Question 3 Pendulum on a Wheel (29 Marks) Consider the following mechanical setup in two dimensions. A bead of mass m is attached to a massless rotating wheel of radius p. The wheel rotates in a vertical plane with an 3 of 5 APPM2007 angular velocity measured with respect to the horizontal. The end of a massless rod of length L is attached to the bead and hangs vertically downward from the bead. The rod is free to swing around the bead. As the rod moves, the angular displacement of the rod from the vertical line passing through the centre of the bead is measured by the angle 0. At the opposite end of the rod is a ball of mass M. At time =0, the bead lies on the x-axis. See the diagram below, and answer the following questions. M [3.1] Determine the position of the bead at timer. (2 Marks) [3.2] Determine the position of the ball with respect to the bead at a time in terms of the angular displacement 0. (2 Marks) [3.3] Determine the position of the ball with respect to the origin. (2 Marks) [3.4] Determine the potential energy of the bead. (2 Marks) [3.5] Determine the potential energy of the ball. (3 Marks) [3.6] Show that the kinetic energy of the bead can be written as Thead = {mp²² [3.7] Show that the kinetic energy of the ball can be written as Thball = M (p² w²+1²0²-2p Lw sin(wt―0)0). APPM2007 [3.8] Show that a compatible Lagrangian for this system is =(m+M)p²²+ML²0²-MLpw sin(wr-0)0 -g(m+M)p sin(wt)-gML cos(0). (2 Marks) (5 Marks) 4 of 5 [3.9] Compute the Euler-Lagrange Equation for this system. (2 Marks) (5 Marks) [3.10] How does the equation of motion change when the mass of the bead is doubled and the mass of the ball is halved? Does this align with your expectation of the behaviour of a pendulum? Explain your answers. (4 Marks)