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Fowles05 - CHAPTER 5 NONINERTIAL REFERENCE SYSTEMS 5.1 (a)...

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CHAPTER 5 NONINERTIAL REFERENCE SYSTEMS 5.1 (a) The non-inertial observer believes that he is in equilibrium and that the net force acting on him is zero. The scale exerts an upward force, N G , whose value is equal to the scale reading --- the “weight,” W’, of the observer in the accelerated frame. Thus 0 0 Nm gm A + −= G G G 0 A G N G mg G (b) (a) 0 5 0 44 g N mg mA N mg m N mg −−= − = 55 WN m g W == = 150 . Wl b ′ = (b) The acceleration is downward, in the same direction as g G 0 4 g  +=   3 W W =− = WW 90 . ′ = b r 5.2 (a) () cent Fm ωω =− × × G GG G G For r ω , 2 ˆ cent r r e = 11 500 1000 ss ωπ −− 2 62 ˆ 10 1000 5 5 cent r Fe π ×= G dynes outward (b) 2 2 4 1000 5 5.04 10 980 cent g r g = × 5.3 (See Figure 5.1.2) 0 mg T mA +− = D G G G ˆˆˆˆ cos sin 0 10 g m g jT im i θθ −+ + = cos Tm g θ = , and sin 10 mg = T 1 tan 10 = , 5.71 = D 1.005 cos mg g 5.4 The non-inertial observer thinks that g G points downward in the direction of the hanging plumb bob… Thus 1
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ˆˆ 10 g ggAg j =− = − D G GG i For small oscillations of a simple pendulum: 1 2 g π = T 2 2 1.005 10 g gg  =+ =   g 11 2 1.995 1.005 ππ == T 5.5 (a) f mg µ =− is the frictional force acting on the b x, so G G o 0 f mA ma −= G ( a G is the acceleration of the box relative to the truck. See Equation 5.1.4b.) Now, f G the only real force acting horizontally, so the acceration relative to the road is 0 A G f G (b) 3 f mg g ag mm = − = (For + in the direction of the moving truck, the – indicates that friction opposes the forward sliding of the box.) 2 g A D (The truck is decelerating.) from above, 0 ma mA ma so (a) 326 ggg aaA =− = − + = D 5.6 (a) () cos sin rixR t j R =+ Ω + D K K t t 2 sin cos ri R t j R =− Ω Ω + Ω ± KK 22 rr v R ⋅= =Ω ±± vR ∴ =Ω circular motion of radius R (b) rr where r ω ′′ =−× xj y K ( ) ˆ ˆ sin cos iR t jR t k i x j y ˆ Ω − × + 2
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ˆˆˆ sin cos iR t jR t jx iy ˆ ω =− Ω Ω + Ω Ω − + sin x yR t ′′ =− Ω Ω ± cos yx Rt +Ω ± (c) Let here uxi y =+ 1 i = ! sin cos y y R t i xiR t ′′′ =+ = ±±± iu ′ = y ix + sin cos Re it uiu R tiR ti += ΩΩ + = ± Try a solution of the form uA e B e −Ω ui A e i B e ′=− + Ω ± iu iA e iB e ωω () uiui B e +=+ ± so R B = + Ω Also at t the coordinate systems coincide so 0 = () () 00 uABx i y xR =+= + D R Ax RBx R =+−=+− + Ω DD so, R Ax + Ω D Thus, RR ux e e  +   D 5.7 The x, y frame of reference is attached to the Earth, but the x-axis always points away from the Sun. Thus, it rotates once every year relative to the fixed stars. The X,Y frame of reference is fixed inertial frame attached to the Sun. (a) In the x, y rotating frame of reference ( ) ( ) cos x tR ε = Ω− ( ) ( ) sin yt R t where R is the radius of the asteroid’s orbit and R E is the radius of the Earth’s orbit. is the angular frequency of the Earth’s revolution about the Sun and ω is the angular frequency of the asteroid’s orbit.
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This note was uploaded on 03/09/2008 for the course PHYS 301 taught by Professor Mokhtari during the Fall '04 term at UCLA.

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Fowles05 - CHAPTER 5 NONINERTIAL REFERENCE SYSTEMS 5.1 (a)...

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