# Ball rolling and slipping with backspin

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1. Jan 24, 2016

### ClassicalMechanist

1. The problem statement, all variables and given/known data

A solid ball of mass M and radius R and moment of inertia I is placed onto a table with an initial velocity v_0 to the right and angular momentum w_0 anticlockwise (i.e the ball has backspin). Due to friction, the angular velocity and linear velocity changes as the ball both rolls and slips along the table.

Find the velocity v_0 (in terms of the other variables) such that the ball will roll some distance and then stops completely.

2. Relevant equations
I. v=v_0+at, where a=-f/m
II. w=w_0+(alpha)*t, where alpha=torque/I=-fR/I

3. The attempt at a solution

Due to friction, the velocity and angular velocity (in the anticlockwise direction) should both decrease. The velocity is zero when t=mv_0/f. The angular velocity is zero when t=Iw_0/(fR).

My confusion is the necessary condition for the ball to stop completely (meaning no linear motion, and no rotation). Naively, at some point in time, t, both the velocity and angular velocity are zero, so we should equate the two expressions above and solve for v_0=Iw_0/(mR).

However I have an uneasy feeling about this, because I seem to get the same answer by assuming that the ball will stop when the no-slip condition is satisfied, i.e v=Rw. Using this equation we get v_0+at=R(w_0+(alpha)*t). Solving for t, we find t=(v_0-Rw_0)/(alpha*R-a)=(v_0-Rw_0)/(-f(R^2)/I+f/m). The velocity is zero when t=mv_0/f, so equating these two expressions, we have:

(v_0-Rw_0)/(-f(R^2)/I+f/m)=mv_0/f

(v_0-Rw_0)/(1/m-(R^2)/I)=mv_0

I(v_0-Rw_0)/(I-(R^2)m)=v_0

I(v_0-Rw_0)=v_0(I-(R^2)m)

Iv_0-IRw_0=v_0I-v_0(R^2)m

IRw_0=v_0(R^2)m

v_0=Iw_0/(mR)

So I get the same answer by both methods, which might seem encouraging, but I'm not sure about my reasoning for the conditions under which the ball stops completely.

2. Jan 24, 2016

### Suraj M

I might be wrong here
But if you're taking alpha as positive when you equate final velocity to zero won't you get your time to be negative so then if you substitute that into this
$“ t=(v_0-Rw_0)/(alpha*R-a)=(v_0-Rw_0)/(-f(R^2)/I+f/m). ”$
Then you wouldn't get the answer you got by the first method would you?

3. Jan 24, 2016

### ClassicalMechanist

alpha is not positive, because alpha=-fR/I.

4. Jan 24, 2016

### Suraj M

In the equation
$V= V_0 + at$
If you equate V to zero then what value of $t$ do you get?

5. Jan 24, 2016

### ClassicalMechanist

You get t=-v_0/a. But a is negative as well, because a=-f/m. I don't see a problem with signs...

6. Jan 24, 2016

### Suraj M

I really don't think you can use the second method, I'm trying to understand how we still get the same answer. I'm not sure of this but isn't V representing the linear velocity of a particle on the surface of the ball( not considering linear velocity of the body)If there is no slipping then that would be equal to the velocity of the body. But here it wouldn't be right?

7. Jan 24, 2016

### Suraj M

If there is no slipping then isn't $a= \alpha R?$

8. Jan 24, 2016

### PeroK

I didn't check all your algebra, but your reasoning in both cases is sound. The second method seems a more roundabout way, but effectively you're saying:

At some point the ball stops slipping and starts rolling without slipping. If the velocity at this point is 0, then the ball can't be spinning, so that is an equivalent way to find the required value for $v_0$.

The first method is better surely?

9. Jan 24, 2016

### haruspex

Both methods seem fine to me. But the easiest way is to consider angular momentum about a fixed point on the ground along the line of travel. Friction has no moment about such a point, so the angular momentum is conserved. The result follows immediately.