Question

In class we considered the case of briefly pushing a thin rod (mass m, length 2L, centered on the origin and laying along the x-axis) perpendicular to the rod, and asked, conceptually, about the subsequent motion. Let's put the answer to that question on firmer mathematical footing. Assume that you push with F=Fj for a short time Δt. a)Describe (mathematically) the subsequent motion, as a function of where you push on the rod (from x=0  to x=L). That is, what is the center of mass velocity and the angular velocity of the rod about that center of mass as a function of where you push?b)Calculate the total energy of the rod as a function of where you push. If this depends on where you push, how could that be given that you are always pushing with the same force

Image transcriptions

EXAMPLE 3.4 A Sliding and Spinning Dumbbell A dumbbell consisting of two equal masses m mounted on the ends of a rigid massless rod of length 26 is at rest on a frictionless horizontal table, lying on the x axis and centered on the origin, as shown in Figure 3.10. At time / = 0, the left mass is given a sharp tap, in the shape of a horizontal force F in the y direction, lasting for a short time Ar. Describe the subsequent motion. There are actually two parts to this problem: We must find the initial motion immediately after the impulse, and then the subsequent, force-free motion. The initial motion is not hard to guess, but let us derive it using the tools of this chapter. The only external force is the force F acting in the y direction for the brief time Ar. Since P = FX, the total momentum just after the impulse is P = F Ar. Since P = MR (with M = 2m), we conclude that the CM starts moving directly up the y axis with velocity Vem = R = F At/2m. While the force F is acting, there is a torque fed = Fb about the CM, and so, according to (3.28), the initial angular momentum (just after the impulse has

Figure 3.10 The left mass of the dumbbell is given a sharp tap in the y direction. ceased) is L = Fb At. Since L = Iw, with / = 2mb, we conclude that the dumbbell is spinning clockwise, with initial angular velocity ( = F At/2mb. The clockwise rotation of the dumbbell means that the left mass is moving up relative to the CM with speed co, and its total initial velocity is &quot;left = Vem + wb = F At/m. By the same token the right mass is moving down relative to the CM, and its total initial velocity is &quot;right = Ucm - wb =0. That is, the right mass is initially stationary, while the left one carries all the momentum F Ar of the system. The subsequent motion is very straightforward. Once the impulse has ceased, there are no external forces or torques. Thus the CM continues to move straight up the y axis with constant speed, and the dumbbell continues to rotate with constant angular momentum about the CM and hence constant angular velocity.