Calculating Speed of a Point on a Rolling Wheel

AI Thread Summary
A 60cm diameter wheel rolling at 20m/s has a point at its front edge moving at 28m/s. This speed is derived by adding the horizontal velocity of the wheel's center (20m/s) and the vertical velocity of the rim relative to the center (also 20m/s) using the Pythagorean theorem. The angular velocity calculated is 66.67 rad/s, but the key is understanding the relative motion of points on the wheel. The discussion emphasizes the importance of vector addition in determining the speed of points on a rolling wheel. Understanding these concepts clarifies the dynamics of rolling motion.
kppc1407
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Homework Statement



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

Answer - 28m/s

Homework Equations



v = r(omega)
Pythagorean theorem

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

Hi kppc1407! Welcome to PF! :smile:
kppc1407 said:
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.

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? :wink:
 
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.

tiny-tim, I am not sure that I follow your drift.
 
How fast are points on the rim going, relative to the centre? :smile:
 
The points on the rim are going much faster than the centre.
 
alright, so we have our circle rolling forward with 20m/s. That means that the perimeter of the circle is also traveling 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.
 
kppc1407 said:
The points on the rim are going much faster than the centre.

Relative to the centre, the centre is stationary.

So, relative to the centre, how fast are points on the rim going? :wink:
 
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 don't hesistate
 
  • #10
Thank you, dacruick!

tiny-tim... 20m/s?
 
  • #11
kppc1407 said:
tiny-tim... 20m/s?

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

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

(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
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
relative velocities

kppc1407 said:
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?

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 :smile:
 
  • #14
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
That's right! :smile:

(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 :wink:)

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
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
kppc1407 said:
One question first... how do we know the horizontal and vertical components are equal?

|Vcg| is given as 20, and you've just worked out that |Vrc| is also 20.
Ok, the angular velocity is 66.7 rad/s. Right? If so, then what?

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
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
kppc1407 said:
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?

erm :redface:
kppc1407 said:
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?

still crazy? :biggrin:
The second part... v = omega(r)?

That's right! :smile:

So in this case, v = … ?​
 
  • #20
20m/s?
 
  • #21
kppc1407 said:
20m/s?

uhh? :confused: 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
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
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
I am not sure.
 
  • #25
(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.
 
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