# Rolling down a slope vs to slide down a slope

1. Jul 23, 2010

### k4ff3

Consider the rolling cylinder in the figure

To describe the cylinders translational motion, I have

$$Ma=G sin \theta - f$$

$$a$$ is the acceleration of the cylinder parallel to the slope.

If you have an object (cylinder or not) sliding down the same slope without friction, with the same mass as the cylinder in the figure. And if you then apply a force through the center of mass equally big as the friction force $$f$$ above, with a direction parallel to the slope upwards (so that it works against the direction of the acceleration) - will this sliding body act exactly as the rolling body above (considering translational motion, not rotational)?

2. Jul 23, 2010

### hikaru1221

Why not?

3. Jul 23, 2010

### k4ff3

Maybe because the fact that the body is rotating will make it accelerate differently. Plus I dont find it intuitive that the friction force working on the edges of the cylinder will be as effective in changing the motion as a force acting directly in the COM. The frictional force creates a torque, while the force in the COM does not.

4. Jul 23, 2010

### hikaru1221

Well, speaking about COM, in my opinion, it's just a mathematical point which is useful for perceiving the motion. Since it's math, then let's talk about math. Math says, the equation of motion of COM is the equation of motion of a mass point, whose mass equal to the mass of the body, experiencing a force equal to the sum of all forces on the body. So the fact that the body rotates and frictional force acts at the edge or anywhere else (which directly affects only torque, and again, only rotation) doesn't affect the motion of COM (points don't rotate!). Trust math!

5. Jul 23, 2010

### k4ff3

I really don't think that's called math.

Anyway, consider this: You have a rod laying down on a frictionless surface.

If you hit the rod with a force near the edges two things will happen: a) it will get translational motion in the direction of the force, b) it will get rotational motion due to the torque given by the force.

Hit the rod again: If you hit the rod in COM, only a) occurs. This means that it will translate faster than the above. The translational acceleration is greater here than above. In my own clumsy words: "The force acting in COM is more effective in giving translational acceleration of the object".

Thus, the rod moves very differently in the two cases. And silly me draw a parallel to the rolling cylinder case. When the cylinder is rolling, you have a (friction) force (f) acting on the edges, and this gives a rotational motion. When the cylinder is sliding, and you apply a force (f), of equal magnitude as the friction force, through the COM, parallel to the direction of the friction force. You thus have no rotational translation, and my intuition then tells me that the force acting through COM is more effective in "stopping" the object from sliding downwards.

Well, I can't make sense out of it. So I asked here so that maybe some kind soul could clarify this for me.

Thanks

6. Jul 23, 2010

### hikaru1221

Yes, it's more "effective" because you gain rotation besides translational motion. Only rotation makes the difference, agree? If you leave rotation aside, as you wrote in the parentheses in post #1, then the rest is translational motion, and there is no difference in translational motion, correct? Plus the fact that translational motion of the body is actually described by the COM's motion, consequently, there is no difference in translational motion between when f is applied at the edge and when f is applied right at the center.

As I have pointed out, this is not true. The downward motion is in fact the translational motion, which has nothing to do with rotation.

Now the main source of your problem is, I think, the way people usually choose to perceive the motion of a body. In a body, there are infinitely points, which have various motions. You may arbitrarily pick one point and describe the other points' motions via that point, if you want. But physicists find it easy exploiting a mathematical point named COM, defined by the formula $$M\vec{r}_{COM}=\Sigma \vec{r}\Delta m$$. Officially, the "translational motion" term actually comes after COM. It is only after COM is defined is the term born. Translational motion of the body, in fact, is COM's motion. Rotational motion of the body is the other points' motion relative to COM. So when talking about translational motion, we talk about COM, and when talking about COM, we talk about a mass point which has advantageous features that I said earlier.

7. Jul 23, 2010

### K^2

hikaru's analysis is absolutely correct.

You really have to go through derivation of laws of motion for rigid bodies at least once to really see this. Try looking at Wikipedia's article on Rigid Body. It seems sufficiently detailed.

8. Jul 23, 2010

### k4ff3

So we have concluded that the rolling cylinder and the sliding cylinder will have the same translational motion. If they where to race down a slope, it would be a dead one.

What about kinetic energy? It arises from both translation and rotation, meaning that the rolling cylinder will have more KE than the sliding one. Where does this energy come from, then? How is the rolling sylinder able to obtain more KE than the sliding one?

9. Jul 23, 2010

### Staff: Mentor

As hikaru1221 has already explained, it doesn't matter where on an object you apply the force: the acceleration of the COM will be the same.

What might be throwing you off is your intuition that it's more difficult to apply the same force if the body turns than if you push it square towards its center of mass. That's perfectly true! The point of application moves away from you, so you have to move faster to continue to exert the same force. But if you are able to apply the same force at the edge of the rod or at the center, you'll get the same acceleration of the COM.

10. Jul 23, 2010

### Staff: Mentor

I'm not sure if I missed something here, but it seems like people are saying that whether or not there is friction, the object will accelerate down the ramp at the same speed. This isn't true. And the OP figured out probably the clearest way to show why:
This is correct: when the cylinder gets to the bottom of the slope, if it is sliding, all of the KE is in the translational motion, but if it is rolling, some is in the translational and some is in the rotational. By conservation of energy, the translational KE (and therefore speed) of the rotating cylinder must be lower than that of the sliding cylinder.

11. Jul 23, 2010

### hikaru1221

For a body, KE = translational KE + rotational KE. Since the net forces in the 2 cases are the same, the translational speeds are the same, and thus, translational KEs are too. Again, the difference lies in rotation. So the difference has something to do with rotation. Here is what rotation does: it affects the velocity of the point A that friction acts on. Since work element done by friction is $$dW=\vec{F_{friction}}\vec{v_{A}}dt$$, the works done by friction in the 2 cases are different. |vA| in the rolling case is smaller, but friction does negative work, so eventually, KE gained in the rolling case greater.

12. Jul 23, 2010

### Staff: Mentor

While I understand this, I'm not sure how it answers the OP's question. The applied force in both cases is the weight of the cylinder....but the reaction forces are different in each case, so the "unbalance" of the cylinder is different. This is not a case like the one you just described.

Specifically, the friction force causes the cylinder to spin while opposing (reducing) its translational acceleration.

13. Jul 23, 2010

### hikaru1221

The OP made up 2 situations:
1 - there is friction, and it's rolling
2 - there is no friction, and it's sliding, but there is another force of the same magnitude of friction, applying at the center, acting upwards.
If you understood the problem that way, you may rethink of the last conclusion you made.

14. Jul 23, 2010

### Staff: Mentor

The KE gained in the two cases cannot be different because conservation of energy requires that the total energy of the system is constant throughout the entire experiment. The energy available is the potential energy at the start of the experiment and at the end of the experiment, the resulting KE must be exactly equal to that PE.

Consider the exaggerated example of a yoyo. By changing the diameter of the shaft, you can change where the force is applied to the cylinder. The smaller the diameter, the more of the energy is converted to rotational KE and the slower the yoyo will fall.

15. Jul 23, 2010

### K^2

I see what russ' complaint is, and if this is exactly what OP was talking about, russ has a point.

The friction force in rolling case will not be equal to friction force if you restricted cylinder to slide, rather than roll. In first case, the force will be determined by angle of the incline and the moment of inertia. In the second, by kinetic friction coefficient.

But the way OP stated it, I got the impression that he simply applies a force equivalent to friction in the rolling case to the cylinder. In which case, hikaru's logic applies.

Might be nice if OP could clarify that.

16. Jul 23, 2010

### Staff: Mentor

Key words in the OP: "...sliding down the same slope without friction..." Obviously, if it is sliding and not rolling, there can be no friction and vice versa - if there is no friction, it will be sliding, not rolling.

17. Jul 23, 2010

### hikaru1221

This is what the OP said in post #1:

"If you have an object (cylinder or not) sliding down the same slope without friction, with the same mass as the cylinder in the figure. And if you then apply a force through the center of mass equally big as the friction force f above"

18. Jul 23, 2010

### k4ff3

These two figures illustrate the rolling case and the sliding case respectively:

Rollin'

---
Slidin'
[PLAIN]http://img213.imageshack.us/img213/8353/frce.png [Broken]

Let me break down my questions:
• Which one wins the race?
• If it's a dead race, as some here claim, then what about the KE?
• If the opposite is true, then it does matter where on an object you apply the force (in contrast to what Doc Al claimed). Which I also find non-intuitive.

Im sorry if I have caused confusion. I hope this clear things up.

--
Credits go to mspaint.exe

Last edited by a moderator: May 4, 2017
19. Jul 23, 2010

### Staff: Mentor

It's a tie.
What about it? In the rolling case, the friction force (static friction) does no work, but it does transform gravitational PE into rotational KE. In the sliding case, the applied force does negative work.
As far as the acceleration of the COM is concerned, it doesn't matter where the force f is applied. But the overall motion of the cylinder is certainly different. In one case it rotates; in the other, it doesn't.

20. Jul 23, 2010

### k4ff3

.. so the PE=mgh does not apply for rotating bodies, because PE is being transformed into KE (so that PE<mgh)??