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 This is the required angular momentum and will be directed into the page. (b) Let us have a look at the following figure. Therefore the weight of the stone makes an angle of 53.1 with the line passing through the point O. As already mentioned above that the rate of change of angular momentum is torque. Therefore, torque due to the weight of the stone = mgrsin 53.1 0 W 2 0 2 2 W = 2.00 kg × 9.8 m/s × 8.00 m × sin 53.1 = 125 kg.m /s The torque due to the weight of the stone is 125 kg.m /s2 2 and will be directed out of the page.