Angular momentum for combined rotation and translationa about a fixed point

[Highwaydude]

One more serious doubt...
In feynman's tips on physics' problems... there is one regarding marble rolling...
"An amusing trick is to press a finger down on the marble, on a horizontal table, in such a way that he marble is projected along the table with initial linear speed v0 (v-naught) and initial backward rotational speed w0(omega-naught)... w0 being about the horizontal axis perpendicular to v0... the coefficient of sliding friction is constant.. the marble has radius R..
What relationship must hold for the marble to slide to a complete stop?"
Now I have done this using both the conservation of angular momentum about the point touching the ground(since torque due to friction about this point is zero) and the retardation in linear and angular velocity due to friction(in which you get two equations and you have to eliminate t(time) to get the relation)...
The angular momentum about a FIXED POINT O = L_cm + M.R.V₀
(where Lcm is ang momentm about the centre of mass about the fixed point O and V is the linear velocity of the CM about O)
So when you do it...
initial angular momentum about the point of sphere touching the ground= -(2/5). MR^2.(w0)+MR.v0
Final ang momentum= 0
therefore equating them v0=2/5 w0(which is the required relation)

BUT THE POINT TOUCHING THE GROUND IS MOVING AND EVEN THE LINEAR VELOCITY OF THE CM IS NOT V0 ABOUT THAT POINT... ITS (V0-W0.R)...(however, the relative angular velocity of CM about the ground point IS -w0)... CAN SOMEONE PLEASE EXPLAIN THIS LONG STANDING DOUBT IN MY MIND ...

 

[A.T.]

v0=2/5 w0(which is the required relation)

I think it's v0=2/5 R w0. I found two solutions for this:

1)
http://www.feynmanlectures.info/solutions/shooting_marbles_sol_1.pdf

When the marble slides to a complete stop, its angular momentum is zero. As the external
force, force of kinetic friction, acts horizontally through the point of contact so angular
momentum about the starting point is conserved. Initial angular momentum, about the
starting point, by virtue of translation of centre of mass is MV0R clockwise and by virtue of
rotation of all the particles of marble about c.m. is Icmω0 anti clockwise where Icm = 2MR
2/5, is the moment of inertia of the spherical marble about its c.m. Hence initial total angular
momentum is Icmω0 - MV0R. As angular momentum is conserved so Icmω0 - MV0R = 0, or,
V0 = 2ω0R/5.

Note that he considers conservation of angular momentum about the static starting contact point, not the moving contact point. This seemed to be the source of your confusion.

2)
http://www.feynmanlectures.info/solutions/shooting_marbles_sol_2.pdf

The kinetic force of friction, fk = µMg, acting horizontally leftward through the point of contact produces leftward translational acceleration µg [Force / mass] and rotational anticlockwise acceleration, α = 5 µg / 2R [torque / Icm] .

If the marble comes to a stop after time t, then V0 = µgt [since, v = u + at] and ω0 = αt [since, ω = ω0 + αt]. Eliminating t, we get V0 = 2 ω0R/5.

My naive solution : The same force applied over the same time cancels both momentums. I simply consider the change of L about the centre of mass, and don't care that the CM is moving.

F = const
F t = p₀
F R t = L₀

p₀ R = L₀
M v₀ R = 2/5 M R² ω₀
v₀ = 2/5 R ω₀

Is my argument valid? If not why do I get the same answer?

 

[Highwaydude]

I too was speculating about the contact point being the ground and not the point of the sphere touching the ground... and on imagining the fact that even if the sphere rolls a bit forward, the torque due to friction remains zero since the line of force intersects the point...so yeah, makes sense.

I think your argument is very correct by the way...
The force acting on the sphere changes the angular momentum about the CM by -Ft.R and this must cancel out the initial angular momentum i.e 2/5 MR^2. w0... (about the CM of course) and also cancel the initial linear momentum over time t.

Anyways thanks...