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Original Post : here

On the accepted answer , it was said that the Normal Force is more on the right side of the centre of mass which provides an anti-torque to the rotation of the body which slows down the rolling.

I also found some similar explanations on "Why a rolling Body Slows Down" in the book "Concepts of Physics by HC Verma"

pic1 pic2

In the second picture , you can see that it is written that the Normal Force is shifted Right of the center of mass because the front part pushes the surface a bit more . Here it is :

In fact, when the sphere rolls on the table, both the sphere and the surface deform near the contact. The contact is not at a single point as we normally assume, rather there is an area of contact.The front part pushes the table a bit more strongly than the back part. As a result the normal force doesnt pass through the center, it is shifted towards the right. This force then has an anticlockwise torque. The net torque causes an angular deceleration.

But it is not explicitly explained(neither in the book , nor in the answer of the above mentioned post) why the front side pushes it a "bit more" than the back side.

Why does this happen?

Jdeep
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  • The front part pushes the table a bit more strongly than the back part What the hell,-What's the reason for that ? If mass of wheel is distributed uniformly, then normal forces should be symmetric in both sides. Also if it is like author says,- then wheel made hole in ground must be asymmetric too and not circular one. Is there any proof for that ? – Agnius Vasiliauskas May 26 '20 at 16:22
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    @AgniusVasiliauskas , Currently I don't have a proof for that. In the OP question also the accepted answer had a similar explanation for anti clockwise torque but it didn't explain the reason for asymmetrical distribution of the Normal Force. – Jdeep May 26 '20 at 16:31
  • @AgniusVasiliauskas In this case we are not assuming rigid bodies/surfaces – BioPhysicist May 26 '20 at 16:41
  • @BioPhysicist ok, then explain how follows from continuum mechanics that normal forces are asymmetric and why forces depicted asymmetrically, author depicts wheel made hole in symmetrical way? It doesn't make sense – Agnius Vasiliauskas May 26 '20 at 16:48
  • @BioPhysicist If somebody will find out that wheel made hole in current movement position is asymmetric too - then I will start to believe in this counter torque explanation. Until then ocams razor cuts this explanation out. – Agnius Vasiliauskas May 26 '20 at 16:55
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    @AgniusVasiliauskas , I have added a possible explanation below. Please see if it has some error. – Jdeep May 27 '20 at 15:18
  • I'm guessing because the writers thought it was obvious. If something is rolling in one direction, there's a component of the rotation force that will increase the forces on the "side" where it is rotating. If the forces were all identical, there would be no motion. – iheanyi May 27 '20 at 15:53

4 Answers4

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I clicked some photos of circular duct tape . I intentionally pressed the duct tape hard so that the deformation can be seen.

At the Normal position : pic 1

Now , In the next infinitesimal time $dt$ , lets say the tape covers a small distance $dx$ .

Here is a picture of it : pic2

As you can see , in the small time interval $dt$ , the back part of the tape was still deformed due to which when the tape was rotated , the point(s) of contact somewhat shifted towards the right ( points of contact was more on the right side of the centre). That might be the same reason why a fully inflated football rolls for a longer time than the one which is partially inflated.

Due to this (I think) , the Normal force is shifted "a bit" right

Note : Since this is only an observation and I do not have any mathematical proof for this , if you feel like there is some error in the observation , then comment below.

Jdeep
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  • first picture is Ok, because deformation is symmetric as it should be, but Im not sure about second. Where does this asymmetry comes from ? I could believe in such asymmetry at $t_0$ time, because body generates acceleration and thus due to acceleration forward part of wheel can experience bigger load. But as it starts to move in constant speed - highest load should return back under COM of wheel (first picture). Besides it unreasonable that contact area after $dt$ time passes gets smaller - area should be the same, just it will be shifting along movement path. Sorry, but unconvincable. – Agnius Vasiliauskas May 27 '20 at 16:13
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    @AgniusVasiliauskas it's not an asymmetry of deformation, it's a tilt of the not yet re-deformed object. – Ruslan May 27 '20 at 16:34
  • @AgniusVasiliauskas , Thanks for the comment! . I have replaced the second image so that the symmetry is visible – Jdeep May 27 '20 at 16:34
  • @AgniusVasiliauskas , I believe the asymmetry is caused because the initially "deformed" part has not yet "reformed" in the time $dt$ . Due to this , the deformed part is not in contact of the surface. and hence experiences no Normal Force – Jdeep May 27 '20 at 16:39
  • @NoahJ.Standerson. I made my own test of rolling toilet paper, check this out here. As you see no any assymetry or tilting, deformed area just slides along movement direction. In last frames you can however see some tilting, but this is just due to changed my elbow angle, thus projected pushing force vector changes and some tilting arises in the end. But normally if you push wheel straight to COM and forward horizontally - no any tilting/asymmetry seen. Convincing ? – Agnius Vasiliauskas May 27 '20 at 16:53
  • @AgniusVasiliauskas , I saw the video. But stop the video at 0:04 . Imagine a line through the centre of the toilet paper. Now compare the bottom part of the toilet paper in contact with the surface along the imaginary – Jdeep May 27 '20 at 17:03
  • @AgniusVasiliauskas Can I show a snapshot of your video in the post? – Jdeep May 27 '20 at 17:07
  • Of course you are welcome, make a chart on top of it / edit it if you wish, cause my imagination is poor ;-) I compare both parts from COM visually to the left and to the right, but see no difference. – Agnius Vasiliauskas May 27 '20 at 17:08
  • @AgniusVasiliauskas , okay I added the image – Jdeep May 27 '20 at 17:13
  • @NoahJ.Standerson Hey, but make also an an image from frames 0:01 or 0:02 seconds - cause there are no such asymmetry which you see starting from 0:04. Why asymmetry is only starting from approx 1/2 of movie ? No you can't explain ? I can - as I said from some 1/2 of movie my elbow starts changing direction, so pushing vector becomes gradually less horizontal, that's why where is such difference. – Agnius Vasiliauskas May 27 '20 at 17:21
  • Let us continue this discussion in chat. – Jdeep May 27 '20 at 17:23
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But it is not explicitly explained(neither in the book , nor in the answer of the above mentioned post) why the front side pushes it a "bit more" than the back side.

It is due to the viscoelastic behavior of the contacting materials.

For purely elastic materials the relationship between stress and strain is linear so that the loading and unloading (compressing and uncompressing) forces are equal. See the diagram at the left below.

Viscoelastic materials behave like elastic materials in that both eventually recover from deformation when the load is removed. See diagram to the right below. However, the viscous behavior of a viscoelastic material is such that the stress (force) during unloading is less than that during loading for the same amount of deformation giving the material a strain rate dependent on time. The area in red between the loading and unloading curves represents the hysteresis heat loss. In contrast with ideal elastic behavior, the deformation when the material is viscoelastic does not recover right after the load is removed. In other words, there is a time delay for the material strain to fully recover, which is not shown in the diagram to the right.

In terms of say a tire rolling, the above means the forces acting on the leading portion of the tire (in the direction of motion) in contact with the road under compression (loading) are greater than the forces acting on the trailing portion of the tire in contact with the road under decompression (unloading). The overall result is the difference between the compression and decompression forces results in a net torque counter to the rotation of the tire.

Hope this helps.

enter image description here

Bob D
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The front of the sphere is moving toward the ground, the back away (as per clockwise rotation). Therefore, momentum from a tiny piece of sphere there is more downward force on the front side than the back, deforming the ball/ground more.

nortcele
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Suppose the ball is moving in the vacuum, the floor surface is perfectly horizontal and ball and floor are made from hard material.

When an horizontal force is applied to start the movement, the ball initially slides, until the torque mentioned in the previous post, caused by the friction force, transforms the sliding in rotating movement.

In that process the ball loses translational kinetic energy (the friction force opposes the velocity), but acquires rotational kinetic energy.

Once the ball is rotating without slip, there is no friction force opposing the velocity.

The only process that can take energy from the ball is elastic deformation in the region of contact, that transform it from a point into an area. I believe it is illustrated in the reference of your post. But that effect is relevant in a soccer ball for example, and very small in a bowling ball.

Claudio Saspinski
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