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 28 2 85 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 21 3 2 5 13 2 2 r a d / y e a r 1.5 10 cm 86400 365 sec/day 7.3 10 1.5 10 0.6 cm/sec 365 r × × × =× × = 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 ff rr = + +×+× ×+ × a r r v ±± ± ωω ω ω aR 333
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334 CHAPTER 10 Now we have 0 00 cos sin 0 f rr aa r Va θθ =− + = == Ri ri va kk ±± ± ωω j Substituting gives 2 0 cos sin f v aaa r + + ai j j i ( 2 0 cos sin 1 f v a r θ  + + +   j ) a (1) We want to maximize f a , or alternatively, we maximize 2 f a : 42 2 22 2 2 2 2 2 2 cos cos 2 sin sin 2 2c o s s i n f v a v a =+ + ++ + =+ + + a 2 2 2 0 0 2 2 cos 2 cos 0 when tan f d av a dr ar v + a (Taking a second derivative shows this point to be a maximum.) 2 0 2 4 0 n implies cos ar v v ar v + ta and 0 4 0 sin ar v = + Substituting into (1) 0 4 4 0 1 f ar v a r v v + + + j This may be written as 24 2 0 f aav 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 mm m m = − −× −×× − × r r v ±± ± ωω ω ω FFR r 0 ω The only forces acting are centrifugal and friction, thus 2 s mg m r µ ω = , or 2 s g r = 10-4. Given an initial position of (–0.5 R ,0) the initial velocity (0,0.5 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 xxy =+ ± (1) 2 2 yyx =− ± (2) These coupled differential equations must be solved with the initial conditions ( ) 0 0 0.5 m xx ≡= , ( ) 0 00 m yy , and ( ) ( ) 1 0 2 m s xyv = = 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. (,) xy 0.47 m > s 0 v 1 53 m s =⋅ . From here, one may continue to solve for different values of v until the wanted crossing is eyeball-suitable. This may be an entirely satisfactory answer, depending on the inclinations of the instructor. An interpolation over several trajectories would show that an accurate answer to the problem is , which exits the merry-go-round at 3.746 s. The figure shows this solution, which was numerically integrated with 200 steps over the time interval.
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This note was uploaded on 11/03/2008 for the course PHYS 3210 taught by Professor Ford during the Fall '08 term at Colorado.

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

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