Finding Velocity of Particle at Wheel Edge

AI Thread Summary
The discussion focuses on calculating the velocity of a particle at the edge of a rolling wheel using the relationship V(cm) = R*w, where w is the angular velocity. To find the velocity of a specific point on the wheel, the velocity of the center of mass (V(cm)) must be combined with the relative velocity of the point (V(i, rel)). It is clarified that the relative velocity is a vector with a magnitude of ωR and a direction tangent to the wheel. For a point at the front edge of the wheel, the overall velocity relative to the ground is derived as V(cm) - V(i, rel), resulting in a speed of √2 V = √2 ωR. To determine the velocity at other points on the wheel, trigonometric methods are necessary to combine the velocities as vectors.
klandestine
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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.
 
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Realize that the velocity of a particle relative to the cm is a vector whose magnitude is given by \omega R and whose direction is tangent to the wheel. So, if the velocity of the cm is V \hat{x}, then the velocity of a point at the front edge relative to the cm is -V \hat{y}. The velocity relative to the ground of that point is V \hat{x} -V\hat{y}, giving a speed of \sqrt{2} V = \sqrt{2} \omega R.
 
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?
 
Right. You'd have to add the two velocities as vectors, which will involve a bit of trig.
 
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