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Name: Dwight
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This is a question about rotation and acceleration. We heard a puzzler on a radio program where they stated that on the tire of a moving vehicle, the point on the tire in contact with the ground is not moving and the point opposite that is moving at twice the speed of the vehicle. It is perfectly clear on the face of it how this can be. But it is difficult to wrap one's brain around. One student noted that every point on the outside surface of the tire would have to accelerate from, say, 0 to 120MPH in the time and distance it takes for the wheel to make half a turn: like less than three feet distance in a few hundredths of a second. This is almost like being shot out of a gun. Why doesn't that kind of acceleration cause the tire to fly apart? Is this constant acceleration the same as centripetal force? Can you give us a formula for calculating the acceleration on a 24 inch diameter wheel going 60 miles an hour?

The wheel does not increase its speed at all. In this case, the outer surface of the tire is moving at 60 mph relative to the axle at all times; only its direction is changing. The acceleration is v*v/r, where v is the speed and r is the radius of the tire.

Tim Mooney

The velocity of a point on the tire relative to the ground is the sum of the velocity of the point on the tire relative to the car plus the velocity of the car relative to the ground. That is

V(tire,ground) = V(tire,car) + V(car,ground)

Since velocities are vectors this could be relatively complicated but for the case of points on top or bottom of the tire the addition is simple since the movement of those points on the tire is parallel to the movement of the car. So, if the car is moving at 60 mph, V(car,ground)=60 mph

The top of the tire is moving 60 mph relative to the car in the same direction as the car. You should be able to convince yourself that this is true.

V(tire top,car)= 60 mph

The bottom of the tire is moving 60 mph relative to the car in the direction opposite to the car's motion so

V(tire bottom,car)= -60 mph.

Thus the top of the tire is moving 60+60=120 mph relative to the ground and the bottom of the tire is moving 60-60=0 mph relative to the ground. This has to be true otherwise the tire would be skidding.

David Kupperman

The problem becomes simpler if you use a different frame of reference. Try using the car axle as the frame of reference and you see that you can view this situation as one of stable linear motion of a center of rotation. From the point of view of the forces on the tire the only motion is a rotation about the axle, and that is at a steady rate. As long as the tire can hold up to the rotational acceleration there is no danger.

Of course, I have ignored many things here, such as deformation of the tire where it contacts the ground, etc, but these are not significant to the question.

Greg Bradburn

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