Classical Mechanics Qualifying Exam Solutions Problem 1.

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1 Jauary 4, Uiversity of Illiois at Chicago Departmet of Physics Classical Mechaics Qualifyig Exam Solutios Prolem. A cylider of a o-uiform radial desity with mass M, legth l ad radius R rolls without slippig from rest dow a ramp ad oto a circular loop of radius a. The cylider is iitially at a height h aove the ottom of the loop. At the ottom of the loop, the ormal force o the cylider is twice its weight. a) Expressig the rotatioal iertia of the o-uiform cylider i the geeral form (IβMR ), express the β i terms of h ad a. h a ) Fid umerical value of β if the radial desity profile for the cylider is give y ρ ( r) ρr ; c) If for the cylider of the same total mass M the radial desity profile is give y ρ where,,,3..., descrie qualitatively how do you expect the value of β to chage with icreasig. Explai your reasoig. ( r) ρr, Solutio: a) Cetripetal acceleratio as the all rolls aroud the circular loop at the ottom of the track is a c v /a, ad could e expressed from free ody diagram equatio: N-W Mv /a, where N is the ormal force ad the W is the weight. We are give that NMg, so Mg Mg Mv /a, i. e. v ga Relatig the agular ad traslatioal velocities y vaω, we ext use the expressio for the total kietic eergy of rollig oject (o slippig) K ½ Mv + ½ Iω

2 Ad apply eergy coservatio for the all etwee its iitial positio at rest ad its positio at the ottom of the loop: Mgh ½ Mv + ½ Iω Sustitutig for v ad for ω (from aove) ad usig I βmr expressio we fid h ½a +½ β a Rearragig: β h/a - ) The momet of iertial aout the rotatioal axis of the cylider is give y I r dm. For the quadratic desity profile dm ρ r lπrdr I R 6 4 R 3 πlr ρ r ρ r lπrdr πρl MR, with M 6 3 ρ ( r) dv πlρ r dr, so R β 3 c) For a aritrary value of : I R R + πlρ R π lρ r dr πlρ, with M π lρ + 4 r dr + R +, so I MR β 4, + 4, e.g., β ;, β

3 Prolem. A particle of uit mass is projected with a velocity v at right agles to the radius vector at a distace a from the origi of a ceter of attractive force, give y f ( r) 4 a k + 3 r r 5 9k For iitial velocity value give y v, fid the polar equatio of the resultig orit. a Solutio: Calculatig the potetial eergy dv 4 a f ( r) k dr r r Thus, V k a + r 4 4 r The total eergy is 9k 9k E To+ Vo vo k + a 4 a a 4 a Its agular mometum is 9k 4 l a v costat r & o θ Its KE is dr dr l T ( r& + r & θ ) + r & θ + r dθ dθ r The eergy equatio of the orit is dr l a T + V + r k dθ r r 4r dr 9 k r k a dθ 4r r 4r dr or ( a r ) dθ 9 dr dφ Lettig r acosφ the asiφ dθ dθ So Thus dφ dθ 9 r acos θ 3 φ θ 3 ( r a@ θ o ) 4 3

4 Prolem 3. A simple pedulum of legth ad mass m is suspeded from a poit o the circumferece of a thi massless disc of radius a that rotates with a costat agular velocity ω aout its cetral axis. Usig Lagragia formalism, fid a) the equatio of motio of the mass m; ) the solutio for the equatio of motio for small oscillatios. y ωt x Solutio: a) Coordiates: x acosωt+ siθ y asiωt cosθ x& aωsiωt+ & θcosθ y& aωcosωt+ & θsiθ θ m L T V m( x& + y& ) mgy m a ω + & θ + & θ a ω si( θ ω t ) mg ( a si ω t cos θ ) d L m θ& + maω( & θ ω) cos( θ ωt) dt & θ L m & θaωcos ( θ ωt) mgsiθ θ L d L The equatio of motio θ dt & is θ ω a g θ cos( θ ωt) + siθ (Note the equatio reduces to equatio of simple pedulum if ω.) ) For small θ the equatio of motio reduces to that of a costat drivig force harmoic oscillator : g ω a θ + θ cos( ωt) 4

5 The geeral solutio for this equatio cosists of two parts, a complemetary fuctio u(t) ad a particular solutio v(t). Complemetary solutio comes from the simple harmoic oscillator equatio: θ g g + θ, ad could e immediately writte as u( t) Acos t ϕ. Give the form of the drivig force, the particular solutio could e defied as v( t) B cos( ωt). Takig the derivatives ad pluggig ack i the equatio of motio, oe fids the expressio ω a for B as B, ad the geeral solutio i form: g ω g ω a θ( t) u( t) + v( t) Acos t φ + cos( ωt) g ω 5

6 Prolem 4. A rigid ody cosists of six particles, each of mass m, fixed to the eds of three light rods of legth a,, ad c respectively, the rods eig held mutually perpedicular to oe aother at their midpoits. Solutio: a) Write dow the iertia tesor for the system i the coordiate axes defied y the rods; ) Fid agular mometum ad the kietic eergy of the system whe it is with a agular velocity ω aout a axis passig through the origi ad the poit (a,,c). ) (,,c) a) I mx y sice either x i or y i is zero for all six xy i i i i particles. Similarly, all the other products of iertia are zero. Therefore the coordiate axes are priciple axes. I m y + z m c + c ( ) ( ) ( ) ( ) ( ) ( ) xx i i i i I m + c xx (,,) Iyy m a + c Izz m a + (a,,) + c t I m a + c a + a r ω ( ) ( ) ω ω aeˆ + eˆ ˆ + ce3 a + + c ( ) a + + c c r t r Usig L Iω, + c a r mω L a c + ( a + + c ) a + c a( + c) r mω L ( a + c ) ( a c ) + + c( a + ) 6

7 Usig T r ω L r a( + c) mω ( ) [ ] T a c ( a + c ) a + + c c( a + ) mω T a + c + a + c + c a + a + + c mω T ( a + a c + c ) a + + c ( ) ( ) ( ) 7

8 Prolem 5. The force of a charged particle i a iertial referece frame i electric field E ad magetic field B is give y F q( E + v B), where q is the particle charge ad v is the velocity of the particle i the iertial system. a) Prove that the trasformatio from a fixed frame to a frame is give y r & r & r ʹ + & r r r r r r ω ʹ + ω vʹ + ω ω ʹ ( ) ) Fid the differetial equatio of motio referred to a o-iertial coordiate system with agular velocity q ω B m, for small B (eglect B ad higher order terms). Solutio: a) Let s cosider a coordiate system ( i ʹ, jʹ, k ʹ ), aout the axis defied y the uit vector with respect to the ( i, j, k ) system with agular velocity ω ω. Positio of ay poit P i space ca e expressed i two systems (i case of commo origi) as The velocity the ca e writte r i x + jy + kz rʹ i ʹ xʹ + jʹ yʹ + k ʹ z ʹ dx dxʹ di ʹ di ʹ v i +... i ʹ xʹ +... vʹ + xʹ +... vʹ + ω rʹ dt dt dt dt This fidig is geerally true for ay vector, e.g. for derivative of the velocity vectors: dv dvʹ dvʹ fixed dv fixed dv + ω v d( vʹ + ω rʹ ) + ω ( vʹ + ω rʹ ) d( ω rʹ ) + + ω vʹ + ω ω rʹ dω drʹ + rʹ + ω + ω vʹ + ω ω rʹ Chagig otatios, oe arrives at the trasformatio equatios from a fixed to a frame: 8

9 r r r r v vʹ + ω ʹ r & r & r ʹ + & r r r r r r ω ʹ + ω vʹ + ω ω ʹ ( ) ) The equatio of motio: r r r mr& qe + q v B r ( ) Trasformig from a fixed frame to a movig frame: r q r r ω B so & ω m r r q r r r mr& r r r r r r ʹ q( B vʹ ) B ( ω ʹ ) qe + q ( vʹ + ω ʹ ) B r r q r r r r mr& r r r r r r ʹ + q( vʹ B) + ( ω ʹ ) B qe + q ( vʹ B) + q ( ω ʹ ) B r r q r mr& r r ʹ qe + ( ω ʹ ) B q r r r q qb ( ω ʹ ) B ( rʹ )( siθ)( B) B m r Neglectig terms i B, mrʹ qe r & (Larmor s theorem) 9

Problem 1. Problem Engineering Dynamics Problem Set 9--Solution. Find the equation of motion for the system shown with respect to:

Problem 1. Problem Engineering Dynamics Problem Set 9--Solution. Find the equation of motion for the system shown with respect to: 2.003 Egieerig Dyamics Problem Set 9--Solutio Problem 1 Fid the equatio of motio for the system show with respect to: a) Zero sprig force positio. Draw the appropriate free body diagram. b) Static equilibrium