Question

Bookmark A person (mass m) is shot out of a cannon with an initial velocity of V1 at an angel theta with respect to the horizontal. Find the angular momentum of the person about the origin when the person is at :

1) the origin

2) the maximum height of it's trajectory

3) The end of it's flight

Answer 1

Solution:

1) About the origin the angular momentum = mvR sin =0 since distance at the origin = 0

2) At the maximum height of the trajectory,the H = v^2 sin^2 /2g

Horizontal component of velocity = vx = v cos

Angular momentum == m v x H = m (v cos ) (  v^2 sin^2 /2g ) = m v³ (sin ² ) ( cos ) / 2g

c) At the end of the trajectory distance = range = v² sin 2 /g

Angular moentum = mv R = m (v sin ) (  v² sin 2 /g)

= ( 2m v ³ sin² cos ) /g = 4 times the angular momentum at the Maxium height.