Question

Aparticlewhosemassis2.0kgmovesinthexyplanewithaconstantspeedof3.0m/s along the direction r = i + j . What is its angular momentum (in kg · m2/s) relative to the point (0, 5.0) meters?

Answer 1

Answer:

$\vec{L}=-30\frac{kgm^2}{s}\hat{k}$

Explanation:

In order to calculate the angular momentum of the particle you use the following formula:

$\vec{L}=\vec{r}\ X\ \vec{p}$       (1)

r is the position vector respect to the point (0 , 5.0), that is:

r = 0m i + 5.0m j    (2)

p is the linear momentum vector and it is given by:

$\vec{p}=m\vec{v}=(2.0kg)(3.0m/s)(\hat{i+\hat{j}})=6\frac{kgm}{s}(\hat{i}+\hat{j})$   (3)

the direction of p comes from the fat that the particle is moving along the i + j direction.

Then, you use the results of (2) and (3) in the equation (1) and solve for L:

$\vec{L}=-30\frac{kgm^2}{s}\hat{k}$

The angular momentum is -30 kgm^2/s ^k