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

This preview shows pages 1–4. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern