Calculating Speed of a Point on a Rolling Wheel

Click For Summary

Homework Help Overview

The discussion revolves around calculating the speed of a point on the edge of a rolling wheel, specifically a wheel with a diameter of 60 cm moving at a speed of 20 m/s. Participants are exploring the relationship between linear and angular velocities in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of angular velocity and its relevance to the problem. There are attempts to apply the Pythagorean theorem to combine the linear velocity of the wheel's center and the velocity of points on the rim. Questions arise about the instantaneous velocity of points in contact with the ground and the reasoning behind vector addition in this scenario.

Discussion Status

The discussion is active, with participants sharing insights and clarifying concepts. Some guidance has been offered regarding the use of relative velocities and the application of the Pythagorean theorem to find the resultant speed of points on the wheel. There is an ongoing exploration of different interpretations and methods to understand the problem better.

Contextual Notes

Participants express confusion regarding the relationship between linear and angular velocities, particularly in the context of rolling motion. There are mentions of assumptions about the center of rotation and the instantaneous velocities of points on the wheel.

kppc1407
Messages
19
Reaction score
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.
 
Physics news on Phys.org
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.
 

Similar threads

Wheel rolling on a horizontal surface
  • · Replies 65 ·
3
Replies
65
Views
6K
Velocities and Acceleration on a rolling wheel
  • · Replies 2 ·
Replies
2
Views
2K
Have a problem with this Rotational Motion question
  • · Replies 5 ·
Replies
5
Views
2K
Friction Required for Billiard Ball to Roll without Slipping
  • · Replies 60 ·
3
Replies
60
Views
5K
Mechanical Energy of a Homogeneous Circular Wheel Rolling at Different Speeds
  • · Replies 1 ·
Replies
1
Views
1K
Motion in one dimension -- Experiments with a Hot Wheels car rolling down a ramp
  • · Replies 12 ·
Replies
12
Views
5K
Shooting an arrow through the spokes of a turning wheel
  • · Replies 7 ·
Replies
7
Views
2K
A wheel rolling on a horizontal flat or inclined surface
  • · Replies 13 ·
Replies
13
Views
3K
Dynamics homework -- Motion of a stick attached to a rolling disk
  • · Replies 6 ·
Replies
6
Views
2K
Rolling Wheel / quick question -- Linear and Angular Velocity
  • · Replies 2 ·
Replies
2
Views
2K