# Homework Help: A 0.5 kg ball sits on a cart

1. Apr 29, 2015

### pgardn

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

I am a substitute today for a physics class. I am baffled by the following and could use some help.

A 0.50 kg ball sits on a cart. When the cart is pulled to the right by a net force of 3N, friction exerts a force of 2.0 N on the bottom of the ball. The ball rolls without slipping, how far will the ball roll relative to the ground in 0.10S.

2. Relevant equations:
I took this to be a 2nd law problem with kinematics as the mass of the cart is not given. And there is nothing about I or the radius of the ball on top of the cart.

F=ma delta x = vot + 1/2at^2

3. The attempt at a solution

I decided there was one force in the horizontal acting on the ball to the left, so F=ma.
2N = 0.5 kg (a) I assumed this 2N force was to the left a = 4 m/s/s

Now delta x = 0 + 1/2at^2 delta x = 1/2(4)(0.1)^2 delta x = 0.02 m as an answer

But then I started to think relative to the ground. And since the ball is rolling on a cart that is accelerating right, I needed to determine how far the cart would move. But I had no info on the mass of the cart. So I thought the ball really just had one force acting on it and would begin to move on the cart like it was frictionless and act like some sort of wind was applying a force of 2N to the left. Then I stopped in complete bafflement.

[Mentor's note: Removed bold formatting from text content]

Last edited by a moderator: Apr 29, 2015
2. Apr 29, 2015

### Suraj M

Do you think this is correct? If the force is to the left then the ball should move to the right as the force is parallel to the cart surface.. is that what happens?
And could you check again if the mass of the cart is given.

Last edited: Apr 29, 2015
3. Apr 29, 2015

### pgardn

Excuse me, the force on the surface of the ball is to the right causing it to rotate counter clockwise.
And no mass was given for the cart. I found the answer key and it turned out that I was correct. But I dont know why for sure. I have reservations about my explanation.

4. Apr 29, 2015

### haruspex

As you note, since we are not told the cart's mass, the force exerted on the cart must be irrelevant. Just consider the forces on the ball.
I don't like this wording:
The ball is rolling relative to the cart, in the sense that its movement relative to the cart consists of rolling. But it doesn't really mean anything to say it is rolling relative to the ground. It is moving relative to the ground, and rotating, but the two are not related by a rolling constraint.
Anyway, from the given answer, it seems they want how far its mass centre moves relative to the ground.
Your method is fine. The rolling turns out to be irrelevant. This feels surprising - you naturally think the ball will go a shorter distance because it is allowed to roll back. But that's because we tend to think of the cart's movement as constant, i.e. the same whether the ball rolls or not. However, it's the frictional force that's given, and to achieve the same frictional force for a ball that's allowed to roll as for a box not sliding, the cart has to accelerate faster.

5. Apr 30, 2015

### pgardn

Thank you very much. I got it. I think anyway.

So the application of the force at the ball/cart surface gives the ball's CM the same acceleration as if was applied along the CM.
This does seem counter intuitive if one thinks of the work done on the ball. So if the ball was in space and the same force was applied along the CM, it would not cause the ball to rotate. So it would have less KE compared to the same force applied to the same ball but at the surface of the ball. So I imagine the Fx d work; well the d part must differ I guess. And the amount of time further confuses me. But that was not in the original question.

I am suspect of my knowledge of rotation now. If you could add anything to the above to slap the stupid out of me I would appreciate it.

And thank you for your time.

6. Apr 30, 2015

### haruspex

Yes, if the force is applied at the perimeter, causing the ball to rotate, then the force acts over a greater distance than the CoM of the ball moves.

7. Apr 30, 2015

### pgardn

Thanks very much.

8. Dec 16, 2017

### nbsmith

It isn't clear to me how this problem can be approached without using rotational dynamics. The friction on the ball exerts a torque which causes the ball to roll relative to the cart. This would affect the ball's motion relative to the ground. I did this problem using rotational dynamics and got an answer of 0.014 m (1/70 m to be exact).

9. Dec 16, 2017

### haruspex

The only horizontal force on the ball is 2N to the right. Applying ΣF=ma, the ball's mass centre accelerates to the right at 4m/s2.

10. Dec 16, 2017

### nbsmith

But that 2 N force is acting on the edge, so that force will result in some translational acceleration, and some angular acceleration. If all of the force were acting at the CM, the ball's acceleration would be 4 m/s2, but since it's not, the translational acceleration will be less than 4, and the angular acceleration will be greater than 0. Right?

11. Dec 16, 2017

### haruspex

No, that is a common error. I tried to explain this at the end of my post #4.
Two ways I can answer you on this:
1. Appeal to Newton's laws. The law is ΣF=ma. If the net force is 2N and mass is 0.5kg then the mass centre acceleration is 4m/s2. Newton made no exceptions forces applied off-centre.
2. Your intuition is that the force is from a source such as a descending mass or a human pulling. In both cases, if the mass resists less then it will be harder to exert the same force, indeed the force will tend to be less, so less mass centre acceleration. But here the force is an absolute given, so the ability of the mass to rotate merely means that the point on the ball contacting the cart accelerates faster. In a given time, the force advances further so does more work. That extra work is what provides the rotational KE.

12. Dec 16, 2017

### nbsmith

I want to clarify what you're saying with a different example. Let's imagine a ball in space (with no other outside forces acting on it). In the top picture of the diagram, it's easy to see that the ball would accelerate at 2 m/s2.

If we take that same ball and wrap a string around it a few times, then pull that string with the same 8 N force, you're saying that the ball would still accelerate at 2 m/s2? (I see this scenario as being identical to the ball on the cart; correct me if there's a difference.)

I think I'm starting to see why. Since the ball would rotate some, to actually apply a constant 8-N force, you'd need to "pull harder" to make up for the otherwise decreasing tension in the string (because of the ball's acceleration). Keeping the tension "higher" (that is, at a constant 8 N), keeps the acceleration at a constant 2 m/s2.

Hmm, interesting. Thanks.

13. Dec 16, 2017

### haruspex

Yes, that is all correct.