# Homework Help: Bowling ball velocity problem

1. Aug 23, 2010

### w3390

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

A bowling ball is initially bowled so that it doesn't roll with a translational velocity Vo. Show that when the velocity of the ball drops to (5/7)Vo, the ball will begin to roll without sliding.

I= (2/5)MR^2
mass= M

2. Relevant equations

V=wR

3. The attempt at a solution

I have had several unsuccessful attempts and my latest attempt has gotten me close to the right answer. The correct answer is (5/7)Vo, but I got sqrt(5/7)Vo. Here's what I did:

KE_T= KE_R
(1/2)MVo^2 = (1/2)MV^2 + (1/2)Iw^2
(1/2)Vo^2 - (1/5)R^2*w^2 = (1/2)V^2
Vo^2 - (2/5)R^2*w^2 = V^2
Vo^2 - (2/5)V^2 = V^2
Vo^2 = (7/5)V^2
V = SQRT(5/7)Vo

I am currently trying to figure out where I went wrong. Maybe I forget to account for friction. Any help would be much appreciated.

2. Aug 23, 2010

### Proofrific

Check how you went from angular velocity (w) to translational velocity (v).

3. Aug 23, 2010

### w3390

I'm sorry. Can you rephrase your question. I do not quite understand what you mean by check how I went from angular to translational.

4. Aug 23, 2010

### Proofrific

Whoops, sorry. I meant $$v = r \omega$$, but you did that right. My bad.

I don't see any problems with what you did. It looks like you did it right. Perhaps the solution is wrong?

5. Aug 23, 2010

### w3390

No. The solution is definitely correct. My instructor told all of us that if we got the answer I got that we made one particular mistake. Unfortunately, I cannot find it right now.

6. Aug 23, 2010

### Herr Malus

It looks like an angular momentum approach would actually be easier here. I recall having tried the energy approach when I had to do this problem, and never being able to get it to work. When I switched to conservation of angular momentum though, it clicked very quickly.

7. Aug 24, 2010

### hikaru1221

@w3390: Why do you think KE_T= KE_R? Energy is not conserved because before the ball reaches to the roll-without-sliding stage, it does slide. And what does it mean by "sliding"? Something about friction
@Herr Malus: Though the angular momentum of the ball about the lowest point is conserved, the reason behind this is not as simple as the "normal" angular momentum conservation law.

8. Aug 24, 2010

### Swap

I tried to take into account of the friction but I got (sqrt5)/3.

9. Aug 24, 2010

### hikaru1221

10. Aug 24, 2010

### Swap

ok distance travelled from the time the ball has been thrown and it starts to roll= (Vo^2-Vx^2)/2a. Vx being the final velocity, a being the accln due to friction.
Now I apply the conservation of energy theorem
0.5mVo^2=0.5mVx^2+0.5I(Vx/R)^2+ma(Vo^2-Vx^2)/2a
after solving this equation I get Vx=(sqrt5)/3.
The limitation of my answer is that I assumed the frictional force to be constant but I m not sure whether it is allowed or not.

11. Aug 24, 2010

### hikaru1221

You can assume so to simplify the problem. You can say, that friction is constant is a special case, so the result got from this case should match the needed result.

I make bold the wrong part. The wrong thing is that the velocity of the point where friction exerts on is not equal to the velocity of the center (remember the formula dW = Fvdt, where v is the velocity of the point where the force F exerts on?). Therefore, the work done by friction is not equal to (friction force x distance traveled by the center).

P.S.: Why don't you try force analysis, and instead of using F=ma, you can use $$X=m\Delta v$$ where X is the impulse.

Last edited: Aug 24, 2010
12. Aug 24, 2010

### w3390

I don't see how using impulse will make the problem any easier. Also, when the ball is sliding why wouldn't the velocity of the point where friction acts be the same as the center of mass. If the velocity of the center of mass was different than the velocity of the contact point, wouldn't the ball begin to roll?

13. Aug 24, 2010

### Staff: Mentor

Mechanical energy is not conserved (as already pointed out).
The ball begins to roll immediately due to friction. (It both rolls and slides.) You are looking for the point where the motion becomes rolling without sliding.

14. Aug 24, 2010

### Herr Malus

The reason I could never get the energy method to work was because I was never able to come up with an expression for the Work done by friction. As for it not being "simple", I beg to differ. You need a point where all external torque can either be countered or is not a factor. You already pointed out that the lowest point of the ball satisfies this, justification would solve the problem.

15. Aug 24, 2010

### w3390

@Herr Malus
So one of my options is to make my point of origin the contact point between the ball and the ground? I think this is what you mean by making torque not a factor because the radius component will be zero. Is that what you meant?

16. Aug 24, 2010

### Herr Malus

Yes, if you are using the point of contact between the ball and the floor you should see that torque is not a factor (hence angular momentum is conserved).

First you need to figure out which forces experienced by the ball may be a source of torque.

17. Aug 24, 2010

### w3390

Well I thought the only torque on the ball would be the friction force, but by making the contact point my origin I thought I eliminated that torque. The only other force acting on the ball would be gravity but that is antiparallel with the r component so it would just be zero. Am I wrong?

18. Aug 24, 2010

### hikaru1221

The work done by friction can be computed: $$dW = (v-\omega R)Fdt = (v-\omega R)mdv$$. From here, we need force analysis to derive the relation between v and omega. I'll leave the force analysis part for someone else.

Torque = 0 is just one reason. The general formula is NOT: torque = d(angular momentum)/dt. Though this formula is popular, it's only true under some certain conditions.

19. Aug 24, 2010

### Herr Malus

Correct, Gravity is acting parallel and produces no torque. Friction acts at the point of contact and produces no torque. This gives you the justification to set up the equation initial angular momentum equals final angular momentum.

@hikaru1221
-Educate me. What are the conditions?

20. Aug 24, 2010

### hikaru1221

I don't remember the exact formula, but the condition is $$\vec{v}_{center-of-mass}\times\vec{v}_{A}=0$$, where A is the point we consider to calculate angular momentum. You can prove this with prudence.
The problem is even more complicated, as there can be 3 kinds of points, which correspond to 3 kinds of motion, at the point of contact. But I'll not discuss this here, as this may confuse the OP. Anyway, caution must be taken when applying the angular momentum conservation law.