Prelim2SolS06

# Prelim2SolS06 - g(s)=mgs sin θ U s(s)=k(s-s 2/2;U(s)=U g(s...

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Solutions: Prelim 2, Spring 2006 1. F=16 N 2. N=3 3. G [2 points for A, 3 for H] 4.(a) W=0J (b) W=-686 J 5. A 6. I=I COM +Md 2 ; d 2 =R 2 /2; I=2MR 2 /3 7.(a) Conservation of Energy: E=mgH=mgD+mv 2 /2 Normal force vanishes at top: W=mv 2 /r Solve for H: H=D+r/2 (b)1. (c) In rolling only a fraction of the potential energy is transformed to translational kinetic energy, so more initial potential energy is needed to end up with the same speed. (d) 2 (e) The hollow ball has a larger moment of inertia so more of the potential energy is turned into rotational kinetic energy. More initial potential energy is therefore needed to end up with the same speed. 8. (a) U
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Unformatted text preview: g (s)=mgs sin( θ ); U s (s)=k(s-s ) 2 /2;U(s)=U g (s)+U s (s) S [m] U g (s) [J] U s (s) [J] U(s) [J] 0.00 4.5 4.5 0.15 1.5 1.125 2.625 0.30 3 3 0.90 9 9 (b) There is one point of equilibrium and it is stable. To find it we set: 0=F= -dU ds =-W sin( θ ) − k(s-s ), which gives s=0.2m. It is stable because d 2 U ds 2 =k>0 (c) 9.(a) v Ai = 8.8cm 0.4s =22cm/s v Af = 6cm 0.3s =20cm/s (c) v cm = 7.2cm 0.4s =18cm/s (d)Momentum is conserved so the COM velocity is constant (e) v Bi = 5.9cm 0.4s =15cm/s v Bf = 3.3cm 0.2s =16.5cm/s (f) K f K i = v Af 2 +v Bf 2 v Ai 2 +v Bi 2 =0.96...
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