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

`\vec{L}=-30(kgm^2)/(s)\hat{k}`

Step-by-step 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(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(kgm^2)/(s)\hat{k}`

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