Mechanics - PHYS 211 Homework 10 Problem 1: An Mm = 45 kg gure skater is spinning on the toes of her skates at a), = 1.0 rev/s. Her arms are outstretched as far as they will go. In this orientation, the skater can be modeled as a cylindrical torso (M = 40 kg, 20 cm average diameter, 160 cm tall) plus two rod-like arms (m : 2.5 kg each, L : 66 cm long) attached to the outside of the torso. The skater then raises her arms straight above the head, where she appears to be a 45 kg, 20-cm-diameter, ZOOcm-tall cylinder. What is her new angular velocity, (of, in rev/s.7 a) Fig] illustrates the model for the gure skater with outstretched arms. Combining the moments of inertia of the cylinder and two thin rods, compute the initial moment of inertia, I,-, of the skater. You need to use the parallel-axis theorem, where the distance d between the center of mass of each rod and the rotation axis is shown in Fig. 1, (Answer: Ii : 1.31 kg m2.) FIG. 1: The scheme for Problem 1 b) Aer the skater raises her arms, she can be modeled by a single cylinder with mass Mm. Compute her nal moment of inertia, I f. c) Assuming the ice is frictionless, the angular momentum of the skater, L = 14, is conserved (here, L is the magnitude of the vector L directed Vertically up or down depending on the direction of the rotation). Using the conservation of angular momentum, compute wf. (Answer: 5.8 rev/s.) Problem 2: Fig 2 shows a cube ofmass m sliding without friction '1 at speed c. it undergoes a perfectly elastic collision with the bottom no ofa rod of length d and mass M = Zn. The rod is pivoted about a frictionless axle through its center, and i ally it hangs straight O l/ down and is at rest. What is the cube's velocity - both speed and direction , aer the collision? in i a) Let's denote the velocity of the cube aer the couision by o, a (at the end, positive as would mean that the cube continues to move to the right. while negative v, would mean that it moves to the left), Using the conservation of angular momentum of the system just before and just alter the collision with respect to the pivot poinl (the center of the rod), show that the angular velocity of the rod after the collision is a) = an. vet. FIG 2 The scheme for Problem 2 b) Formulate the conservation of energy of this system before and aer the collision (it should involve the kinetic energy of the cube and the rotational kinetic energy of the rod) Using the ronnula for o from the previous step. show that the energy conservation law can be reduced to the following quadratic equation for vx' so: 6001),, +o = o. c) Solve this quadratic equation to get u, in terms of (/9.' ' One of the two roots or the equntion should be u, : no, which would correspond to the case oi no collision nt nu, and it is or no interest to us. You are interested in the second mo| Page 3 of 3 - ZOOM + 3 Problem 3: Two spherical asteroids have the same radius R. Asteroid 1 has mass M and asteroid 2 has mass 2M. The two asteroids are released from rest with distance 10R between their centers. What is the peed of each asteroid just before they collide? a) Write down the conservation of energy for this system, Uj + Ki = Uf + Kf, and show that the final kinetic energy of both asteroids taken together is Kf = . (One of the most common mistakes made in this problem is to double count the potential energy. Note that UG = - Simmz is the gravitational potential energy of the system as a whole, both masses included.) b) Use the conservation of linear momentum of the system to get the relation between the final speeds, v1 and v2, of the two asteroids. c) The total final kinetic energy you found in a) is the sum of the final kinetic energies of the two asteroids. Using this fact and the relation between of and v2 from part b), find the speed of each asteroid. (Partial answer: 01 = 4 95k.) 2 Potential energy is the negative integral of the force along the motion of the object experiencing the force. You may imagine one of the masses being fixed in space, and in this case the integral will be solely due to the motion of the second mass. Or, you can think about both masses being in motion with respect to some origin of coordinate system and use Newton's third law stating that the forces with which the masses are acting on one another have the same magnitudes and opposite directions. Independently on what way you choose to evaluate the integral, it will result in UG = - Gmimz