CP? A meter stick with a mass of 0.180 kg is pivoted about one end so it can rotate without friction about a horizontal axis. The meter stick is held in a horizontal position and released. As it swings through the vertical, calculate (a) the change in gravitational potential energy that has occurred; (b) the angular speed of the stick; (c) the linear speed of the end of the stick opposite the axis. (d) Compare the answer in part (c) to the speed of a particle that has fallen 1.00 m, starting from rest.
Solution 81P Step 1 of 5: Step 2 of 5: a) we know that , U = mgy U = mgy and U = mgy cm 2 2 2 1 The center of the mass of the stick is at its center, socm 1= 1 m and ycm 2= 0.50 m The change in potential energy of the rope is U = U 2U 1 = mg(y y ) cm 2 cm 1 = (0.180 kg)(9.8 m/s )(0.50 m 1 m) = 0.882 J Step 3 of 5: b)Conservation of energy is, K 1 U +1W other K 2 U 2 only the gravity that does work on meter stick, so W = 0. K = 0 other 1 Thus K = U U = U 2 1 2 K 2 U 1 2 2I 2 U 2(U) 2 I = 6(0.882 2) (0.180 kg)(1 m) = 5.42 rad/s