ch11-p001

# ch11-p001 - 1. The velocity of the car is a constant v = +...

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which yields 2 v = +44 m/s. This is consistent with Fig. 11-3(c). (i) We can proceed as in part (h) or simply recall that the bottom-most point is in firm contact with the (zero-velocity) road. Either way – the answer is zero. (j) The translational motion of the center is constant; it does not accelerate. (k) Since we are transforming between constant-velocity frames of reference, the accelerations are unaffected. The answer is as it was in part (e): 1.5 × 10 3 m/s 2 . (1) As explained in part (k), a = 1.5 × 10 3 m/s 2 . 1. The velocity of the car is a constant () ˆˆ 80 km/h (1000 m/km)(1 h/3600 s) i ( 22m s)i, v =+ = + G 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 towards the rear at G vv road ms =− =− 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
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## This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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