# Angular momentum for combined rotation and translationa about a fixed point

1. Aug 20, 2011

### 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 = Lcm + 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 ...

2. Aug 20, 2011

### A.T.

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
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

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?

3. Aug 20, 2011

### 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....

Last edited: Aug 21, 2011