CDIM10 - CHAPTER 10 Motion in a Noninertial Reference Frame...

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CHAPTER 10 Motion in a Noninertial Reference Frame 10-1. The accelerations which we feel at the surface of the Earth are the following: (1) Gravitational : 2 980 cm/sec (2) Due to the Earth’s rotation on its own axis: ( ) ( ) ( ) 2 2 8 2 8 5 2 rad/day 6.4 10 cm 86400 sec/day 6.4 10 7.3 10 3.4 cm/sec r π ω = × × = × × × = 2 (3) Due to the rotation about the sun: ( ) ( ) 2 2 13 2 5 13 2 2 rad/year 1.5 10 cm 86400 365 sec/day 7.3 10 1.5 10 0.6 cm/sec 365 r π ω = × × × × = × × = 10-2. The fixed frame is the ground. y a θ x The rotating frame has the origin at the center of the tire and is the frame in which the tire is at rest. From Eqs. (10.24), (10.25): ( ) 2 f f r r = + + × + × × + × a r r v ±± ± ω ω ω ω a R 333
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334 CHAPTER 10 Now we have 0 0 0 cos sin 0 f r r a a r V a r r θ θ = − + = = = = = R i r i v a k k ±± ± ω ω j Substituting gives 2 0 cos sin f v a a a r θ θ = − + + a i j j i ( 2 0 cos sin 1 f v a r θ θ = − + + + a i j ) a (1) We want to maximize f a , or alternatively, we maximize 2 f a : 4 2 2 2 2 2 2 2 2 2 0 0 4 2 2 2 2 2 0 0 2 cos cos 2 sin sin 2 2 cos sin f v av a a a r r v av a a r r a θ θ θ θ θ = + + + + + = + + + a θ 2 2 2 0 0 2 2 cos 2 cos 0 when tan f d av a d r ar v θ θ θ θ = − + = = a (Taking a second derivative shows this point to be a maximum.) 2 0 2 2 2 4 0 n implies cos ar v v a r v θ θ = = + ta and 0 2 2 4 0 sin ar a r v θ = + Substituting into (1) 2 2 0 2 2 4 2 2 4 0 0 0 1 f ar v av a r a r v a r v = − + + + + + j a i This may be written as 2 4 2 0 f a a v r = + + a
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MOTION IN A NONINERTIAL REFERENCE FRAME 335 θ A This is the maximum acceleration. The point which experiences this acceleration is at A : where 0 2 ar v tan θ = 10-3. We desire . From Eq. (10.25) we have eff 0 = F ( ) eff 2 f r m m m m = × × × × r r v ±± ± ω ω ω ω F F R r 0 ω The only forces acting are centrifugal and friction, thus 2 s mg m r µ ω = , or 2 s g r µ ω = 10-4. Given an initial position of (–0.5 R ,0) the initial velocity (0,0.5 ω R ) will make the puck motionless in the fixed system. In the rotating system, the puck will appear to travel clockwise in a circle of radius 0.5 R . Although a numerical calculation of the trajectory in the rotating system is a great aid in understanding the problem, we will forgo such a solution here. 10-5. The effective acceleration in the merry-go-round is given by Equation 10.27: 2 2 x x y ω ω = + ±± ± (1) 2 2 y y x ω ω = ±± ± (2) These coupled differential equations must be solved with the initial conditions ( ) 0 0 0.5 m x x = − , ( ) 0 0 0 m y y = , and ( ) ( ) 1 0 0 0 2 m s x y v = = ± ± 0 v , since we are given in the problem that the initial velocity is at an angle of 45° to the x -axis. We will vary over some range that we know satisfies the condition that the path cross over . We can start by looking at Figures 10-4e and 10-4f, which indicate that we want . Trial and error can find a trajectory that does loop but doesn’t cross its path at all, such as 0 v 1 0. 0 0
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