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Department of Physics University of Maryland College Park, Maryland PHYSICS 410 Prof. S. J. Gates Fall 2005 Final Exam Dec. 16, 2005
This is a OPEN book examination. Read the entire examination before you begin to work. Be sure to read each problem carefully. Any questions should be directed 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 including your name. Show all calculations needed to support your answers, where necessary. Most importantly, THINK before you start to calculate.
Problem (1.)
(a.) Using the inverse of the radial distance, (i.e. r = u−^1 ) and θ as the indepen- dent variable, leads to
K =
m
{ ( d r dθ )^2 + r^2
} ( d θ dt )^2 , L = m r^2 d θ dt
K =
m
{ 1 u^4
d u dθ
)^2 +
u^2
} ( d θ dt )^2 , L = m u−^2 d θ dt
and thus K L^2
2 m
{ ( d u dθ
)^2 + u^2
}
This equation then implies 1 2 m
{ ( d u dθ )^2 + u^2
} = A 0 exp[ 2 θ ]
Since the right hand side of the equation involves an exponential, it is natural to make the ansatz u(θ) = α 0 exp[ θ ]
where α 0 is a constant. The equation will be solved it α 0 =
m A 0
(b.) From the solution above
r(θ) =
m A 0 exp[ − θ ]
If the angular momentum is given by L 0 exp[−2(t/τ 0 ) ] then it must be the case that
L 0 exp[−2(t/τ 0 ) ] =
2 A 0
} exp[ − 2 θ ] ( d θ dt
exp[−2(t/τ 0 ) ]d t =
2 A 0 L 0
} exp[ − 2 θ ] d θ
− (τ 0 /2)d { exp[−2(t/τ 0 ) ]} = − d
{ exp[ − 2 θ ] 4 A 0 L 0
}
d { exp[−2(t/τ 0 ) ]} =
2 τ 0 A 0 L 0
} d { exp[ − 2 θ ] } ∫ d { exp[−2(t/τ 0 ) ]} =
2 τ 0 A 0 L 0
} ∫ d { exp[ − 2 θ ] }
exp[−2(t/τ 0 ) ] − 1 =
2 τ 0 A 0 L 0
} { exp[ − 2 θ ] − exp[ − 2 θ 0 ] }
To make further progress, it is useful to choose θ 0 such that
exp[ θ ] =
√^1
2 τ 0 A 0 L 0 exp[ (t/τ 0 ) ]
which leads to θ(t) = (t/τ 0 ) − 12 ln[2 τ 0 A 0 L 0 ]
r(t) =
√ 2 τ 0 L 0 m
(^) exp[ − (t/τ 0 ) ]
Now the if there is a potential U (r, θ) it must satisfy
∂ U
∂r ) = m
[ d^2 r dt^2 − r ( d θ dt
)^2
]
and when the expressions for r(t) and θ(t) are used this implies
(
∂ U
∂r
and thus for the second wire R^ ~(2) cm =
[ (^2) π
] r 0 ̂z
Finally to find the center of mass of the system
R^ ~( cmT ot )=
M^ w^ R~(1) cm + M (^) w R~(2) cm 2 M (^) w
(^) = 1 2
[ R~(1) cm + R~(2) cm
]
[ (^1) π
] r 0 [ ŷ + ẑ ]
(b.) To find the moment of inertia tensor for the system of wire we start with the definition of the moment of inertia tensor for the first wire I(1) i j =
∫ dρ ρdφ dz μ(~r)
[ |~r|^2 δi j − ri rj
]
∫ (^) π 0
d φ M πw
[ |~r|^2 δi j − ri rj
]
[ (^) Mw π
] ∫^ π 0
d φ
y^2 − x y 0 − y x x^2 0 0 x^2 + y^2
[ Mw r^20 π
] ∫ (^) π 0 d φ
sin^2 φ − cosφ sinφ 0 − sinφ cosφ cos^2 φ 0 0 0 1
= Mw r^20
1 2 0 0 0 12 0 0 0 1
.
This implies for the second wire
I(2) i j = Mw r^20
1 2 0 0 0 1 0 (^0 0 )
.
and thus for the total moment
I( i jT ot ) = I(1) i j + I(2) i j = 12 Mw r^20
.
(c.) The rotational kinetic energy of the wire system is thus
T (^) Rot = 14 Mw r^20
[ 2(ωx)^2 + 3 (ωy)^2 + 3 (ωz )^2
]
Problem (3.)
Each captain states in their frame of reference the frequency of their running light is 430 trillion Hz. The data of the Vulcan scientist reads
Table 1: Sensor Data
Mass Frequency Length Enterprise 190 million kg 680 trillion Hz 1000 m Warbird 200 million kg 720 trillion Hz 1250 m
(a.) If the scientist observed speed of the approach vE of the Enterprise to Vulcan and the speed of the approach vW of the Warbird to Vulcan, she could deduce the speed of approach of the Warbird observed from the deck of the Enter- prise. vA =
[ vE + vW 1 + vE c^ v 2 W
] , βA = v cA
The formulae for the relativistic Doppler Effect is given by
f ′^ = f
√ 1 ± β 1 ∓ β If we define the ratio f /f ′^ = F this leads to
β =
∣∣ ∣∣ ∣
1 − F 2
1 + F 2
∣∣ ∣∣ ∣ =
∣∣ ∣∣ ∣
(f ′)^2 − f 2 (f ′)^2 + f 2
∣∣ ∣∣ ∣
FE = (68/43) , FW = (72/43) ,
vE = c
∣∣ ∣∣ ∣
(43)^2 − (68)^2
(43)^2 + (68)^2
∣∣ ∣∣ ∣ ,^ βE^ =^
vE c
vW = c
∣∣ ∣∣ ∣
(43)^2 − (72)^2
(43)^2 + (72)^2
∣∣ ∣∣ ∣ ,^ βW^ =^
vW c
(b.) The Vulcan scientist is not in the rest from of the Enterprise, so the mass she observes ( ME )′^ = 190 × 106 kg is not the rest mass of the ship M (^) E^0. The relation between these is
M (^) E^0 = ( ME )′^
√ 1 − (βE )^2
The mass ( ME )′′^ observed from the deck of the Warbird is related to the
ˆe 3 = − sinϕ 0 ω 0 [ sin(ω 0 t)x̂ + cos(ω 0 t)ŷ ] + cosϕ 0 ẑ
(d.) To write Newton’s Second Law in your frame of reference, it is important to note
d dt eˆ 1 = − ω 0 cosϕ 0 ˆe 2 d dt eˆ 2 = ω 0 [ cosϕ 0 eˆ 1 + sinϕ 0 ˆe 3 ] d dt
eˆ 3 = − ω 0 sinϕ 0 eˆ 2
The position vector for an object in your reference frame takes the form ~ξ = U eˆ 1 + V ˆe 2 + W ˆe 3
for some coordinates U, V, W. If the object has a mass of M you write
F^ ~ = M d
2 dt^2
~ξ = M d dt
d dt
~ξ
= M
d dt
{ dU dt ˆe 1 + dV dt ˆe 2 + dW dt eˆ 3
}
− M
d dt
{ U ω 0 cosϕ 0 ˆe 2 }
- M d dt { V ω 0 [ cosϕ 0 ˆe 1 + sinϕ 0 eˆ 3 ] }
− M
d dt { W ω 0 sinϕ 0 eˆ 2 }
F^ ~ = M
{ d^2 U dt^2 ˆe 1 + d^2 V dt^2 ˆe 2 + d^2 W dt^2 eˆ 3
}
− 2 M
{ dU dt ω 0 cosϕ 0 ˆe 2
}
+ 2 M
{ dV dt
ω 0 [ cosϕ 0 ˆe 1 + sinϕ 0 ˆe 3 ]
}
− 2 M
{ dW dt ω 0 sinϕ 0 eˆ 2
}
− M (ω 0 )^2 { U cosϕ 0 + W sinϕ 0 } [ cosϕ 0 ˆe 1 + sinϕ 0 ˆe 3 ] − M (ω 0 )^2 V ˆe 2
Problem (5.)
A bead of mass M is constrained to slide along the frictionless surface of a sphere of radius R 0. There is a potential energy associated with the position of the given by
U (~r) = M A 0 [ 1 x + 2 y + ` 3 z ]
(a.) To find the Lagrangian for this system we note spherical coordinate are perfect to use x = R 0 cosφ sinθ , y = R 0 sinφ sinθ , z = R 0 cosθ.
so that T = 12 M (R 0 )^2
[ ( d θ dt )^2 + sin^2 θ ( d φ dt
)^2
]
U = M A 0 R 0 [ 1 cosφ sinθ + 2 sinφ sinθ + ` 3 cosθ ] L = T − U
(b.) For the equation of motion of this system via the Euler-Lagrange equations we find [ ∂ L ∂( θ˙)
] = M (R 0 )^2 ( θ˙) ,
[ ∂ L ∂( φ˙)
] = M (R 0 )^2 sin^2 θ ( φ˙) , [ ∂ L ∂θ
] = M (R 0 )^2 sinθ cosθ( θ˙)^2
− M A 0 R 0 [ 1 cosφ cosθ + 2 sinφ cosθ − ` 3 sinθ ] , [ ∂ L ∂φ
] = − M A 0 R 0 sinθ [ − 1 sinφ + 2 cosφ ]
d dt
[ ∂ L ∂( θ˙)
] −
∂ L
∂θ
d dt
[ M (R 0 )^2 ( θ˙)
] − M (R 0 )^2 sinθ cosθ( θ˙)^2
- M A 0 R 0 [
1 cosφ cosθ + 2 sinφ cosθ − ` 3 sinθ ] = 0 , d dt
[ ∂ L ∂( φ˙)
] −
∂ L
∂φ
d dt
[ M (R 0 )^2 sin^2 θ ( φ˙)
]
- M A 0 R 0 sinθ [ −
1 sinφ + 2 cosφ ] = 0.
(c.) To find the Hamiltonian of this system, we first note.
pθ =
[ ∂ L ∂( θ˙)
] = M (R 0 )^2 ( θ˙) ,
pφ =
[ ∂ L ∂( φ˙)
] = M (R 0 )^2 sin^2 θ ( φ˙) ,
given by F^ ~ = − M A 0 [ 1 xˆ + 2 yˆ + ` 3 zˆ ]
= − M A 0 |`|^2
(^) √[^ ^1 xˆ^ +^^2 ˆy^ +^ ^3 ˆz^ ] ( 1 )^2 + (2 )^2 + ( 3 )^2
= − M A 0 |`|^2 nˆ
where |`|^2 is defined by
√ (1 )^2 + ( 2 )^2 + (3 )^2. This is a constant force with magnitude of M A 0 ||^2 directed along the direction 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 = zˆ · nˆ =
(^) √ ^3 ( 1 )^2 + (2 )^2 + ( 3 )^2
This implies that we can use generalized coordinates to simplify the problem
β 1 = θ 1 − μ , β 2 = θ 2 − μ
and the Lagrangian for the system using the new coordinates takes the form
L = 12 M (R 0 )^2
[ ( d β 1 dt
)^2 + sin^2 (β 1 + μ) ( d φ 1 dt
)^2
]
+ 12 M (R 0 )^2
[ ( d β 2 dt )^2 + sin^2 (β 2 + μ) ( d φ 2 dt
)^2
]
− M A 0 R 0 |`|^2 [ cosβ 1 + cosβ 2 ]
−
kA R^20 (6β 1 − 5 β 2 + μ)^2
−
kB R^20 (β 1 + μ)^2 −
kC R^20 (β 2 + μ)^2
Now the first benefit of the coordinate change is apparent. The potential is independent of φ 1 and φ 2! To make further progress it is useful to make the small angle approximation.
L ≈ 12 M (R 0 )^2
[ ( d β 1 dt )^2 + sin^2 (μ) ( d φ 1 dt
)^2
]
+ 12 M (R 0 )^2
[ ( d β 2 dt )^2 + sin^2 (μ) ( d φ 2 dt
)^2
]
− M A 0 R 0 |`|^2 [ 2 − 12 (β 1 )^2 − 12 (β 2 )^2 ]
−
kA R^20 (6β 1 − 5 β 2 + μ)^2
−
kB R^20 (β 1 + μ)^2 −
kC R^20 (β 2 + μ)^2
which makes it clear that only the β-angles are involved in the normal modes. The equations of motion for these takes the form d dt
[ M (R 0 )^2 ( d β 1 dt
] = − M A 0 R 0 |`|^2 β 1 − 6 kA R^20 (6β 1 − 5 β 2 + μ)
− kB R^20 (β 1 + μ) d dt
[ M (R 0 )^2 ( d β 2 dt
] = − M A 0 R 0 |`|^2 β 2 + 5 kA R^20 (6β 1 − 5 β 2 + μ)
− kC R^20 (β 2 + μ)
or more simply d dt
[ d β 1 dt
] = − A^0 |`|
2 R 0 β^1 −^6
kA M (6β^1 −^5 β^2 +^ μ)
− k MB (β 1 + μ) d dt
[ d β 2 dt
] = − A^0 |`|
2 R 0 β^2 + 5^
kA M (6β^1 −^5 β^2 +^ μ)
− k MC (β 2 + μ)
and after further simplification d^2 β 1 dt^2
[ A 0 |`|^2 R 0 +^
( (^36) kA + kB M
) ] β 1 + (^30) MkA β 2
−
( (^6) k A +^ kB M
) (^) k B M μ d^2 β 2 dt^2 = (^30) MkA β 1 −
[ A 0 |`|^2 R 0 +^
( (^25) kA + kC M
) ] β 1
−
( (^5) k A +^ kC M
) (^) k B M μ
The terms that are independent of the β’s can be eliminated via a redefinition. Now it is convenient to define three frequencies
ΩA =
√ 30 kA M ,
ΩAB =
√ 36 kA + kB M +^
A 0 |`|^2 R 0 ,
ΩAC =
√ 25 kA + kC M +^
A 0 |`|^2 R 0 ,
