# Homework Help: Rolling with Slipping

1. Dec 10, 2015

### Raios168

1. The problem statement, all variables and given/known data
A billiard ball is imparted a horizontal impulse of 2 N*s at a height 4 cm above the center of the ball. The mass of the ball is 0.02 kg and it has radius 5 cm. Find velocity of the center of mass right after the impulse.

Note: I simplified the question, I was actually given that a linearly increasing force was applied from 0 to 40 000 N in 0.001 seconds and then was decreased linearly from 40 000 N to 0 in 0.001 seconds. So this created a triangle when I created a Force-time graph, calculating the area of the triangle yielded 2 N*s

2. Relevant equations

L = Iw

3. The attempt at a solution
Okay so the impulse causes more rotational motion than translational meaning Rw is greater than Vcm since it is hit above 2/3 its radius.This means friction (not sure if kinetic or static) will point opposite the direction of rotation. This makes sense since it will create a torque in the opposite direction of rotation causing rotation to decrease and translation to increase (will eventually start pure rolling). Alright so this means the impulse (2) simultaneously creates angular momentum and also linear momentum. With this I got:

2 = MVcm + Iw

I also have the net force equation:

F = Ma
f = Ma --> f is friction
μMg = Ma ----> I don't know μ so this doesn't get me anywhere

Then I looked at the torque about the center of mass of the ball:

∑τ = fR ----> I once again do not know μ so not helpful

Alright so this is where I've gotten. I have no clue where to go from here, could really use some assitance. Thanks in advance.

2. Dec 10, 2015

### ehild

Check the value of the impulse. The force of friction is much less then the average applied force, it can be ignored.

3. Dec 10, 2015

### Raios168

Sorry I don't quite understand what you mean. Do you mean that I made an error in calculating impulse and why can I ignore friction?

If I ignored friction, what would be causing rotational momentum to decrease?

4. Dec 11, 2015

### ehild

Yes, the numerical value for the impulse is not 2 Ns.
The problem asks " Find velocity of the center of mass right after the impulse" Friction will decrease the rotational motion gradually after the applied force ceased.

5. Dec 11, 2015

### Raios168

Uhm okay. I see my error now. I wrote down the wrong number for time. It's actually 10^-4 seconds. But that still doesn't solve my problem. Even with the correct value of impulse (which I am now pretty sure is 4 Ns), how could I find the velocity of the center of mass?

6. Dec 11, 2015

### ehild

What is the relation between impulse and change of momentum of a rigid body? What forces act on the ball?

7. Dec 11, 2015

### Raios168

Okay so the relation is this initial momentum plus impulse equals final momentum. In this case final momentum would be when pure rolling occurs correct?

So I'm getting:

For translational:
Mvi + μmgt = Mvf --(1)

For rotational:
Iωi - μmgRt = I(vf/R)

Dividing this by R: Iωi/R -μmgt = Ivf/R^2 ---(2)

(1) + (2) yields:

Mvi + Iωi/R = Mvf + Ivf/R^2

I now have 3 unkowns: vi, ωi, and vf. And I'm stuck. Where to go from here?

8. Dec 11, 2015

### ehild

No, the problem asks " Find velocity of the center of mass right after the impulse"
Till there is force, impulse is imparted to the body. When the force becomes zero, it is the end of imparting impulse. You need the velocity of the CM at that moment.
You overcomplicate the problem.

9. Dec 11, 2015

### Raios168

Okay!! I think I understand you. So, I know the change in momentum is equal to the external force which is equal to Ma. Change in momentum is equal to impulse which is 4 if I calculated it correctly. I also know the time it took for this was 0.0002 seconds. So I setup the following equation:

4/0.0002 = Ma
4/0.0004 = M(vcm/0.0004)
4 = Mvcm
4/0.02 kg = vcm
200 m/s = vcm

Could you confirm if I got the correct answer?

10. Dec 11, 2015

### ehild

You need only one equation: the impulse transmitted is equal to the change of momentum.
The force is given, and it is not 4/0.0002, it changes with time. The end result is correct, but the derivation is not.

11. Dec 11, 2015

### Raios168

But isn't change in momentum divided by time equal to the net force which is equal to Ma? I did 4 (the change in momentum) divided by time (0.0002) to equal Ma.

12. Dec 11, 2015

### ehild

Ma=F. If the force changes with time, so does the acceleration. You have the equation $\int{Fdt}=\Delta(mv)$. You get v directly from this equation, and it is the question, not the average acceleration during the impulse.