# Wheel rolling on a horizontal surface

1. Nov 9, 2017

### Jahnavi

1. The problem statement, all variables and given/known data

2. Relevant equations

3. 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 ?

Last edited: Nov 9, 2017
2. Nov 9, 2017

### kuruman

One is to assume that you are looking for the components of the linear acceleration. Angular acceleration is not one of them.

3. Nov 9, 2017

### Jahnavi

Is radial acceleration the net linear acceleration of point P ?

4. Nov 10, 2017

### Abhishek kumar

From where u got 4/3

5. Nov 10, 2017

### Jahnavi

The point whose acceleration is to be calculated is at a distance (4/3)m from the bottom most point .

6. Nov 10, 2017

### haruspex

The linear acceleration consists of the radial acceleration (which you found) and the tangential acceleration (which follows swiftly from the angular acceleration).

7. Nov 10, 2017

### Jahnavi

Tangential acceleration = 4m/s2

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

Is that correct ?

8. Nov 10, 2017

### haruspex

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. Nov 10, 2017

### Jahnavi

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. Nov 10, 2017

### collinsmark

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:

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]

Last edited: Nov 10, 2017
11. Nov 10, 2017

### collinsmark

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?

12. Nov 10, 2017

### Abhishek kumar

Pattern you have shown in figure is when we work from frame other than wheel frame

13. Nov 10, 2017

### Jahnavi

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 ?

14. Nov 10, 2017

### collinsmark

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 in to 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. Nov 10, 2017

### PeroK

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. Nov 10, 2017

### haruspex

Yes, I do, belatedly.

17. Nov 10, 2017

### PeroK

That is acceleration relative to point O, whose acceleration you would then need to calculate.

18. Nov 10, 2017

### Jahnavi

19. Nov 10, 2017

### Jahnavi

Isn't wheel undergoing pure rotation about O ?

20. Nov 10, 2017

### PeroK

Yes, but point O is accelerating. The bottom of the wheel is instananeously at rest, but it is accelerating!