A point mass 10 g moves in the xy plane with a velocity `2 ms^(-1)` along a straight line. Find the magnitude and direction of its angular momentum (i) with respect to the origin O and (ii) with respect to the origin O. The magnitude of the position vector `vecr` is 0.5 m and the angle `theta = 30^(@)`.

Explanation

`m = 10g = 0.010 kg`
`v= 2ms^(-1) ,r = -0.5 m`
OO. =` r sin theta`
`=0.5 sin 30`
=0.25 m
Angular momentum about O
`=mvr sin theta`
`=0.01 xx 2 xx 0.25`
`=5 xx 10^(-3) J.s.`
Angular momentum about O. = 0, because the direction of the velocity passes through O..