# Caculating velocity on a different points on a rim of a wheel

1. Oct 17, 2009

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

Motorcycle is accelerating from the rest with a constant acceleration a = 2 m/s². We are observing its front wheel (see picture) after a time t = 10 s. This wheel is not slipping while accelerating. How much are velocities of points A, B, and C on rim of wheel? How much are the accelerations of these points?

2. Relevant equations

v= v0 + at

3. The attempt at a solution

A: v = v0 + at → v= -20 m/s
B: v= sqrt(vA² + vB²) → 28.3 m/s
C: v= v0 + at → v= 20 m/s

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• ###### FI.bmp
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2. Oct 17, 2009

### Staff: Mentor

Hint: Find the velocity of the center of the wheel with respect to the ground, the velocity of the rim with respect to the center, then the velocity of each part of the rim with respect to the ground.

3. Oct 17, 2009

### turin

There is a rotational part that you seem to be neglecting, and when you include it, you may be quite surprised, especially about point C. Also, don't forget that the problem asks for the accelerations, too.

4. Oct 17, 2009

So, that means that I have to take into the consideration also centripetal acceleration?

If so, is equal to:

ac= 4πr / t²
v= 2πr / t

Because I don't have the r, I combined the above formulas together to get:

ac= 2πv / t
ac= 12.56 m/s²

Is this what you meant by "There is a rotational part that you seem to be neglecting"?

But how do I go on from here?

5. Oct 17, 2009

I did some research and this is what I found out:

1. A wheel rolling over a surface has both a linear and a rotational velocity.
2. The linear velocity of any point on the rim of the wheel is given by vcm= ωr.
3. Because when the wheel is in contact with the ground, its bottom part is at rest with respect to the ground, the wheel experiences a linear motion with a velocity equal to + vcm besides a rotational motion (picture).
4. Conclusion: the top of the wheel moves twice as fast as the center and the bottom of the wheel does not move at all.

The only problem now is: How to calculate r?

#### Attached Files:

• ###### Motion of wheel sum.gif
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Last edited: Oct 17, 2009
6. Oct 18, 2009

### turin

That's a good question. I'm sorry, I didn't notice that the problem doesn't give you the radius. I have no idea how to do it without knowing the radius (or a way to determine it, which I also don't see). Are you sure that the problem doesn't tell you the size of the wheel, or some other info that you're neglecting?

7. Oct 18, 2009

No, there is no mentioning of the quantity of r, except if knowing that we are talking about motorcycle, we just put an average size for the r of motorcycle tire? Can this option be possible?

8. Oct 18, 2009

### turin

Sure, in the "real world". However, in the "real world", if the solution were really important, you would still go and determine r. It just depends on what kind of a course this is.

9. Oct 18, 2009

College physics, I'm preparing for my first colloquium. I'm studying pharmacy in Slovenia.

10. Oct 18, 2009

### Integral

Staff Emeritus
Just leave the answer in terms of r.

11. Oct 18, 2009

### Staff: Mentor

I agree with turin. While you can easily find the speed of the points, without the radius I see no way to determine their accelerations.

12. Oct 18, 2009

### RoyalCat

I disagree.

You can find all those velocities in terms of the velocity of the center of mass. Taking that expression and differentiating it with respect to time should give you an answer solely in terms of the acceleration of the center of mass. Therefore, the radius of the wheel is irrelevant.

13. Oct 18, 2009

### Staff: Mentor

Differentiating the speed as a function of time will give you one component of the acceleration. What about the change in angle?

14. Oct 18, 2009

So, how can I write down the acceleration in terms of r for the three point A. B, and C?
Any hint?

I know that formula for tangential acceleration is a_t= rα.

Thank you.

15. Oct 19, 2009

### RoyalCat

I think that's irrelevant to this question. It asks about the mathematical points A, B and C. Those at the bottom, top and right side of the wheel.

What their acceleration is, in this case, is how their velocity changes with respect to time.
To know the change in angle over time would, of course, require we know the radius.

16. Oct 19, 2009

### Staff: Mentor

Velocity can change in magnitude or direction.
Precisely the point.

17. Oct 19, 2009

### RoyalCat

And again, that is irrelevant as you are asked for the linear acceleration, and not the angular acceleration.

The direction of the velocity does not change, since the acceleration of these points is always directed in the same direction as the velocity.

You even used that as a premise for solving for the velocities.

For rolling motion without sliding:

$$V_{CM}=\omega R$$

$$a_{CM}=\alpha R$$

With respect to the center of mass:
$$V_{tangential}=\omega R$$
$$a_{tangential}=\alpha R$$

Vector addition can help us find the velocities and accelerations with respect to the original frame of reference in which the wheel does not slide.

18. Oct 19, 2009

### turin

A changing angle leads to a linear (radial) acceleration. The angle and radius together provide the linear magnitude, which is why we either need to know the radius, or how to determine it.

This is incorrect. The essence of rotation is a contantly changing direction.

19. Oct 19, 2009

### Staff: Mentor

The question asks for the acceleration of the specified points.

Why would you think that? It's rolling!

No I didn't.

20. Oct 19, 2009

### RoyalCat

You assumed that the velocity of the top-most point is pointing in the direction of velocity of the center of mass. The acceleration would therefore have to be in the direction of the movement, otherwise you would have no way to tell the direction of the velocity.

Oh! I get it now! You need the radius to find the radial acceleration, which you would then sum up with the known tangential acceleration to find the acceleration of the specified points.

Heh, the magic of italics made it click for me.

Is that what you meant, or do I still not understand?