# Initiation of rolling without an opposite force.

1. Sep 6, 2008

### sganesh88

I'm driven crazy by this question these days. so here it goes.
Consider a ball sliding on a frictionless surface with some velocity U. If it encounters a surface with friction after some time, then it starts rolling (thats what my intuition says). Meaning that a new rotation component has been added to its motion along with translation. But we know that rotation can be initiated only in the presence of an opposite force thats not inline with friction (non zero lever arm). So how can it roll? It should just keep sliding with decreasing velocity. I know the question could appear silly but please bear. :-)

2. Sep 6, 2008

### atyy

The correct statement is that rotation about any particular point cannot be initiated unless the sum of (force X lever arm about that point) is non-zero. Friction has a lever arm about the centre of the ball, so it can rotate the ball about its centre.

3. Sep 7, 2008

### sganesh88

@atyy
The correct statement is that rotation about any particular point cannot be initiated unless the sum of (force X lever arm about that point) is non-zero
Consider a tall block on a frictionless surface. If we apply a force F at its top most point, do you think the block will trip over since we have a non-zero moment of F about the point in contact with the surface? No. It'll just have linear acceleration of magnitude F/m.

4. Sep 7, 2008

### HallsofIvy

You may be thinking of an object spinning about its axis in place, without any translation. That would require equal and opposite forces at diametrically opposite points. Here, you will have both turning about the axis and translation.

5. Sep 7, 2008

### Staff: Mentor

Whether it tips depends on how hard you push it, but that force definitely exerts a torque about the center of mass. If it were the only force exerting such a torque the block would rotate. But the normal force from the surface exerts a counter torque.

A simpler case that gets to your point would be to imagine that tall block lying on its side on a frictionless surface. If you push one end of it (perpendicular to its axis), does it rotate or translate or what?

6. Sep 7, 2008

### ibc

Try to consider 2 balls attached by a string, floating in space or free-falling, if you push one of the balls (vertically to the line of the string), in this case, it's very intuitive to see both effects: that the 2 balls as one must have a linear translation movement in the force's direction, and that the 2 balls must rotate around one another.
Try to see the resemblance between the two cases, and that it won't change alot if the string would be a rigid rod.

The fact is that any torque will cause an object to rotate, not only such torque that prevents the object from doing translation movement

7. Sep 7, 2008

### sganesh88

@Doc Al
If it were the only force exerting such a torque the block would rotate. But the normal force from the surface exerts a counter torque.
The normal force cannot exert a counter torque as it passes through the COM of the block.

A simpler case that gets to your point would be to imagine that tall block lying on its side on a frictionless surface. If you push one end of it (perpendicular to its axis), does it rotate or translate or what?
This would not result in rotation(toppling precisely) if the surface is perfectly frictionless. We just can't think of such a surface. Experiences drive intuition. Thats the problem! :-)

8. Sep 7, 2008

### Staff: Mentor

Nope. When you push the top of the block, the point of application of the normal force shifts to the front.
Note that this example was chosen because it has nothing to do with toppling. (Think of a stick lying on ice.) And it would both rotate and translate.

9. Sep 7, 2008

### firavia

these 2 attached figures , explain the phenomena , I hope it may help , ( friction force = Mu x normal force ).

Mu is the coefficient of friction and it is dependant on the floor type and its roughness.

#### Attached Files:

File size:
25.7 KB
Views:
80
• ###### friction force.jpg
File size:
28.5 KB
Views:
78
10. Sep 8, 2008

### ibc

http://ocw.mit.edu/OcwWeb/Physics/8-01Physics-IFall1999/VideoLectures/index.htm [Broken]
and watch lecture 21, if you want to get the whole rigid-body idea, watch lectures 19,20,21, and I'm sure you'll understand why 1 torque is enough to make an object rotate, but in the middle of lecture 21 there is an experiment, and then you'll SEE that indeed a rigid body rotates when there's only 1 force on 1 of it's side.

(and beside you might want to watch all the other lectures on that page, and other walter lewin's courses, they're very good)

Last edited by a moderator: May 3, 2017
11. Sep 10, 2008

### sganesh88

@Doc Al
Nope. When you push the top of the block, the point of application of the normal force shifts to the front.
Why should it? The application of a horizontal force shouldn't have varied the gravitational force on the block and hence no change in the magnitude, direction or point of application normal force as long as there is no other third party culprit like friction. :-)
or am i missing something here?

12. Sep 10, 2008

### atyy

Consider a pin balanced on its point on a surface without friction. Will you be able to tip it over?

13. Sep 10, 2008

### Staff: Mentor

You seem to think that the normal force always acts as if it were applied at the center of the block. Not so. (Imagine a block about to tip. Where does the normal force act then?)

What you're missing is the fact that the applied force exerts a torque about the center of mass, thus tending to rotate the object, which causes the effective position of the normal force to shift in order to prevent such rotation.

(The normal force is not simply the reaction force to gravity, by the way.)

14. Sep 10, 2008

### sganesh88

You seem to think that the normal force always acts as if it were applied at the center of the block. Not so.
No. I think that the point of application of the normal force wouldn't alter due to the horizontal force. Can you explain the reason?

What you're missing is the fact that the applied force exerts a torque about the center of mass, thus tending to rotate the object,
You can calculate torque w.r.t any point(or axis). That doesn't mean that the torque will cause rotation w.r.t it right? All we need is a F vector and an r vector.

The normal force is not simply the reaction force to gravity, by the way
Yes yes. A beautiful revelation when i started learning mechanics. :-)

15. Sep 10, 2008

### Staff: Mentor

How do you calculate the normal force? By applying the conditions for translational equilibrium perpendicular to the surface. For simple cases where you can model the objects as point masses, that's all you need. But for extended objects you also need to include the conditions for rotational equilibrium. (You certainly need the latter if you want the effective point of application of the normal force.)
If there's no other force (such as the normal force) providing a counter torque, sure it will begin to rotate.

16. Sep 10, 2008

### atyy

The above is correct!

But this is not. Yes, you can calculate the torque about any point, even a point that's not in the body. If there is a torque, the body will "rotate" about that point. This is indeed unintuitive - if I accelerate an object in a straight line - I can calculate my torque about any point, and surely it will be non-zero about some point. How can an object going in a straight line be rotating about some point? But think about an object orbiting the sun. The larger its orbit, the more nearly a straight line some small segment of it will be. So rotating about a distant point and straight line motion are not that different. What you should do to see if this makes any sense is treat a force causing a linear acceleration as a rotational problem and see whether you reproduce the answer you would get the normal way.

Edit: oops - torques are unintuitive - there's no torque when a planet orbits the sun!