Rotational Angular Momentum of a Bowling Ball

[KatlynEdwards]

Homework Statement

A 7 kg bowling ball whose radius is 0.13 m is rolling directly toward you turning three times per second. What are the magnitude and direction of its rotational angular momentum about you? What is its translational angular momentum about you?

Homework Equations

L = r x p = r x (m*v)
L = I\omega
L of a moving object = L_cm + L_rot

The Attempt at a Solution

So I was thinking the rotational angular momentum is L = i\omega which is 2/5*m*r²\omega
Which would be (2/5)*7*(0.13)²*(6*pi / 1 sec)
But I don't know how to get the translational angular momentum.

 

[tiny-tim]

Hi Katlyn!

(have an omega: ω and a pi: π )

A 7 kg bowling ball whose radius is 0.13 m is rolling directly toward you turning three times per second. What are the magnitude and direction of its rotational angular momentum about you? What is its translational angular momentum about you?
…
But I don't know how to get the translational angular momentum.

That's a rather weird question.

I think what they mean is that when you measure the angular momentum of a body about an axis, it's the angular momentum about a parallel axis through the centre of mass, plus the angular momentum of the whole mass as if it were at the centre of mass, and moving with the same velocity as the centre of mass …

I think they're calling the second one translational angular momentum.

(Though then they must be calling the first one "the rotational angular momentum about you" when it quite clearly isn't about you )