Question

A 1.10-kg particle moves in the xy plane with a velocity of v with arrow = (3.80 î − 4.10 ĵ) m/s. Determine the angular momentum of the particle about the origin when its position vector is r with arrow = (1.50 î + 2.20 ĵ) m.

Answer 1

Answer:

15.961 kgm^2 / s along negative z axis

Explanation:

m = 1.10 kg,

v = (3.80 i - 4.10 j) m/s

r = (1.50 i + 2.20 j) m

The formula for the angular momentum is given by

$\overrightarrow{L}=m (\overrightarrow{r}\times \overrightarrow{v})$

$\overrightarrow{L}=1.10 (1.50 i +2.20j)\times (3.80i-4.10j)$

$\overrightarrow{L}=-15.961 \hat{k}$

Magnitude of angular momentum is 15.961 kgm^2 / s