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Rolling motion

  1. Mar 25, 2005 #1
    I know that because of the rolling constraint, V(cm)=R*w [w=omega]. Also, if I want to find the velocity of particle i, i can sum V(cm) and V(i, relative to the center of mass). But how do I determine V(i, rel)? For example, if I am given the speed and radius of a wheel, how fast is a point at the very front edge of the wheel going? I keep trying to use these equations, but I keep coming up with 2wR, which I know is the velocity for the TOP of the wheel.
     
  2. jcsd
  3. Mar 25, 2005 #2

    Doc Al

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    Staff: Mentor

    Realize that the velocity of a particle relative to the cm is a vector whose magnitude is given by [itex]\omega R[/itex] and whose direction is tangent to the wheel. So, if the velocity of the cm is [itex]V \hat{x}[/itex], then the velocity of a point at the front edge relative to the cm is [itex]-V \hat{y}[/itex]. The velocity relative to the ground of that point is [itex]V \hat{x} -V\hat{y}[/itex], giving a speed of [itex]\sqrt{2} V = \sqrt{2} \omega R[/itex].
     
  4. Mar 25, 2005 #3
    ah-ha!

    thank you! so, if i wanted to know the velocity of a point at a more arbitrary location, i would have to use trigonometry, right?
     
  5. Mar 25, 2005 #4

    Doc Al

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    Staff: Mentor

    Right. You'd have to add the two velocities as vectors, which will involve a bit of trig.
     
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