Department of Physics University of Maryland College Park, Maryland. Fall 2005 Final Exam Dec. 16, u 2 dt )2, L = m u 2 d θ, ( d θ

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1 Department of Physics University of arylan College Park, arylan PHYSICS 4 Prof. S. J. Gates Fall 5 Final Exam Dec. 6, 5 This is a OPEN book examination. Rea the entire examination before you begin to work. Be sure to rea each problem carefully. Any questions shoul be irecte to the proctor. There is an hour & fifty minute time limit. Show all of your work. Use the backs of pages if necessary or request an extra booklet. Be sure to complete the front page of the examination booklet incluing your name. Show all calculations neee to support your answers, where necessary. ost importantly, THINK before you start to calculate. Problem (.) (a.) Using the inverse of the raial istance, (i.e. r u ) an θ as the inepenent variable, leas to K m ( r } θ ) + r ( θ ), L m r θ, K m u ( u 4 θ ) + } ( θ u ), L m u θ, an thus K L m ( u } θ ) + u This equation then implies ( u } m θ ) + u A exp θ Since the right han sie of the equation involves an exponential, it is natural to make the ansatz u(θ) α exp θ where α is a constant. The equation will be solve it α m A

2 (b.) From the solution above r(θ) m A exp θ If the angular momentum is given by L exp (t/τ ) then it must be the case that } L exp (t/τ ) exp θ ( θ A ) } exp (t/τ ) t exp θ θ A L } exp θ (τ /) exp (t/τ ) } 4 A L } exp (t/τ ) } exp θ } τ A L } exp (t/τ ) } exp θ } τ A L } exp (t/τ ) exp θ exp θ } τ A L To make further progress, it is useful to choose θ such that exp θ τ A L exp (t/τ ) which leas to θ(t) (t/τ ) ln τ A L r(t) τ L m exp (t/τ ) Now the if there is a potential U(r, θ) it must satisfy ( U r ) m r r ( θ ) an when the expressions for r(t) an θ(t) are use this implies ( U r )

3 This implies that U(r, θ) U(θ). Next there is the equation ( U θ ) m r θ m r m r r τ m τ τ exp θ m A exp θ τ A exp θ exp θ (τ ) A exp θ m A an this has the solution U(r, θ) (τ ) A exp θ Problem (.) (a.) To fin the location of the center of mass, we first fin the mass/length for each wire. The mass of each is w an the raius of each semi-circle is r 4. So that mass/length w /πr. This means that we have using cylinrical coorinates V µ( r) ρ ρφ z µ( r) π φ w π. where µ( r) is the mass per unit volume. So the center of mass for the wire in the x-y plane is given by R () cm ρ ρφ z r µ( r) w R () cm π w w π φ r π π φ r On the first piece of wire, we have r r cosφ x + sinφ ŷ so that R () cm r π π r π ŷ π φ cosφ x + sinφ ŷ φ sinφ π r ŷ 3

4 an thus for the secon wire R () cm π r ẑ Finally to fin the center of mass of the system (T ot) R cm w R () cm + w R () cm w π r ŷ + ẑ R () cm + R () cm (b.) To fin the moment of inertia tensor for the system of wire we start with the efinition of the moment of inertia tensor for the first wire I () i j π ρ ρφ z µ( r) r δ i j r i r j w π π φ w π r δ i j r i r j φ w r π π w r φ y x y y x x x + y sin φ cosφ sinφ sinφ cosφ cos φ. This implies for the secon wire I () i j w r. an thus for the total moment (T ot) I i j I () i j + I () i j w r 3 3. (c.) The rotational kinetic energy of the wire system is thus T Rot 4 w r (ωx ) + 3 (ω y ) + 3 (ω z ) 4

5 Problem (3.) Each captain states in their frame of reference the frequency of their running light is 43 trillion Hz. The ata of the Vulcan scientist reas Table : Sensor Data ass Frequency Length Enterprise 9 million kg 68 trillion Hz m Warbir million kg 7 trillion Hz 5 m (a.) If the scientist observe spee of the approach v E of the Enterprise to Vulcan an the spee of the approach v W of the Warbir to Vulcan, she coul euce the spee of approach of the Warbir observe from the eck of the Enterprise. ve + v W v A + v E v W, β A v A c c The formulae for the relativistic Doppler Effect is given by f f ± β β If we efine the ratio f/f F this leas to β F + F (f ) f (f ) + f F E (68/43), F W (7/43), (43) (68) v E c (43) + (68), β E v E c (43) (7) v W c (43) + (7), β W v W c (b.) The Vulcan scientist is not in the rest from of the Enterprise, so the mass she observes ( E ) 9 6 kg is not the rest mass of the ship E. The relation between these is E ( E ) (β E ) The mass ( E ) observe from the eck of the Warbir is relate to the 5

6 rest mass of the ship E via ( E ) E (β A ) ( E ) (β E) (β A ) Problem (4.) (a.) To fin the acceleration an velocity vectors of the airplane we see V p ρ ω sin(ω t) x + cos(ω t)ŷ v ẑ A p ρ (ω ) cos(ω t) x sin(ω t)ŷ (b.) The airplane lans when ẑ R p an this occurs at the time t H /v. If n enotes the number of complete rotations an f the fractional part then n + f ω H π v (c.) The orthogonal unit vectors for your x-irection, y-irection an z-irection may be enote by ê, ê an ê 3. For an observer on the groun these are written as ê ρ ω sin(ω t) x + cos(ω t)ŷ + v ẑ (ρ ω ) + (v ) ê cos(ω t) x sin(ω t)ŷ ê 3 v sin(ω t) x + cos(ω t)ŷ + ρ ω ẑ (ρ ω ) + (v ) It is convenient to efine ϕ by tanϕ v ρ ω } so that ê cosϕ ω sin(ω t) x + cos(ω t)ŷ sinϕ ẑ ê cos(ω t) x sin(ω t)ŷ 6

7 ê 3 sinϕ ω sin(ω t) x + cos(ω t)ŷ + cosϕ ẑ (.) To write Newton s Secon Law in your frame of reference, it is important to note ê ω cosϕ ê ê ω cosϕ ê + sinϕ ê 3 ê3 ω sinϕ ê The position vector for an object in your reference frame takes the form ξ U ê + V ê + W ê 3 for some coorinates U, V, W. If the object has a mass of you write F ξ ξ U ê + V ê + W ê 3 } U ω cosϕ ê } + V ω cosϕ ê + sinϕ ê 3 } W ω sinϕ ê } F } U ê + V ê + W ê 3 } U ω cosϕ ê } V + ω cosϕ ê + sinϕ ê 3 } W ω sinϕ ê (ω ) U cosϕ + W sinϕ } cosϕ ê + sinϕ ê 3 (ω ) V ê Problem (5.) 7

8 A bea of mass is constraine to slie along the frictionless surface of a sphere of raius R. There is a potential energy associate with the position of the given by U( r) A l x + l y + l 3 z (a.) To fin the Lagrangian for this system we note spherical coorinate are perfect to use x R cosφ sinθ, y R sinφ sinθ, z R cosθ. so that T (R ) ( θ ) + sin θ ( φ ) U A R l cosφ sinθ + l sinφ sinθ + l 3 cosθ L T U (b.) For the equation of motion of this system via the Euler-Lagrange equations we fin L ( θ) (R ) ( θ) L, ( φ) (R ) sin θ ( φ), L (R ) sinθ cosθ( θ θ) A R l cosφ cosθ + l sinφ cosθ l 3 sinθ, L A R sinθ l sinφ + l cosφ φ L ( θ) L θ (R ) ( θ) (R ) sinθ cosθ( θ) + A R l cosφ cosθ + l sinφ cosθ l 3 sinθ, L ( φ) L φ (R ) sin θ ( φ) + A R sinθ l sinφ + l cosφ. (c.) To fin the Hamiltonian of this system, we first note. L p θ ( θ) (R ) ( θ), L p φ ( φ) (R ) sin θ ( φ), 8

9 an thus H p θ ( θ) + p φ ( φ) L (p θ ) + (R ) (p φ ) (R ) sin θ + A R l cosφ sinθ + l sinφ sinθ + l 3 cosθ Problem (6.) Given two particles of mass an with coorinates (x, y, z ) an (x, y, z ) are constraine to the surface of the same sphere, we can introuce spherical coorinates for both an use angular coorinates for both. The potential energy of the new system is given by U T otal U( r ) + U( r ) + k A R (6θ 5θ ) + k B R (θ ) + k C R (θ ) (a.) What is the form of Newton s secon law? (R ) ( θ ) (R ) sinθ cosθ ( θ ) A R l cosφ cosθ + l sinφ cosθ l 3 sinθ 6 k A R (6θ 5θ ) k B R (θ ) (R ) sin θ ( φ ) A R sinθ l sinφ + l cosφ. (R ) ( θ ) (R ) sinθ cosθ ( θ ) A R l cosφ cosθ + l sinφ cosθ l 3 sinθ + 5 k A R (6θ 5θ ) k C R (θ ), (R ) sin θ ( φ ) A R sinθ l sinφ + l cosφ. (b.) To escribe the equation of motion for this incluing a iscussion of normal moes, eigenmoes an eigenfrequencies. it is first important to look at the potential in problem five. This potential implies a force 9

10 given by F A l ˆx + l ŷ + l 3 ẑ A l l ˆx + l ŷ + l 3 ẑ (l ) + (l ) + (l 3 ), A l ˆn where l is efine by (l ) + (l ) + (l 3 ). This is a constant force with magnitue of A l irecte along the irection of ˆn. But this is exactly like the force of gravity! It follows that the angle µ with which the force meets with the z-axis is given by cosµ ẑ ˆn l 3, (l ) + (l ) + (l 3 ) This implies that we can use generalize coorinates to simplify the problem β θ µ, β θ µ an the Lagrangian for the system using the new coorinates takes the form L (R ) ( β ) + sin (β + µ) ( φ ) + (R ) ( β ) + sin (β + µ) ( φ ) A R l cosβ + cosβ k A R (6β 5β + µ) k B R (β + µ) k C R (β + µ) Now the first benefit of the coorinate change is apparent. The potential is inepenent of φ an φ! To make further progress it is useful to make the small angle approximation. L (R ) ( β ) + sin (µ) ( φ ) + (R ) ( β ) + sin (µ) ( φ ) A R l (β ) (β ) k A R (6β 5β + µ) k B R (β + µ) k C R (β + µ)

11 which makes it clear that only the β-angles are involve in the normal moes. The equations of motion for these takes the form (R ) ( β ) A R l β 6 k A R (6β 5β + µ) (R ) ( β ) k B R (β + µ) A R l β + 5 k A R (6β 5β + µ) k C R (β + µ) or more simply β β A l R β 6 k A (6β 5β + µ) k B (β + µ) A l R β + 5 k A (6β 5β + µ) k C (β + µ) an after further simplification β A l R + ( 36 k A + k B ( ) 6 k A + k B kb µ β 3k A β A l R ( 5 k A + k C ) kb µ ) β + 3k A β + ( 5 k A + k C ) β The terms that are inepenent of the β s can be eliminate via a reefinition. Now it is convenient to efine three frequencies 3k Ω A A, 36 k Ω AB A + k B 5 k Ω AC A + k C + A l R, + A l R,

12 This leas to the eigenvalue conition (ω) (Ω AB ) (ω) (Ω AC ) (Ω A ) 4 (ω) 4 (Ω AB ) + (Ω AC ) (ω) + (Ω AB ) (Ω AC ) (Ω A ) 4 (ω) 4 b (ω) c (ω) b ± b + 4 c (ω) (ΩAB ) + (Ω AC ) ± (Ω AB ) (Ω AC ) + 4 (Ω A ) 4 an this is the stanar two couple oscillator system so the eigenvectors are E, E+