In problem 4 of "Problem Solving Circular Motion Kinematics," Lewin asks for the speed of each point, a, b, c, and d, on the circumference of the wheel with respect to the ground. He uses the equation for angular velocity, omega = v/r, but uses the speed of the center of the wheel (for v), not a point on the rim. Isn't this equation only valid for a point on the circumference? To illustrate this point, think of a wheel spinning in midair but not going anywhere; the center of the wheel wouldn't have any speed, but the wheel would have angular speed. What am I doing wrong here?

Question

anonymous · Answer

In my solution I tried to relate the speed of a point on the circumference to the speed of the center and found that Vt (speed of pt. on circ.) = Vo*pi/2. I then solved the rest of the problem in the same way as prof. Lewin, but obviously with different results (which I can post if needed).

Any help is appreciated.

vincent-lyon.fr · Answer

Can you draw a picture ? Where are points A, B C and D?

If the wheel is rolling without slipping on a fixed base, then the contact point with the base (denoted I) is the instantaneous centre of rotation.

Speed will be directly proportional to distance of point (A, B, C or D) to point I.

For instance, if v is the speed of centre of the wheel, then the point at the top will have a speed equal to 2v.

anonymous · Answer

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and here's the original question:

A bicycle tire rolls without slipping along the ground with its center of mass moving with 
speed  v0. You may neglect any rolling resistance. What is the speed with respect to the 
ground of each of the four marked points on the tire shown in the figure below?