Question

3. 25 points A uniform hoop of mass M and radius r has a uniform rod of mass m and length r welded to it as illustrated. a) Given that the rod is at an angle θ from the horizontal, and that the hoop is rotating at a rate ω and rolling on the floor without slipping, what is the total Kinetic Energy, and total Potential Energy (use the height of the center of the hoop as your datum)? Hints: The total KE is the sum of the KEs of each part, and each part has a KE that can be written as the sum of its center- of-mass KE and its KE of rotation around its CM. Recall the moment of inertia of a hoop around its center is M r2 and the moment of inertia of a rod around its center is (1/12) m L2 You will need the distance b (from the contact point at the floor to the center of the rod) as a function of 0; you can get it using trig. b) If the object starts from rest at θ = 0 and then we let go, what is the speed V of the center of the hoop after the hoop has rotated 90 degrees, i,e when θ-π/2 and b r/2?

Answer 1

Casine rule in delta ABO b^2 = r^2 + (1/2)^2 - 2(r) (2/2) cos (90 - theta) b^2 = r^2 + r^2 + r^2/4 - r^2 sin theta (r = L) b^2 = 5 r^2/4 - r^2 sin theta moment of Inertia of whole syste, bout Instranteneous center of Rotation I_o = (mr^2 + mr^2) + mL^2/12 + mb^2 (r = L) I_0 = 2 Mr^2 + mr^2/12 + m(5r^2/4 - r^2 sin theta) I_0 = 2Mr^2 + mr^2/12 + 5 mr^2/4 - mr^2 sin theta I_0 = 2 Mr^2 + 16 mr^2/12 - mr^2 sin theta Total i.cE of the system = 1/2 I_0 w^2 K E = 1/2 (2 Mr^2 + 16 mr^2/12 - mr^2 sin theta) PE of the system (center of Hoop is datum) PE = PE of hoop + PE of bar = 0 + - mg 2/2 sin theta (r = L) PE = -mg r/2 sin theta \r\n (b) from energy balance