# Homework Help: Rolling Motion

1. Jan 21, 2010

### kppc1407

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

A 60cm diameter wheel is rolling along at 20m/s. What is the speed of a point at the front edge of the wheel?
Diameter - .6m
Velocity - 20m/s

2. Relevant equations

v = r(omega)
Pythagorean theorem

3. The attempt at a solution

I tried to get omega which I found to be 66.67 rad/s. Then I tried to use that as the vertical component and 20m/s as the horizontal component to get get the total velocity from the Pythagorean theorem.

2. Jan 21, 2010

### dacruick

I'm not sure what your question is. However I can tell you why the answer is 28 if you would like?

3. Jan 21, 2010

### tiny-tim

Welcome to PF!

Hi kppc1407! Welcome to PF!
I take it you're treating the centre of the wheel as the centre of rotation, and then adding the velocity of the centre of the wheel?

Then why bother to use angular velocity … how fast are points on the rim going, relative to the centre?

4. Jan 21, 2010

### kppc1407

Yes, dacruick, I cannot figure out what I am doing wrong because I do not get that answer so I would love to know why it is 28m/s.

5. Jan 21, 2010

### tiny-tim

How fast are points on the rim going, relative to the centre?

6. Jan 21, 2010

### kppc1407

The points on the rim are going much faster than the centre.

7. Jan 21, 2010

### dacruick

alright, so we have our circle rolling forward with 20m/s. That means that the perimeter of the circle is also travelling at 20m/s. At the front edge of the circle, you can draw one vector radially outward with magnitude 20m/s. You can also draw one vector tangent to the circle at the front edge. That vector will face downwards with a magnitude of 20m/s as well. Then you just add those vectors together using Pythagorean Theorem, and you get roughly 28 m/s in a direction 45 degrees below the horizontal.

8. Jan 21, 2010

### tiny-tim

Relative to the centre, the centre is stationary.

So, relative to the centre, how fast are points on the rim going?

9. Jan 21, 2010

### dacruick

Taking this one step further, at least conceptually this is what you can think about. Think about the instantaneous velocity of the point on the circle that is in contact with the ground. We know that when somthing rolls, it undergoes Static friction, not kinetic friction. So if you apply the same idea that I just used above, you will find that the point in contact with the surface has an instantaneous velocity of 0.

The vector diagram for this will go along these lines:
You will have a 20m/s vector point in the direction of motion.(lets assume to the right).
Since this circle will be rotating clockwise, at the point on the ground, we will have another vector with magnitude 20m/s pointing antiparallel to the first one. They cancel out and give 0

If you have any questions dont hesistate

10. Jan 21, 2010

### kppc1407

Thank you, dacruick!

tiny-tim... 20m/s?

11. Jan 21, 2010

### tiny-tim

Yup! (and the angular velocity doesn't matter does it?)

ok, so now add the velocity of that point relative to the centre, to the velocity of the centre

(and try it for some other points on the rim, also, to see how it all works, and to see where the centre of rotation is).

12. Jan 21, 2010

### kppc1407

dacruick, this is how we would analyze it in class so I like the last post. Some reason this problem is hard for me to picture. I cannot grasp the idea of the vectors cancelling. I can understand in my head why the instantaneous velocity of the point on the ground, but mathematically I cannot map it out. And why is the translational speed equal to the rotational speed?

tiny-tim, if you said the centre is stationary then if I add those two using the Pythagorean theorem wouldn't I just end up with 20m/s?

13. Jan 21, 2010

### tiny-tim

relative velocities

No, the idea is to add relative velocities …

you add the velocity of the rim relative to the centre, to the velocity of the centre relative to the ground, and that gives you the velocity of the rim relative to the ground …

Vrg = Vrc + Vcg

14. Jan 21, 2010

### kppc1407

tiny-tim, I think I got it. So add it in components? Add the vertical component (which would be the velocity of rim relative to the center) and the horizontal component (which is velocity of center relative to the ground) using Pythagorean theorem. Right?

Both of which are 20m/s?

15. Jan 21, 2010

### tiny-tim

That's right!

(you can either add it in components, which of course is easiest in this example, or you can do it by drawing a vector triangle )

ok, now that's sorted, try an alternative method …

calculate the angular velocity, then use the fact that the bottom of the wheel is stationary, relative to the ground.​

16. Jan 21, 2010

### kppc1407

One question first... how do we know the horizontal and vertical components are equal?

Ok, the angular velocity is 66.7 rad/s. Right? If so, then what?

17. Jan 21, 2010

### tiny-tim

|Vcg| is given as 20, and you've just worked out that |Vrc| is also 20.
Well, you know where the centre of rotation is, so what is the equation relating velocity (both magnitude and direction), radius, and angular velocity?

18. Jan 21, 2010

### kppc1407

The first part... I know you probably think I am crazy because I cannot grasp this but when did I work out that the other one was 20?

The second part... v = omega(r)?

19. Jan 21, 2010

### tiny-tim

erm
still crazy?
That's right!

So in this case, v = … ? ​

20. Jan 21, 2010

### kppc1407

20m/s?

21. Jan 21, 2010

### tiny-tim

uhh? how did you get that?

v = ωr, so what is ω, and what is r (the distance from the centre of rotation, at the bottom of the wheel)?

22. Jan 21, 2010

### kppc1407

Oh, I think I might be confused where we are...

at the bottom if the center of rotation is there too, the r is 0m so v is 0m/s. Right?

23. Jan 21, 2010

### tiny-tim

No, I meant what are ω, and r, for the front edge of the wheel, using the bottom of the wheel as the centre of rotation.

24. Jan 21, 2010

### kppc1407

I am not sure.

25. Jan 22, 2010

### tiny-tim

(just got up :zzz: …)

ok, the instantaneous centre of rotation is the point, B, at the bottom of the wheel.

Temporarily, the whole wheel is rotating about that point.

So the velocity of every point, P, on the wheel is perpendicular to the line BP, and of magnitude ω|BP|.

For example, the centre of the wheel has velocity ωr horizontally, and the top of the wheel has speed 2ωr horizontally.