# 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 N mg mA + = G G G 0 A G N G mg G (b) (a) 0 5 0 4 4 g N mg mA N mg m N mg = = = 5 5 4 4 W N mg W ′ = = = 150 . W lb ′ = (b) The acceleration is downward, in the same direction as g G 0 4 g N mg m + = 3 4 4 W W ′ = = W W 90 . W l ′ = b r 5.2 (a) ( ) cent F m ω ω = − × × G G G G G G G For r ω , 2 ˆ cent r F m r ω e = 1 1 500 1000 s s ω π = = ( ) 2 6 2 ˆ 10 1000 5 5 cent r F e π π = × × = G dynes outward (b) ( ) 2 2 4 1000 5 5.04 10 980 cent g F m r F mg π ω = = = × 5.3 (See Figure 5.1.2) 0 mg T mA + = D G G G ˆ ˆ ˆ ˆ cos sin 0 10 g mg j T j T i m i θ θ + + = cos T mg θ = , and sin 10 mg θ = T 1 tan 10 θ = , 5.71 θ = D 1.005 cos mg T m θ = = 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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ˆ ˆ 10 g g g A g j ′ = = D G G G i For small oscillations of a simple pendulum: 1 2 g π = T 2 2 1.005 10 g g g ′ = + = g 1 1 2 1.995 1.005 g g π π = = 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 a g m m µ µ = = − = − = − (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) 3 2 6 g g g a a A ′ = = − + = D 5.6 (a) ( ) ˆ ˆ cos sin r i x R t jR = + + D K K t t 2 ˆ ˆ sin cos r i R t j R = − Ω Ω + ± K K 2 2 r r v R = = Ω ± ± v R = Ω circular motion of radius R (b) r r where r ω = × K K K K ± ± ˆ ˆ r ix jy = + K ( ) ˆ ˆ ˆ ˆ sin cos i R t j R t k ix jy ω ˆ = − Ω Ω + Ω − × + 2
ˆ ˆ ˆ sin cos i R t j R t j x i y ˆ ω ω = − Ω Ω + Ω − + sin x y R t ω = − Ω ± cos y x R t ω = − + Ω ± (c) Let here u x iy = + 1 i = ! sin cos u x iy y R t i x i R t ω ω = + = − Ω Ω − + Ω ± ± ± i u ω ′ = y ω = − i x ω + sin cos Re i t u i u R t i R t i ω + = −Ω Ω + Ω Ω = Ω ± Try a solution of the form i t i t u Ae Be ω ′ = + i t i t u i Ae i Be ω ω ′ = − + Ω ± i t i t i u i Ae i Be ω ω ω ω ′ = + ( ) i t u i u i Be ω ω ω + = + Ω ± so R B ω = + Ω Also at t the coordinate systems coincide so 0 = ( ) ( ) 0 0 u A B x iy x R = + = + = + D R A x R B x R ω = + = + + Ω D D so, R A x ω ω = + + Ω D Thus, i t i t R R u x e e ω ω ω ω ′ = + + + Ω + Ω D 5.7 The x, y frame of reference is attached to the Earth, but the x-axis always points away from the Sun.

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• Fall '04
• mokhtari

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