How Do You Calculate Angular Momentum for Different Pivot Points?

[AriAstronomer]

Homework Statement

A particle with a mass of 0.400 kg is attached to the
100-cm mark of a meter stick with a mass of 0.100 kg. The
meter stick rotates on a horizontal, frictionless table
with an angular speed of 4.00 rad/s. Calculate the angu-
lar momentum of the system when the stick is pivoted
about an axis (a) perpendicular to the table through
the 50.0-cm mark and (b) perpendicular to the table
through the 0-cm mark.

Homework Equations

The Attempt at a Solution

I only need help with part a): Basically, I know I can get the right answer using L = Iw:
L = (I_rod + I_mass)w = (1/12ML^2 + MR^2)w = .433 = correct.
If however, I use the other definition L = rxp:
L = rxp_rod + rxp_mass = rMwr_rod + mwr^2 = 0.5 = incorrect. What am I missing? Seems like it's in the rod term. I guess r implies radius, and would work if it was a disk perhaps? Should I always go with the L = Iw definition if working with non radial objects?

Thanks,
Ari

 

[gneill]

L = rxp is easy to use when the object associated with the momentum p is a point mass. When it's some extended shape then you have to break it down into mass elements (dm) each with its own bit of momentum (dp) over all of the object, and apply L = r x dp to all of them, summing as you go. Yes, it's an integral

 

[tiny-tim]

Hi Ari!

(have an omega: ω and try using the X² icon just above the Reply box )

If however, I use the other definition L = rxp:
L = rxp_rod + rxp_mass = rMwr_rod + mwr² = 0.5 = incorrect. What am I missing? Seems like it's in the rod term. I guess r implies radius, and would work if it was a disk perhaps? Should I always go with the L = Iw definition if working with non radial objects?

Yes, you're right, it is the rod term.

L = r x p will always work for a point mass, but you need to use https://www.physicsforums.com/library.php?do=view_item&itemid=31" for anything else.

So you can use r x p for the mass in either case, and add it to the Iω for the rod, or you can use Iω for the rod-plus-mass.

btw, L = Iω only works for an axis through the centre of rotation (as in boht these cases) or the centre of mass …

for any other point, you need L = I_c.o.m.ω + r x mv_c.o.m.