Freefall to rolling with no slip

[qwanzaden]

Homework Statement

A uniform sphere with know mass m, moment of inertia I, and radius R_s 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.

Homework Equations

F_x = m*a_x
a_x = (v₂-v₁)/t
F*R = I*alpha
alpha = (w₂-w₁)/t
impulse = F*t
*during no slip rolling*
v_x = w*R_s

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 (v_x = w_s*R_s) therefore:
(1) F = m*a_x => (F*t)/m = v₂-v₁
(2) F*R_s = I_rolling*alpha => (F*t)*R_s/I_rolling = w₂-w₁
(3) v₂ = w₂*R_s
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*[/B]
substituting (3) into (1):
(F*t)/m = w₂*R_s-v₁ => (F*t) = (w₂*R_s-v₁)*m (equation: 4)
substituting (4) into (2):
(w₂*R_s-v₁)*m*R_s/I_rolling = 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 I_rolling = I + m*R_s² (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?

 

[Nathanael]

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

 

[qwanzaden]

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

 

[haruspex]

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

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.

 

[Satvik Pandey]

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

 

[qwanzaden]

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

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.

 

[qwanzaden]

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.

With the term I_rolling = I + m*R_s² (located in the smaller text) the moment is being taken about the point of contact.

 

[haruspex]

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.

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.

 

[dean barry]

The ratio of linear KE rotational KE of a non slipping sphere is the same regardless of velocity.
Start with the original linear KE, this will be conserved in your case.

 

[haruspex]

Start with the original linear KE, this will be conserved in your case.

I see no reason why that would be true.
Also, remember that before impact the rotation has no particular relationship with the linear movement.