Linear Velocity at Top (Rotational Kinematics)

AI Thread Summary
A bike traveling at X m/s has tires with a diameter of 0.60 m, leading to discussions about the linear velocity of a point at the rim of the wheel. The relationship v = r omega and the circumference formula C = 2 pi r were referenced to understand the motion. Initially, there was confusion about whether the inquiry was regarding angular velocity or linear velocity relative to the ground. It was clarified that the speed of the topmost point of the wheel is actually 2X m/s, while the instantaneous velocity at the point of contact with the ground is zero. This highlights the complexities of rotational kinematics and the differences between linear and angular velocities.
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1. A bike is moving at X m/s. Its tires are .60 m in diameter. How fast is a point at the rim of the top of one of the wheels moving relative to the ground?



2. v = r omega, C = 2 pi r (?)



3. At first we thought he just wanted angular velocity. Then when we realized that his given velocity is for the axle, and that he wanted velocity compared to the ground. Then we thought perhaps it is the same, but this wouldn't be true since linear velocity would depend on the radius. When I looked at a figure in the textbook, it appeared that the linear velocity for the top at anyone instant would be 0 m/s, no matter what the angular velocity was, since it opposes the bottom with static friction. So ultimately, we are just very confused.
 
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v=rw, (w is omega) where v is the speed of the bike and also the axle. The topmost point has a horizontal speed v wrt the axle. So, the speed of the topmost point is 2v wrt the ground (or 2X m/s as given by you). This is only the instantaneous velo, mind you.

The instantaneous velo of the pt of contact is zero.
 
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