Chapter11 - 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 ˆ (2 2m / s ) i . v G (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 2ms ) i . v =− G 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 ω = constant, which means the only acceleration for points on the rim is radial (centripetal). Therefore, the magnitude of the acceleration is 22 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 ) i . v G (h) In part (b), we found G top,car and we use Eq. 4-39: top,ground top,car car,ground ˆ ii 2 i v v = + = GG G
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2. The initial speed of the car is () 80 km/h (1000 m/km)(1 h/3600 s) 22.2 m/s v == . 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. 11-2 leads to com0 0 22.2 m/s 59.3 rad/s. 0.375 m v R ω = (b) With θ = (30.0)(2 π ) = 188 rad and = 0, Eq. 10-14 leads to 2 22 2 0 (59.3 rad/s) 29 . 3 1 r a d / s . 2 188 rad ωω α =+ ¡ (c) Eq. 11-1 gives R = 70.7 m for the distance traveled.
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For a uniform disk (relative to its center of mass) Im R = 1 2 2 (Table 10-2(c)). Since the wheels roll without sliding ω = v / R (Eq. 11-2). Thus the numerator of our fraction is 44 1 2 2 22 2 2 R v R mv = F H G I K J F H G I K J = and the fraction itself becomes () 2 210 1 fraction 0.020. 2 2 1000 50 mv m Mv mv M m == = = = ++ The wheel radius cancels from the equations and is not needed in the computation.
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Chapter11 - 1. The velocity of the car is a constant ^ v =...

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