A 2.00-kg rock has a horizontal velocity of magnitude 12.0 m/s when it is at point P in ?Fig. E10.35?. (a) At this instant, what are the magnitude and direction of its angular momentum relative to point ?O?? (b) If the only force acting on the rock is its weight, what is the rate of change (magnitude and direction) of its angular momentum at this instant? Fig. E10.35

Solution 37E The numerical expression for angular momentum is given by, L = mvrsin …..(1) Here, m is the mass of the object, v is its velocity and rsin is the perpendicular distance of the axis of rotation to the object. The rate of change of angular momentum is torque. Given in the question, m = 2.00 kg v = 12.0 m/s r = 8.00 m 0 Therefore, the perpendicular distance is = rsin = 8.00 m × sin 36.9 = 4.80 m (a) Substituting these values in equation (1), Angular momentum L = 2.00 kg × 12.0 m/s × 4.80 m × 4.80 m 2 L = 115 kg.m /s...