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Unformatted text preview: Chapter 11 1. The initial speed of the car is v = (80.0)(1000/3600) = 22.2 m/s. The tire radius is R = 0.750/2 = 0.375 m. (a) The initial speed of the car is the initial speed of the center of mass of the tire, so Eq. 112 leads to ϖ 22 2 0 375 59 3 = = = v R com0 rad s. . . . (b) With θ = (30.0)(2 π ) = 188 rad and ϖ = 0, Eq. 1014 leads to ϖ ϖ αθ α 2 2 2 2 59 3 2 188 9 31 = + ⇒ = = . . . b g rad s 2 (c) Eq. 111 gives R θ = 70.7 m for the distance traveled. 2. The velocity of the car is a constant ( 29 ( 29 ˆ 80 1000 3600 ( 22m s)i, v = + = + r 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 r 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. 1018 gives top ˆ ( 22m/s)i. v = + r (c) The bottommost 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 =  r This also follows from Eq. 1018. (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 ϖ = constant, which means the only acceleration for points on the rim is radial (centripetal). Therefore, the magnitude of the acceleration is 489 CHAPTER 11 a v r top 2 m s = = = × 2 2 3 22 0 33 15 10 . . . (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 ( 22m s)i. v = + r (h) In part (b), we found r v v top,car = + and we use Eq. 439: top, ground top, car car, ground ˆ ˆ ˆ i i 2 i v v v v v v = + = + = r r r which yields 2 v = +44 m/s. This is consistent with Fig. 113(c). (i) We can proceed as in part (h) or simply recall that the bottommost point is in firm contact with the (zerovelocity) 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 constantvelocity frames of reference, the accelerations are unaffected. The answer is as it was in part (e): 1.5 × 10 3 m/s 2 ....
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This note was uploaded on 04/19/2008 for the course PHYS 1100, 1200 taught by Professor Many during the Fall '08 term at Rensselaer Polytechnic Institute.
 Fall '08
 Many
 Physics, Center Of Mass, Mass

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