Answer
(a) The magnitude of the angular momentum is $L = 115~kg~m^2/s$, and it is directed in the clockwise direction (into the page).
(b) The magnitude of the rate of change of angular momentum is $125~N~m$, and it is directed in the counterclockwise direction (out of the page).
Work Step by Step
(a) $L = mvr~sin(\theta)$
$L = (2.00~kg)(12.0~m/s)(8.00~m)~sin(36.9^{\circ})$
$L = 115~kg~m^2/s$
The magnitude of the angular momentum is $L = 115~kg~m^2/s$, and it is directed in the clockwise direction (into the page).
(b) The magnitude of the rate of change of angular momentum is equal to the torque.
$\tau = r\times F$
$\tau = (8.00~m)(2.00~kg)(9.80~m/s^2)~sin(53.1^{\circ})$
$\tau = 125~N~m$
The magnitude of the rate of change of angular momentum is $125~N~m$, and it is directed in the counterclockwise direction (out of the page).
