Rolling down a slope vs to slide down a slope

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SUMMARY

The discussion centers on the differences in motion between a rolling cylinder and a sliding object down an incline. It establishes that while both objects experience the same translational motion, the rolling cylinder has additional rotational kinetic energy due to friction acting at its edges. The key takeaway is that the force applied at the center of mass (COM) results in different translational accelerations compared to forces applied at the edges, primarily due to the effects of torque and rotational motion. Ultimately, the rolling cylinder will have a greater total kinetic energy than the sliding object, as it converts some potential energy into rotational energy.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with concepts of translational and rotational motion
  • Knowledge of kinetic energy equations for rigid bodies
  • Basic principles of torque and friction in physics
NEXT STEPS
  • Study the derivation of the laws of motion for rigid bodies
  • Explore the concept of center of mass (COM) in detail
  • Learn about the effects of torque on rotational motion
  • Investigate the relationship between potential energy and kinetic energy in rolling versus sliding objects
USEFUL FOR

Physics students, educators, and anyone interested in understanding the dynamics of motion involving rolling and sliding objects on inclines.

  • #31
hikaru1221 said:
I'll borrow your picture for a while. Thanks in advance :biggrin:
It's Okay - you can keep it ;)

hikaru1221 said:
CASE 1: Rolling.
(...)
By the energy conservation law: PE = KE = translational KE + rotational KE
If this is a typo, and that you really mean PE => KE = (...) where arrow means "is being transformed to", I agree. Because PE = KE does not follow from the energy conservation law.

hikaru1221 said:
CASE 2: Sliding.
(...)
Now do you agree with me that total KE in case 2 (sliding) is smaller than in case 1 (rolling)?
Yes, I do :)

hikaru1221 said:
For the above questions of yours:
_ #1 and 2: Maybe you mean "rotational KE" instead of "rotational PE"?
Yes, I did :)

hikaru1221 said:
_ #3: Nature only knows static friction doesn't do any work (i.e. \vec{v}_A=0)
That's true! No energy is therefore lost to the environment. Mech. Energy is conserved.Thanks a bunch! :)
 
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  • #32
k4ff3 said:
I have corrected the rotational PE typo, I meant of course rotational KE.
Yes, I caught that.

What does it mean that the point of application of the force do not move? Why does it move when you apply the same force on the COM?
The instantaneous speed of the bottom of the cylinder as it rolls down the incline is zero. (Otherwise it would be sliding.) The speed of the COM of the cylinder is not zero, of course.

By leaking I meant that the body was constantly losing PE, i.e PE was transformed. Anyway, I understood you explanation - see if I got this right:

(Arrow here means "is being transformed to")
a) Rolling case: PE -> translational KE & rotational KE
b) Sliding case: PE -> translational KE & internal E to the "thing" that exerts the "braking force" f
c) Sliding case without any other force than gravity acting: PE -> KE

In a) and b) the translational KE is the same. However, in a) the PE transforms to energy that stays within the system, namely rotational energy. Thus, the total Mechanical Energy in a) will be conserved - as opposed to what happens in b): here energy is being transferred out to the environment (like the book). Therefore: Mechanical Energy is conserved in a), not in b) - but the translational KE are in both cases the same! The race will be a tie.
In c) the Mechanical Energy is clearly conserved, the sliding body will gradually lose PE in favor of gaining more (trans.) KE. No potential energy is lost to the environment, or to making rot. KE. Since the trans. KE and trans. velocity are connected, the cylinder in c) obviously wins the race!
Sounds good to me!

Can you please elaborate on this? My English isn't too good.. "In order for an external force to perform work, the point of application of that force must be displaced". Can you say this sentence with other words?
Work requires force X displacement, not just force. Since, as pointed out above, there's no instantaneous motion of the point of contact of the friction force, there's no instantaneous displacement--and no work done. This is a subtle point.

Here's another, totally different, example where a force acts to change the motion of something yet no work is done. Crouch down, then leap into the air. The ground exerts a force on your feet, propelling you upward, yet the ground doesn't move and no work is done. So where does your KE come from? (Answer: You've transformed internal energy--chemical energy in your muscles--into translational KE.)

Another example, involving static friction. Driving along, you step on the gas to accelerate your car. Assuming no slipping of the tires, the friction is static friction. And, again, the instantaneous speed of the tire patch in contact with the ground is zero so no work is done, even though, obviously, the ground is what propels the car forward. Where does the energy come from? The car converts chemical energy into mechanical energy--it burns gas.

Thank you very much! I feel like I'm on the verge of an epiphany :)
I think you are very close. As I said, some of these concepts are subtle, so don't be so hard on yourself.

Just for fun, why not use Newton's law to fully analyze the motion of the rolling cylinder? You can solve for the static friction required and the acceleration of the cylinder. And you can use the 'rolling without slipping' condition to relate rotation to translation.
 
  • #33
hikaru1221 said:
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"
Yeah...that second part was confusing to me. Really, it seems contradictory: if it is sliding without friction, then the friction force "f" must be zero.

In order to make it non-contradictory, we would have to completely eliminate the sliding without friction case. I see these cases right now:

1. Sliding without friction.
2. Rolling under the influence of gravity alone.
3. Rolling with a force equal to the friction force applied to the cog along the slope.

#1 & #3 are a tie, #2 is slower.

If #3 (vs #1) really is what the OP was looking for, it seems like a confusing and highly contrived scenario to me, but in that case it seems the other answers are correct.

I'm thinking it could also be a reading comprehension issue for me or organizational issue with the OP: the two sentences looked to me to be part of the same case, but maybe they are intended to be the two separate case #1 & #3.
 
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  • #34
russ_watters said:
Yeah...that second part was confusing to me. Really, it seems contradictory: if it is sliding without friction, then the friction force "f" must be zero.
In the sliding case, the force "f" is not friction, but a force equal in magnitude to the friction that exists in the rolling case that acts through the center. The point being to compare situations where the same force acts at different points.

In order to make it non-contradictory, we would have to completely eliminate the sliding without friction case. I see these cases right now:

1. Sliding without friction.
2. Rolling under the influence of gravity alone.
3. Rolling with a force equal to the friction force applied to the cog along the slope.
The three cases are:
1. Sliding without friction and no force besides gravity (and the normal force).
2. Rolling without slipping; the friction force happens to equal "f".
3. Sliding without friction and with an applied force equal to "f" acting through the center of mass.

The main comparison is between cases #2 and #3.
 
  • #35
k4ff3 said:
If this is a typo, and that you really mean PE => KE = (...) where arrow means "is being transformed to", I agree. Because PE = KE does not follow from the energy conservation law.

Okay, you may write PE => KE, no problem. But people usually write: PE = KE. Initially, it has PE, and finally it gets KE. No external work done on it, so energy is conserved: PE = KE. Of course, writing in your way shows much about the physics, but when dealing with math and calculation, you cannot avoid PE = KE.
 
  • #36
You could also have rolling with slipping... But I suppose there is enough confusion going about without it.
 
  • #37
russ_watters said:
(...)
If #3 (vs #1) really is what the OP was looking for, it seems like a confusing and highly contrived scenario to me, but in that case it seems the other answers are correct.
.

Doc Al's explanation of the cases is correct. Anyway, since I'm so good at asking questions, what is "a highly contrived scenario"? I've tried to look it up, but I can't make sense out of it in my native language. An online dictionary says: "Contrived = Obviously planned or calculated; not spontaneous or natural; labored". A "highly calculated scenario" just sounds wrong to me..
 

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