A metre stick is positioned vertically with one end resting on a rough horizontal floor. It is gently released to fall onto the floor. Assuming the end on the floor does not slip, determine the angular speed of the rod just as it strikes the floor. (Take g = 9.8 m/s ^2 )

Step-by-step solution

Using conservation of mechanical energy: Loss in potential energy = Gain in rotational kinetic energy mg L 2 = 1 2 I ^2 Since the rod rotates about its lower end, its moment of inertia is I = mL^2 3 . mg L 2 = 1 2 ( mL^2 3 ) ^2 ^2 = 3g L Given L = 1 m and g = 9.8 m/s ^2 : = 3 9.8 1 = 29.4 5.4 rad/s