Question

Problem 3. (24 points) A uniform rod of mass M and length d is free to pivot about one end. The moment of inertia of the rod about the pivot is I = Md2/3, and the rod's center of mass is at its midpoint. The rod is released from rest at angle above the horizontal, then rotates downward under the influence of gravity. d x e When the rod reaches angle below the horizontal, determine (a) (4 points) the rotational kinetic energy of the rod, (b) (8 points) the magnitude of the rod's angular momentum, (c) (6 points) the angular acceleration of the rod, (d) (6 points) the radial and tangential components of the rod's center of mass acceler- ation. For all quantities, treat the rod's pivot point as the rotation axis.

Answer 1

solution Što 2 2 & I cd Mgosino (a), From the conservation of mechanical energy ng a sing-(-mg d sino) Mods sino Rotational Kinetic energy. Mga sino (6) I2w2 = 21 Mgosino → ID= magnitude of the red's angular momentum 21 Mgosino 12 Md x Mga sina Md gasimo Ma 3 gosino @ Torque on the rad about the Piot Point = T = > x = Mg of Caso x 8 d2 = 3goso ESO Qd mg을 IQ = motza 38coso Qy = radial acceleration = 002 The aingasine) = mgdisine Q el (d) ar Md2 I at 3gsin tangentiel acceleration 1.2 3g cosa ad 3gcosa 3g sino 3gcose 4 ar at 4