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# ch11 - Chapter 11 1 The velocity of the car is a constant v...

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469 Chapter 11 1. The velocity of the car is a constant   ? 80 km/h (1000 m/km)(1 h/3600 s) i ( 22m s)i, v     and the radius of the wheel is r = 0.66/2 = 0.33 m. (a) In the car’s reference frame (where the lady perceives herself to be at rest) the road is moving toward the rear at v v road m s     22 , and the motion of the tire is purely rotational. In this frame, the center of the tire is “fixed” so v center = 0. (b) Since the tire’s motion is only rotational (not translational) in this frame, Eq. 10-18 gives top ˆ ( 22m/s)i. v   (c) The bottom-most point of the tire is (momentarily) in firm contact with the road (not skidding) and has the same velocity as the road: bottom ˆ ( 22m s)i. v   This also follows from Eq. 10-18. (d) This frame of reference is not accelerating, so “fixed” points within it have zero acceleration; thus, a center = 0. (e) Not only is the motion purely rotational in this frame, but we also have = constant, which means the only acceleration for points on the rim is radial (centripetal). Therefore, the magnitude of the acceleration is 2 2 2 3 top (22 m/s) 1.5 10 m s . 0.33 m v a r (f) The magnitude of the acceleration is the same as in part (d): a bottom = 1.5 10 3 m/s 2 . (g) Now we examine the situation in the road’s frame of reference (where the road is “fixed” and it is the car that appears to be moving). The center of the tire undergoes purely translational motion while points at the rim undergo a combination of translational and rotational motions. The velocity of the center of the tire is ˆ v   (h) In part (b), we found v v top,car   and we use Eq. 4-39: top, ground top, car car, ground ? ? i i 2 i v v v v v v

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