|
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?
-
-
, \;\;y=0.5b\cos(3vt/b)$)
-
, \;\;y=0.5b\cos(3vt/b)$)
-
, \;\;y=0.5b\cos(6vt/b)$)
-
, \;\;y=0.5b\cos(6vt/b)$)
|
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 ^2) .
Thus, the angular momentum equation gives ^2 \omega =m b/2 (v + 2v) \Rightarrow 2(b/2) \omega = 3v \Rightarrow \omega = 3v/b) , since the cross-products point in the same direction. Now that one has the angular velocity, one eliminates all but choices (B) and (C).
Now, take the time-derivative of x for choices (B) and (C), then evaluate it at  .
For B, one has  \rightarrow 2.5v) for .
For C, one has  \rightarrow 2v) for .
Since the top skater is initially at =2v) , 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 |
| There are no Alternate Solutions for this problem. Be the first to post one! |
|
|
Comments |
Ning Bao
2008-01-30 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 y-coordinate to stay fixed. Thus, C. |  | mygoldfleece
2007-11-01 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
2007-10-21 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,
^{2}+\frac{1}{2}mv^{2}=\frac{1}{2}I\omega^{2}+\frac{1}{2}(2m)(\frac{v}{2})^{2}) ,
where the moment of inertia ^{2}) . | | Richard
2007-09-16 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
2007-10-21 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
2008-09-23 16:16:11 |
shouldn't  =  ?
|
| |
|
| Post A Comment! |
|
|
Bare Basic LaTeX Rosetta Stone
|
LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces  .
|
| type this... |
to get... |
| $\int_0^\infty$ |
 |
| $\partial$ |
 |
| $\Rightarrow$ |
|
| $\ddot{x},\dot{x}$ |
 |
| $\sqrt{z}$ |
 |
| $\langle my \rangle$ |
 |
| $\left( abacadabra \right)_{me}$ |
_{me}) |
| $\vec{E}$ |
 |
| $\frac{a}{b}$ |
 |
|
|
|
|
The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it... |
|
|
|
