The way to think about this problem is to imagine the tire rotating about a fixed point at its circumference, which in this case is the point where the tire touches the ground, and NOT think of the tire rotating about its center. The reason we would need to think of this problem this way is because the tire is rolling on the ground and not rotating freely around it center axis.
So given the above picture, it's easy to see that the bottom of the tire, the fixed point about which the tire is rotating/rolling has to have a velocity of ZERO, just like the center of a freely rotating wheel would have a transverse velocity of zero as well.
And since we were given that the transverse (horizontal) velocity of the center of the wheel as 60 mph, we then can calculate the the transverse velocity of the top of the tire:
ω(center) = v(center)/r, where r is the radius of the tire
ω(top) = v(top)/2r, the reason it's 2r is because the top of the wheel is rotating at a distance of 2r from the bottom of the tire
ω(center) = ω(top), since the rotational velocities are the same
v(center)/r = v(top)/2r => v(top) = 2v(center) = 120 mph!
2006-09-25 18:45:00
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answer #1
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answered by PhysicsDude 7
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The bottom is standing still. The center is moving at 60mph. The top of the tire is moving at 120mph.
2006-09-25 13:39:48
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answer #2
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answered by Anonymous
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If the bottom of the tire is in contact with the pavement, and you are not skidding, the bottom of the tire is not moving with respect to the pavement. Think about it.
2006-09-25 13:44:46
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answer #3
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answered by gp4rts 7
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Depends on the size of the tire. The center of the tire is moving 60mph, of course.
2006-09-25 13:40:40
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answer #4
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answered by Anonymous
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60 even the spare is goin 60
2006-09-25 13:40:19
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answer #5
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answered by cwb63ss 6
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At bottom dead center (bdc) it will be going 0mph,relative to the road
2006-09-25 14:00:09
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answer #6
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answered by Anonymous
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60 m.p.h. if I'm not utterly mistaken
2006-09-25 13:38:26
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answer #7
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answered by Anonymous
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