A bob of mass 200 gm is attached to a string of length 100 cm and is revolved at the rate of 2 revolutions per second. What is its angular momentum ? If after 10 s it makes only on revolutions per second calculate the mean torque.

A bob of mass 200 gm is attached to a string of length 100 cm and is r

1.	A bob of mass 200 gm is attached to a string of length 100 cm and is revolved at the rate of 2 revolutions per second. What is its angular momentum ? If after 10 s it makes only on revolutions per second calculate the mean torque.
Answer» Solution :m = 200 gm = 0-200 kg,r = 100 cm = 1.00 m Initial angular velocity `=omega_(0) = 2pi xx 2` rad/s Angular MOMENTUM `=L =mvr = mr^(2) omega_(0)` `=0.2 xx 1^(2) xx 4pi` `=2.51` J.s perpendicular to the PLANE of the motion Torque = `("CHANGE of angular momentum")/("Time")` `tau = mr^(2) (domega)/(dt) = (0.2 xx 1^(2) xx (2pi xx 1))/10` =0.125 Nm.