Prelim2SolS06

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This note was uploaded on 03/27/2008 for the course PHYS 1112 taught by Professor Leclair,a during the Spring '07 term at Cornell.

Ask a homework question - tutors are online