Question

A uniform crate C with mass mC is being transported to the left by a forklift with a constant speed v1. What is the magnitude of the angular momentum of the crate about point A, that is, the point of contact between the front tire of the forklift and the ground

Answer 1

The angular momentum of the crate is
`M_(C) V_(1) d`

Step-by-step explanation:

mass of the crate =
`M_(C)`

speed of forklift =
`V_(1)`

The distance between the center of the mass and the point A = d

Recall that the angular moment is the moment of the momentum.

`L = P*d` ..... equ 1

where L is the angular momentum,

P is the momentum of the system,

d is the perpendicular distance between the crate and the point on the axis about which the momentum acts. It is equal to d from the image

Also, we know that the momentum P is the product of mass and velocity

P = mv ....equ 2

in this case, the mass =
`M_(C)`

the velocity =
`V_(1)`

therefore, the momentum P =
`M_(C)`
`V_(1)`

we substitute equation 2 into equation 1 to give

`L = M_(C) V_(1) d`