# Freefall to rolling with no slip

1. Jan 26, 2015

1. The problem statement, all variables and given/known data
A uniform sphere with know mass m, moment of inertia I, and radius Rs is traveling through free space with initial horizontal linear velocity v1 and rotational velocity w1. It then makes tangential contact with a horizontal surface and instantaneously starts rolling without slipping. What is the rotation/linear velocity the instant after it starts rolling?
Assume positive linear velocity is to the left, and positive rotation is counter-clockwise.
EDIT: assume a gravitational force sufficient enough to induce a frictional force to cause no sliding.

2. Relevant equations
Fx = m*ax
ax = (v2-v1)/t
F*R = I*alpha
alpha = (w2-w1)/t
impulse = F*t
*during no slip rolling*
vx = w*Rs

3. The attempt at a solution
I would try to solve this problem by trying to apply a horizontal impulse to the sphere, changing the linear velocity and the angular velocity, until the new linear velocity matched the linear velocity induced by rolling (vx = ws*Rs) therefore:
(1) F = m*ax => (F*t)/m = v2-v1
(2) F*Rs = Irolling*alpha => (F*t)*Rs/Irolling = w2-w1
(3) v2 = w2*Rs
This is 3 equations and three unknows (v2, w2, F*t). Solving this we get:
*only here for you to check my math, actual reading not necessary*

substituting (3) into (1):
(F*t)/m = w2*Rs-v1 => (F*t) = (w2*Rs-v1)*m (equation: 4)
substituting (4) into (2):

(w2*Rs-v1)*m*Rs/Irolling = w2-w1 => and I'm not going to finish typing that because it takes a long time and the equation formatter is buggy. I realize that that I need to do a substitution of Irolling = I + m*Rs2 (parallel axis theorem because while the sphere is on the ground it is rotating about its edge, not its center of mass)

My main question is: is this the correct way to attack the problem? (apply a impulse until the linear and rotational speeds match). This concept will later be applied to a much more complex system (dynamics simulation). Is it still valid to apply this to a system with complex internal forces, but who's total mass is m, total linear velocity is v1, effective radius is R, and total moment of inertia is I?

Last edited: Jan 26, 2015
2. Jan 26, 2015

### Nathanael

If there is no normal force, there will be no frictional impulse to make the ball begin rolling.

3. Jan 26, 2015

*assume a gravitational force sufficient enough to induce a frictional force to cause no sliding

4. Jan 26, 2015

### haruspex

True, but you can take it as a very small normal impulse with a colossal coefficient of friction.

It's a bit easier if you take moments about the point of contact.

5. Jan 27, 2015

### Satvik Pandey

Try to solve it by conserving angular momentum about the point of contact.

6. Jan 27, 2015

I have thought about conserving angular momentum but because there is an outside force (friction) being applied to the system (sphere of interest) I believe that momentum is not conserved. This is my reasoning behind trying to do a impulse-driven method of solving it.

Last edited: Jan 27, 2015
7. Jan 27, 2015

With the term Irolling = I + m*Rs2 (located in the smaller text) the moment is being taken about the point of contact.

8. Jan 27, 2015

### haruspex

The friction acts through the point of contact, so has no moment about that point. Angular momentum about that point is conserved. This is the same as the point I was trying to make.

9. Jan 27, 2015

### dean barry

The ratio of linear KE rotational KE of a non slipping sphere is the same regardless of velocity.