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Point on wheel with constant velocity

  1. May 24, 2007 #1
    Let a bicycle run straight on a plane past you from left to right with constant speed. As a representative point of the bicycle take the center of the back axle. Since the bicycle has constant speed with regard to the plane, so has the axle (trivial). In addition a point on the rim of the wheel, say R, has constant angular velocity. Even more, with regard to the axle center, the absolute value of the velocity of R, i.e. [itex]|v_R|[/itex] is constant.

    But, with regard to the plane, the velocity of R is not constant. As an example lets look at the horizontal speed of R between the angles 12 o'clock (top) and 3 o'clock (right). R moves [itex]2\pi r/4+r[/itex] if the radius of the wheel is r. Why? At the start, R is on top of the axle, at the end, the axle moved a quarter circumference and R is a distance of r in front of it. With a similar argument we find that between angles from 3 to 6 o'clock, R only moves [itex]2\pi r/4-r[/itex]. Since the time for both moves is the same, the horizontal speed with regard to the plane is obviously different. In both cases the vertical distance travelled by R is r. Consequently the absolute value of the velocity vector of R with regard to the plane is not constant.

    Now the questions:
    1. Is it possible to let the bicycle move such that the absolute value of the velocity of R with regard to the plane is constant? How does the mathematical function look like and how does the bicycle move then?
    2. Can we have constant speed of the bicycle as well as a constant absolute value of the velocity of R at the same time, if necessary drop the requirement of circular motion of R?

    Every hint appreciated,
    Harald.​

    (No, not a homework assignment :frown:)
     
  2. jcsd
  3. May 24, 2007 #2

    russ_watters

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    Sure, all you have to do is lock the wheels and make the bicycle slide along the ground...
     
  4. May 24, 2007 #3

    daniel_i_l

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    I think that the answer is no.
    Lets look at the horizontal component of the speed. You can pretend that the point is connected to a spring and it is ossilating back and forth horizontaly to the ground, the other end of the spring is connected to the axle.
    Now if what you say is possible then simply adding a constant velocity to the ossilating point would make the speed of the point always be 0. But that means that the speed of the point is constant which is false.
     
  5. May 24, 2007 #4
    I wish I could find the picture online, it was on in my physics textbook. It was a shot of a bicycle wheel in motion. The top is very blurred and the bottom is very crisp. This is because a point at the top of the wheel is spinning much faster than one at the bottom (relative to the ground).
     
  6. May 25, 2007 #5

    rcgldr

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    Yes, as long as the point is inside the radius of the wheel. See below.
    It speeds up and slow down relative to 1 / (R + r cos(angle)), where 0 angle means vertical, and increasing clockwise.
    No, unless the point is on the axle (distance from axle to point is zero).

    The shape of the path of the point relative to the ground is a curtate cycloid. If the bicycle was upside down and rolling on the roof, and the point was on the perimeter, the path would be a brachistrone. (Do a web search for graphs and more info).

    Someone here check my math, just to be sure.

    [tex]\mbox{let}\ R = \mbox{radius of wheel}[/tex]

    [tex]\mbox{let}\ r = \mbox{radius to point on wheel}[/tex]

    [tex]\mbox{let}\ \theta = \mbox{angle to point, with 0 meaning vertical and increasing clockwise, like a hand on a clock.}[/tex]

    [tex]x = R\ \theta\ +\ r\ sin(\theta)[/tex]

    [tex]y = R\ +\ r\ cos(\theta)[/tex]

    [tex]\mbox{let}\ \omega = \mbox{angular velocity}= (d \theta \ /\ dt) = \dot \theta [/tex]

    Velocity of point

    [tex]\dot x = R\ \omega \ +\ r\ \omega \ cos(\omega\ t\ ) = \omega \ (\ R\ +\ r\ cos(\omega \ t)\ ) = \omega \ (\ R\ +\ r\ cos(\theta)\ )[/tex]

    [tex]\dot y = -r\ sin(\omega \ t) = -r\ sin(\theta)[/tex]

    Let V = constant velocity of point.

    [tex]\omega = V\ /\ (\ R\ +\ r\ cos(\omega \ t)\ ) = V\ /\ (\ R\ +\ r\ cos(\theta)\ )[/tex]
     
    Last edited: May 25, 2007
  7. May 25, 2007 #6

    Danger

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    Math aside (since I don't know any) the thing that bothers me about the original question is probably just a matter of semantics. As far as I know, there is no such thing as a 'constant angular velocity'. Velocity is a vector involving both speed and direction. On a wheel, that direction is constantly changing, which means that the point in question on the wheel is undergoing constant acceleration.
    Mind you, I've had an awful lot of beer tonight, so I might have missed something. :redface:
     
  8. May 25, 2007 #7

    daniel_i_l

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    Oh, stupid me, I was only thinking of constant velocity.
     
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