Posted by emath on Thursday, March 4, 2010 at 9:56pm.
You do not provide or describe the figure.
You do not identify "the following locations"
You do not tell us what parts (b) and (c) of the question are.
Your answer to (a) does not have the correct dimensions for angular momentum, and therefore cannot be right.
It is not possible to provided assistance because of these omissions
My apologies
The "following locations"
a) at the origin
b)at the highest point of its trajectory
c)just before it hits the ground
the dimensions of angular momentum: ? kg*m^(2)/s
the picture is simply a parabola, a projectile launched with some angle theta with respect to the horizontal
and for part a I realized that it is 0 because the particle is at the origin and I have to find the angular momentum of the particle about the origin:
a = vi x= r
r = 0
emath:
Thank you for taking the time to clarify and complete your question. Your answer to part (a) is correct.
There is a shortcut you can take for calculating the angular momentum with respect to the origin. Write the position in vector form as r = x i + y j, and the momentum as
p = m Vx i + m Vy j. The cross product is then
r = p x r = m(Vx*y - Vy*x) k
(k is a unit vector perpendicular to i and j . 'i' is horizontal along the path and 'j' is vertical.
Next, write down the equations for x, y, Vx and Vy in terms of t
x = Vi cos theta * t
y = Vi sin theta * t - (g/2) t^2
Vx = Vi cos theta
Vy = Vi sin theta - gt
L = m [Vi^2 sincos theta *t -Vi cos theta *t - Vi^2 sincos theta *t + Vi cos theta * g t/2]
At t=0, L obviously = 0, in agreement with your result
For parts (b) and (c) of your question, calculate the corresponding t and then calculate L using the equation above.
For example, at the top of the trajectory, t = (Vi sin theta)/g
At the end of the trajectory, t is twice that amount.
The resulting equation will look a bit messy. I leave the final steps to you
Hey drwls, can you go ahead and finish the rest of the equations, please? I'm having much difficulty with the final equation.