25th February 2008 - 05:04 AM
You are right that there must be a force on the base of the pole when the cart stops. Otherwise, the pole would continue moving forward with no change in its motion, which contradicts the fact that the hinge must stop. To stop the bottom end of the pole, there must be a horizontal impact force directed backward along the cart's path. This force is exerted by the hinge on the bottom of the pole.
Now since there IS a force on the pole, its linear momentum is not conserved. This force will also exert a torque around the center of the pole, so its angular momentum around the center is not conserved either, which makes it seem as if the problem will be hard to solve. It can be solved based on four facts:
1: the base of the pole stops,
2: the speed of the center = (angular velocity) * (1/2 of length),
3: (the applied force) * time = mass * (change of center velocity), and
4: (the applied force) * (1/2 of length) * time = mass/12 * (length)^2 * (change of angular velocity).
Fact 1 is given, fact 2 is the kinematic relation v = v_axis + omega*r for rotational motion (where the center of mass is of course at a distance length/2 from the base), fact 3 is the impulse momentum theorem, and fact 4 is the angular impulse angular momentum theorem, where the moment of intertia of the pole is (mL^2)/12.
By fact 1, v_axis = 0, so by fact 2, v_c = -wL/2, where v_c is the center of mass velocity, w is the angular speed, and L is the length. The minus sign is because a negative (i.e. clockwise) angular velocity will give a forward ( i.e. positive) velocity for the center of mass. By fact 3, Ft = m(v_c - v_0), where v_0 is the original speed. Therefore, by the previous equation, Ft = m(-wL/2 - v_0). Now fact 4 says FtL/2 = (mL^2)/12 * (w-0), because the initial value of w is 0. This equation simplifies to Ft = mwL/6. Now there are two equations for Ft, so their right sides must be equal, therefore m(-wL/2 - v_0) = mwL/6. Cancelling m and solving for w gives w = -(3/2)*(v_0/L). Therefore, the center of mass velocity of the rod after the collision is given by fact 2 as v_c = -wL/2 = (3/4)*v_0. The top of the rod will be going twice this fast, which is (3/2)*v_0, so the top of the rod actually does go faster after the collision, although the center of the rod slows down. The rod loses exactly 1/4 of its original linear momentum.
There is also a shortcut method: look at the angular momentum of the pole around the hinge instead of around its center. Since the force that stops the bottom end is exerted at the hinge, it makes no torque around the hinge, so the angular momentum around that axis IS conserved. (That's an important thing to understand: angular momentum may be conserved around one axis but not another for the same object.) Before the collision, the angular momentum of the pole is (its momentum) * (distance from axis to center of mass), which is -mv_0L/2. The minus sign is because the center of mass passes the hinge while moving tangent to a clockwise circle, which is the negative direction for rotation. After the collision, the motion is now a rotation around the hinge, so the angular momentum is now (1/3)mL^2*w. (This uses the moment of inertia around the end of the rod, instead of the center, so it is (1/3)mL^2.) Since these must be equal, you get -mv_0L/2 = mL^2w/3, which gives w = -(3/2)v_0/L, just as before.
By the way, Limon will disagree with me here, because he has a different view of conservation of angular momentum than most of us do. You will have to decide whom to believe. Personally, I would conduct the experiment and then believe the POLE. You can't go wrong with reality.
Hope that helps!