Calculating Speed of a Point on a Rolling Wheel

In summary: I misread your post …I thought you were talking about a different problem, from the start of the thread … sorry again! :redface:So, back to this problem then.When the wheel is rolling, what is the relationship between the velocity of the centre of the wheel, and the velocity of a point on the rim?I don't think we ever explicitly stated it, but I would think that they are equal in magnitude but opposite in direction. Correct?I don't think we ever explicitly stated it, but I would think that they are equal in magnitude but opposite in direction. Correct?Yes, that's right. :smile:So if the centre of the wheel is
  • #1
kppc1407
19
0

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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  • #2
I'm not sure what your question is. However I can tell you why the answer is 28 if you would like?
 
  • #3
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:
 
  • #4
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.
 
  • #5
How fast are points on the rim going, relative to the centre? :smile:
 
  • #6
The points on the rim are going much faster than the centre.
 
  • #7
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.
 
  • #8
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:
 
  • #9
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.
 

1. What is rolling motion speed of a wheel?

The rolling motion speed of a wheel refers to the speed at which a wheel moves forward while also rotating around its axis. It is the combination of linear and angular velocity.

2. How is the rolling motion speed of a wheel calculated?

The rolling motion speed of a wheel can be calculated by dividing the distance traveled by the radius of the wheel. This is known as the tangential speed, which is the linear velocity at the edge of the wheel.

3. What factors affect the rolling motion speed of a wheel?

The rolling motion speed of a wheel can be affected by the size and shape of the wheel, the surface it is rolling on, and any external forces acting on the wheel such as friction or air resistance.

4. How does the rolling motion speed of a wheel impact its performance?

The rolling motion speed of a wheel is an important factor in its performance, as it determines how quickly the wheel can move and how smoothly it can roll. A higher rolling motion speed can result in faster acceleration and a smoother ride.

5. Can the rolling motion speed of a wheel be changed?

Yes, the rolling motion speed of a wheel can be changed by altering the factors that affect it, such as the size and shape of the wheel or the surface it is rolling on. Additionally, external forces like friction can be managed to increase or decrease the rolling motion speed.

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