find the frequency . Conservation of angular momentum (about the center of the rod) says
But
which gives , as in answer (C).  

Comments 
livieratos 20111109 04:38:54  can someone explain why I = 2m(b/2)^2 please?
neo55378008 20120818 11:44:18 
I = by definition

netphysics 20121101 19:02:11 
and it is the moment of inertia of c.m.

  Niko 20101111 21:47:06  Throw out E because you know the thing is going to rotate.
One should notice that at t=0 the xcomponent of the velocity should be 2v.
Take the time derivative of all of the answers (shouldn't take more than 30s) and realize that only c satisfies.
Good rule of thumb:
When you have to choose the correct relation from a set of equations always look at the limits and initial conditions. Applying this and dimension analysis usually eliminates 2 or more of the choices.
kap09c 20131017 18:33:09 
I think you mean A, and no, it should be .5 v ...

NoPhysicist3 20170323 13:29:00 
yes, he should have said \"throw out A\"\r\nwhy would it be o.5? It is 2.

  Niko 20101111 21:46:27  Throw out E because you know the thing is going to rotate.
One should notice that at t=0 the xcomponent of the velocity should be 2v.
Take the time derivative of all of the answers (shouldn't take more than 30s) and realize that only c satisfies.
Good rule of thumb:
When you have to choose the correct relation from a set of equations always look at the limits and initial conditions. Applying this and dimension analysis usually eliminates 2 or more of the choices.   flyboy621 20101022 20:58:00  There are 2 conserved quantities here: linear momentum and angular momentum.
Conservation of linear momentum tells us that the CM moves at the same velocity after the "collision" as before, namely
so it must be (C) or (E). To narrow it down, we need to find the frequency . Conservation of angular momentum (about the center of the rod) says
But
which gives , as in answer (C).
dc771957 20161018 23:36:59 
Hey, I think you dropped your \'m\' terms in where you first calculated the angular momentum as r x p. You kept the \'m\'s in your moment of inertia term, and you need them in the angular momentum jaunt to cancel out. \r\n\r\nGreat solution btw.

  wikiwert 20100930 07:36:30  I am afraid the GRE solution is wrong. The w in yosunīs solution is the angular velocity "with respect to the center of mass". It is the angular speed about the rotation axis of a rigid body ( L = I * w with I the moment of inertia tensor). But some may not be very familiarized with this Classical Mechanics result; so just consider the center of mass (CM) frame.
Velocities are 3/2 * v and  v/2 in the CM frame. In this system there is only rotation (the CM does not move); this should make clear that the following result will be correct. Angular momentum is conserved and it is initially 3/2 *v * m * b/2 + 1/2 * v * m * b/2 = m * b * v. And in this system (CM does not move) is that we have L = I * w. As I = 2 * m * (b/2)^2, we have that w = 2 * v/b.
Transforming to the original system, a v/2 * t term has to be added to the x ( t ) term.
wikiwert 20101001 19:16:44 
Ignore the previous comment. I donīt know if any body is modertaing the comments any longer...but please erase the previous one. I thought it was very strange that the GRE solution was wrong...but it wasnīt until I told somebody how I solved the problem, that I realized I calculated v+v/2 and not v v/2 : P. Sorry.

  Ning Bao 20080130 07:31:32  I don't think it needs to be that complicated: this is more of a boundary values problem.
Taking derivatives of everything immediately eliminates all but A and C, as v(0)=2v in the x direction. A fails because there is no reason for the ycoordinate to stay fixed. Thus, C.
flyboy621 20101022 18:31:51 
Sorry, but I don't think this comment makes sense at all, except that it's true that y can't be constant.

liliapunto 20130715 15:44:46 
Ning makes total sense and this is the easiest method.

  mygoldfleece 20071101 02:44:07  Why does the energy conserve?
Instead of energy conservation, I use angular momentum conservation (with reference point at (0,b/2))   lambda 20071021 21:46:06  I did it a simpler way than the official solution, although not as quick as Richard's solution.
The center of mass of the two masses will travel to the right with a velocity through conservation of linear momentum (the same way Richard got it). So the expression for will have a term.
The angular velocity is calculated through conservation of energy:
With more math,
,
where the moment of inertia .   Richard 20070916 22:53:32  This problem doesn't have to be complicated.
You need to know two things: the linear velocity and the angular velocity.
The linear velocity is easily obtained by conservation of momentum:
The angular velocity is almost as easy. Note that
The tangential velocity is the sum, , since both contribute to rotation acting at the poles in opposite directions:
and are out of phase by .
The answer is C.
lambda 20071021 21:28:08 
I'm not really seeing why the tangential velocity of the two masses together is equal to the sum of their original linear velocities. Why can they be added this way?

sblmstyl 20080923 16:16:11 
shouldn't = ?

Albert 20091021 11:21:44 
Yes, V(tangential) should be (b/2)*w since the radius is b/2 not b.

cbb6x2 20101108 22:42:43 
Why doesn't it work when you use velocity=b/2*omega? You would get (E) as the answer using this method

 

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