Physics Inertia and Angular Momentum Question?

Question

Hi! 

We're having trouble understanding this problem:
7.  A 3 kg particle is moving at a velocity of 5 m/s parallel to the x axis in the negative x direction. The line of action (path of the particle) is a distance of 4 m from the origin in a given coordinate system, P, and a distance of 8 m from the origin of a second coordinate system, P¢.  The angular momentum of the particle about P¢ as compared to P is
(a) the same
(b) larger by a factor of 2
(c) smaller by a factor of 2
(d) larger by a factor of 5
(e) smaller by a factor of 5

L = Iw
I = .5mr^2
w = 5
Because everything but r is constant, the only difference is r^2
So we have 4^2 and 8^2 which yields 16 and 64, so the difference is 4x. However, the correct answer is 2x, why is this? It seems that it is just the 4 versus 8, but this doesn't make sense to me.

Thanks!

debydete · Accepted Answer

The angular momentum of an object with linear momentum "p" a distance "r" from some origin is, in general;
L = prSin() = mvrSin()

where () is the angle between the direction of "p" and the position vector "r" .
Whe the object is moving parallel to the x-axis a quick sketch will show that rSin() = y
So for this special case the angular momentum is;
L = mvy

I*n your problem the "line of action" is 4m then 8m above the origin. This is essentially "y".

This problem is a little more complicated because the particle is moving along negative x-direction. If it were positive direction the trig. would be easy to see, but its the same magnitude either direction.

Physics Inertia and Angular Momentum Question?

1 Answer