How Do You Calculate the Ratio h/R for a Spinning Billiard Ball?

[AFlyingKiwi]

1. Homework Statement
A spherical billiard ball of uniform density has mass m and radius R and moment of inertia about the center of mass ( ) 2 cm I = 2/ 5 mR^2 . The ball, initially at rest on a table, is given a sharp horizontal impulse by a cue stick that is held an unknown distance h above the centerline (see diagram below). The coefficient of sliding friction between the ball and the table is µk . You may ignore the friction during the impulse. The ball leaves the cue with a given speed v0 and an unknown angular velocity ω0 . Because of its initial rotation, the ball eventually acquires a maximum speed of 9 / 7 v0 . The point of the problem is to find the ratio h / R.

Homework Equations

L=Iw
p=mv
L_i = L_f

The Attempt at a Solution

I have the solution in this link (http://web.mit.edu/8.01t/www/materials/InClass/IC-W15D2-5.pdf) but I don't get a certain part of it. When they give us initial L, it says L_i = mv_0(R) + I_cm(w_0). Why is the mv_0(R) there? Also, why can we use conservation of angular momentum if there is a net external force?

 

[haruspex]

When they give us initial L, it says L_i = mv_0(R) + I_cm(w_0). Why is the mv_0(R) there? Also, why can we use conservation of angular momentum if there is a net external force?

The answer to both questions is that they are taking angular momentum about some point at the level of the table. Friction therefore has no moment, and angular momentum is conserved. The total angular momentum consists of that due to its rotation about its centre plus the contribution from its linear motion, mvR.

 

[AFlyingKiwi]

The answer to both questions is that they are taking angular momentum about some point at the level of the table. Friction therefore has no moment, and angular momentum is conserved. The total angular momentum consists of that due to its rotation about its centre plus the contribution from its linear motion, mvR.

So just to clarify, for any object that is rolling, it's total angular momentum is its rotational component and it's linear component times the radius?

 

[haruspex]

So just to clarify, for any object that is rolling, it's total angular momentum is its rotational component and it's linear component times the radius?

It is not to do with whether it is rolling.
In general, angular momentum (and torque, and moment of a force...) is only meaningful in the context of a given axis. You can decompose the motion of a rigid body into the sum of a linear motion and a rotation about its mass centre. Given the axis, you can find the angular momentum contributions from those two motions and add them to get the total angular momentum about the axis.
E.g. consider a mass m traveling at speed v along y=h, z=0, rotating at rate ω about an axis parallel to the z axis and through its mass centre. Let the MoI about that rotation axis be I. Pick the z axis through the origin as the reference axis.
The linear motion contributes mhv, while the rotation contributes Iω.