Wheel rolling on a horizontal surface

In summary, when a wheel is rolled on a horizontal surface, it experiences both linear and rotational motion. The motion of the wheel can be described using the principles of physics, such as Newton's laws of motion and the conservation of energy. Friction also plays a crucial role in determining the speed and direction of the rolling wheel. Depending on the initial conditions and external forces, the wheel can roll smoothly, slip, or even skid on the surface. Overall, the motion of a rolling wheel on a horizontal surface is complex yet predictable, making it a fascinating topic for study in physics.
  • #1
Jahnavi
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Homework Statement


wheel.jpg

Homework Equations

The Attempt at a Solution



Since the wheel is rolling , vcm=ωr , acm = αr

The wheel is in pure rotation about the point O where it touches the surface .

The point P is a distance 4/3 from the bottom point .

Angular acceleration of the point = 3 rad/s2

vP = 4m/s

Radial acceleration of the point P about bottom point O = vP2/(4/3) = 12 m/s2 .

Are the two components of acceleration correct ?
 

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  • #2
One is to assume that you are looking for the components of the linear acceleration. Angular acceleration is not one of them.
 
  • #3
Is radial acceleration the net linear acceleration of point P ?
 
  • #4
Jahnavi said:

Homework Statement


View attachment 214723

Homework Equations

The Attempt at a Solution



Since the wheel is rolling , vcm=ωr , acm = αr

The wheel is in pure rotation about the point O where it touches the surface .

The point P is a distance 4/3 from the bottom point .

Angular acceleration of the point = 3 rad/s2

vP = 4m/s

Radial acceleration of the point P about bottom point O = vP2/(4/3) = 12 m/s2 .

Are the two components of acceleration correct ?
From where u got 4/3
 
  • #5
Abhishek kumar said:
From where u got 4/3

The point whose acceleration is to be calculated is at a distance (4/3)m from the bottom most point .
 
  • #6
Jahnavi said:
Is radial acceleration the net linear acceleration of point P ?
The linear acceleration consists of the radial acceleration (which you found) and the tangential acceleration (which follows swiftly from the angular acceleration).
 
  • #7
haruspex said:
The linear acceleration consists of the radial acceleration (which you found) and the tangential acceleration (which follows swiftly from the angular acceleration).

Tangential acceleration = 4m/s2

Radial acceleration = 12 m/s2

Net acceleration = √(122+42) = 4√10 m/s2

Is that correct ?
 
  • #8
Jahnavi said:
Tangential acceleration = 4m/s2

Radial acceleration = 12 m/s2

Net acceleration = √(122+42) = 4√10 m/s2

Is that correct ?
Yes, except that what you called net acceleration is the magnitude only. Acceleration is a vector, so I would think you should give the components separately in the answer. Either that or magnitude and direction.
 
  • #9
haruspex said:
Yes, except that what you called net acceleration is the magnitude only. Acceleration is a vector, so I would think you should give the components separately in the answer. Either that or magnitude and direction.

Ok.

I suppose the question is asking for the magnitude of linear acceleration of the point .

But unfortunately , this answer is wrong .

Do you see any mistake ?
 
  • #10
Jahnavi said:
The point P is a distance 4/3 from the bottom point .

Angular acceleration of the point = 3 rad/s2

vP = 4m/s

Radial acceleration of the point P about bottom point O = vP2/(4/3) = 12 m/s2 .

I think there might be a problem with the way you are obtaining the radial acceleration. By the logic you used, the center of the wheel is 1 m above point O, so the center of the wheel has a radial acceleration of 9 m/s^2 (i.e., [3^2]/1). That's not right. The center of the wheel should have a radial acceleration of 0.

Now consider a point at the edge of the wheel. It will travel in a cycloid pattern:
Cycloid_f.gif

As you can see, the radius of curvature when the point is on top is significantly greater than 2R. I think the flaw in your logic is the assumption that the entire wheel is rotating around point O. It doesn't really work that way. Even though things are also accelerating linearly (in addition to rotational acceleration), things are still "rotating" (strictly speaking) about the center of the wheel.

The point you are concerned about isn't at the center or the edge, but rather somewhere in between. So as a sanity check, I would expect the radial acceleration to be somewhere in between what you would get for those other two points (center or edge).

[Edit: Source of animated gif: https://en.wikipedia.org/wiki/Cycloid#/media/File:Cycloid_f.gif]
 

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  • #11
As another point of reference to ponder, a point at the very bottom of the rotating wheel (at point O) has a instantaneous velocity of 0 m/s. But is its instantaneous, radial acceleration also zero?

[Hint: The answer is "no."]
 
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  • #12
Jahnavi said:
The point whose acceleration is to be calculated is at a distance (4/3)m from the bottom most point .
Radius is 1m given
collinsmark said:
I think there might be a problem with the way you are obtaining the radial acceleration. By the logic you used, the center of the wheel is 1 m above point O, so the center of the wheel has a radial acceleration of 9 m/s^2 (i.e., [3^2]/1). That's not right. The center of the wheel should have a radial acceleration of 0.

Now consider a point at the edge of the wheel. It will travel in a cycloid pattern:
View attachment 214738
As you can see, the radius of curvature when the point is on top is significantly greater than 2R. I think the flaw in your logic is that the entire wheel is rotating around point O. It doesn't really work that way. Even though things are also accelerating linearly (in addition to rotational acceleration), things are still "rotating" (strictly speaking) about the center of the wheel.

The point you are concerned about isn't at the center or the edge, but rather somewhere in between. So as a sanity check, I would expect the radial acceleration to be somewhere in between what you would get for those other two points (center or edge).

[Edit: Source of animated gif: https://en.wikipedia.org/wiki/Cycloid#/media/File:Cycloid_f.gif]
collinsmark said:
I think there might be a problem with the way you are obtaining the radial acceleration. By the logic you used, the center of the wheel is 1 m above point O, so the center of the wheel has a radial acceleration of 9 m/s^2 (i.e., [3^2]/1). That's not right. The center of the wheel should have a radial acceleration of 0.

Now consider a point at the edge of the wheel. It will travel in a cycloid pattern:
View attachment 214738
As you can see, the radius of curvature when the point is on top is significantly greater than 2R. I think the flaw in your logic is that the entire wheel is rotating around point O. It doesn't really work that way. Even though things are also accelerating linearly (in addition to rotational acceleration), things are still "rotating" (strictly speaking) about the center of the wheel.

The point you are concerned about isn't at the center or the edge, but rather somewhere in between. So as a sanity check, I would expect the radial acceleration to be somewhere in between what you would get for those other two points (center or edge).

[Edit: Source of animated gif: https://en.wikipedia.org/wiki/Cycloid#/media/File:Cycloid_f.gif]
Pattern you have shown in figure is when we work from frame other than wheel frame
 
  • #13
collinsmark said:
I think the flaw in your logic is the assumption that the entire wheel is rotating around point O.

Isn't the wheel rotating around an axis passing through the bottom most point O ?

collinsmark said:
I think there might be a problem with the way you are obtaining the radial acceleration.

What exactly is the problem with my approach ? I have done few problems considering the bottom most point as the instantaneous axis and obtained correct answers .

So why exactly is it failing this time ?
 
  • #14
Abhishek kumar said:
Pattern you have shown in figure is when we work from frame other than wheel frame

Yes, the pattern in the animated gif shows a frame corresponding to the flat, horizontal surface. I'm guessing that the values in the problem statement (the 3 m/s velocity of the center and the 3 m/s^2 horizontal acceleration of the wheel's center) are all relative to the flat, horizontal surface.

But for the moment, it might help to use a non-rotating, accelerating frame that corresponds to the center of the wheel, by subtracting out the 3 m/s^2 in the horizontal direction. This way, the center of the wheel isn't accelerating in the new frame. (You can always add it back into the x-component later, before your final answer). Now everything is rotating around the center. What's the instantaneous, radial acceleration of the specified point now? What direction is it in?

So when we change back to the frame of reference of the flat surface, is the acceleration that we add back in perpendicular to the point's instantaneous, radial acceleration?
 
  • #15
Jahnavi said:
Isn't the wheel rotating around an axis passing through the bottom most point O ?

What exactly is the problem with my approach ? I have done few problems considering the bottom most point as the instantaneous axis and obtained correct answers .

So why exactly is it failing this time ?

The bottom of the wheel is accelerating relative to the centre of the wheel. So, its acceleration is not the same as the centre.

Acceleration must be the same in all inertial frames. What happens if you consider a frame moving at ##3m/s##?
 
  • #16
Jahnavi said:
Ok.

I suppose the question is asking for the magnitude of linear acceleration of the point .

But unfortunately , this answer is wrong .

Do you see any mistake ?
Yes, I do, belatedly.
 
  • #17
Jahnavi said:
Tangential acceleration = 4m/s2

Radial acceleration = 12 m/s2

Net acceleration = √(122+42) = 4√10 m/s2

Is that correct ?

That is acceleration relative to point O, whose acceleration you would then need to calculate.
 
  • #18
haruspex said:
Yes, I do, belatedly.

Please explain the mistake ?
 
  • #19
PeroK said:
That is acceleration relative to point O, whose acceleration you would then need to calculate.

Isn't wheel undergoing pure rotation about O ?
 
  • #20
Jahnavi said:
Isn't wheel undergoing pure rotation about O ?

Yes, but point O is accelerating. The bottom of the wheel is instananeously at rest, but it is accelerating!
 
  • #21
PeroK said:
Yes, but point O is accelerating. The bottom of the wheel is instananeously at rest, but it is accelerating!

What does it mean when it is said that the bottommost point is the instantaneous axis ?
 
  • #22
Jahnavi said:
What does it mean when it is said that the bottommost point is the instantaneous axis ?

You can calculate the instantaneous acceleration of this axis of rotation. It is not zero.

If you consider a simpler problem where the wheel is not (linearly) accelerating, then every point on the rim of the wheel is accelerating. In fact, every point except the centre is accelerating towards the centre.

You cannot ignore this acceleration.
 
  • #23
PS In fact, a good test of your method would be to calculate the acceleration of the centre of the wheel.
 
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  • #24
I would like to understand what does it mean when a rolling object is said to be undergoing pure rotation about instantaneous axis at the bottom most point ?

If it's pure rotation then why is radial acceleration of point P w.r.t O wrongly calculated ?
 
  • #25
Jahnavi said:
I would like to understand what does it mean when a rolling object is said to be undergoing pure rotation about instantaneous axis at the bottom most point ?

If it's pure rotation then why is radial acceleration of point P w.r.t O wrongly calculated ?

It's not. You correctly calculated the acceleration of point P relative to point O. Your problem is that point O is accelerating.
 
  • #26
Jahnavi said:
Isn't wheel undergoing pure rotation about O ?
You could, hypothetically, choose an accelerating frame where point O is the center (where O is on the bottom of the wheel) but it's not very useful in solving this problem.

When using the frame of the flat surface, then no, the wheel is not rotating around point O, not even instantaneously.

Instantaneously, every point on the wheel is rotating around the center of the wheel. (The center of the wheel is accelerating linearly too, meaning that that bit of acceleration carries over to all points on the wheel.)

Let me put it another way. For every point on the wheel, you can break up its acceleration vector in the following ways:
The linear acceleration corresponding to the acceleration of the wheel's center, plus
its rotational acceleration about the wheel's center.

You can further break up the point's linear acceleration (which matches the acceleration of the wheel's center, for all points on the wheel) into its
x-component and
y-component.

You can further break up the point's rotational acceleration (depends on the point's distance from the wheel's center and the wheel's angular acceleration) into
tangential component and
radial component.

Once you know all 4, then you can add things (vector sum) back together at the end, once you know how the angles line up (how tangential&radial correspond to x&y).
 
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  • #27
PeroK said:
It's not.

Sorry . Isn't wheel undergoing pure rotation about the bottom most point O ?
 
  • #28
collinsmark said:
When using the frame of the flat surface, then no, the wheel is not rotating around point O, not even instantaneously.

Then why is an axis through the bottom most point called as the instantaneous axis of rotation ?
 
  • #29
Jahnavi said:
Sorry . Isn't wheel undergoing pure rotation about the bottom most point O ?

I'm not going to discuss this any more until you have calculated, using your method, the acceleration of the centre of the wheel.

In fact, I'll do it for you:

Radial acceleration of centre of wheel relative to point O is ##9m/s^2## (vertically downward).

Is the centre of the wheel accelerating downward? Or, is the bottom of the wheel accelerating vertically upward at ##9m/s^2##? In the inertial reference frame of the surface?
 
  • #30
PeroK said:
Is the centre of the wheel accelerating downward? Or, is the bottom of the wheel accelerating vertically upward at ##9m/s^2##? In the inertial reference frame of the surface?

The bottom is accelerating vertically upward w.r.t the center . Since center is accelerating horizontally , in the frame of the surface , net acceleration is the resultant of the two .
 
  • #31
Jahnavi said:
The bottom is accelerating vertically upward w.r.t the center . Since center is accelerating horizontally , in the frame of the surface , net acceleration is the resultant of the two .

Exactly. The best approach is to analyse the problem relative to the centre of the wheel, as its acceleration is known.

Also, note that the instantaneous point of contact with the surface is accelerating upward absolutely (not just relative to the centre). The centre has no vertical acceleration.
 
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  • #32
I pretty much agree with you and @collinsmark .

I am confused about something else .

My doubt is - Is wheel undergoing pure rotation about an axis passing through the bottom most point ?

If not , then why is it called instantaneous axis of rotation ?

Please address this .
 
  • #33
Jahnavi said:
I pretty much agree with you and @collinsmark .

I am confused about something else .

My doubt is - Is wheel undergoing pure rotation about an axis passing through the bottom most point ?

If not , then why is it called instantaneous axis of rotation ?

Please address this .
I've never heard the bottom point of a wheel on a flat surface called the "instantaneous axis of rotation." Can you provide a reference of it being called that?

(I could imagine it being called that if a rigid wheel was moving over a sharp rock, or maybe a wheel slowly rolling off of a sudden cliff. But I don't think it's an accurate description for a wheel on a flat surface.)
 
  • #35
Jahnavi said:
I pretty much agree with you and @collinsmark .

I am confused about something else .

My doubt is - Is wheel undergoing pure rotation about an axis passing through the bottom most point ?

If not , then why is it called instantaneous axis of rotation ?

Please address this .

It all depends what you mean by "pure" rotation. In this case, the centre of the rotation is actually an accelerating reference frame. In fact, in the reference frame of any point on the wheel, the motion is a "pure" rotation about that point. You can always decompose the motion of a rigid body as the motion of any single point and a rotation of the body about that point. The most useful case is often to consider the motion of the COM, because the motion of the COM is determined by the external forces.

In the case of a rolling wheel, you could pick any point on the rim and decompose the motion into the motion of that point and a rotation of the wheel about that point. Then, when that point hits the ground it is instantaneously at rest and the motion becomes "pure" rotation. But, that is still relative to an accelerating reference frame.

Look at it another way. Take the two cases. a) The rolling wheel. b) The wheel is (really) rotating about a fixed point on the rim.

If you take a snapshot of the motion, then in both cases the position and velocity of very point on the wheel could be the same - assuming you catch the rotating wheel just at the right moment But, the motions are different because of the acceleration. So, a) and b) are not equivalent, even instantaneously, as the accelerations are different. Each point may have the same position and the same velocity, but not the same acceleration.

You have a simpler case with a ball reaching its highest point under gravity, compared with a ball at rest with no net force acting on it. Both are stationary but they are not in the same state of rest as one is accelerating.

PS In summary: a rolling wheel is instantaneously a pure rotation about the lowest point in terms of position and velocity; but not in terms of acceleration and internal and external forces.
 
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