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PHYS 203 Princeton University Fall 2006

1. (30 points) Ski jump trackConsider the motion of a skier of mass m down a track with a constant slope . Thefriction of the skies is negligible, but the motion is subject to air resistance. The

resistance force is given by F = kv2. The initial velocity of the skier is equal to zero.

l

a) What is the velocity of the skier as afunction of time?

)/sintanh(/sin

/sintanh

/sin

1

/sin

/sin

sin

1

0 2

2

2

mkgtkmgv

tkmg

v

kmgk

m

tvkmg

dv

k

m

dtmkvg

dv

kvmgdt

dvm

v

=

=

=

=

=

b) What is the distance l traveled by the skier as a function of time? How far does hetravel by the time his velocity reaches approximately 90% of its maximum

possible value?

)/sincosh(ln)/sintanh(/sin0 0

mkgtkmdtmkgtkmgvdtl

t t ===

Tanh(x) reaches 90% of its maximum for 47.1)9.0(tanh 1 == x

k

m

k

ml

mkgt

83.0)47.1cosh(ln

47.1/sin

90

90

==

=

c) The coefficient of air resistance is approximately given by k= A/2, where isthe density of air and A is the cross-sectional area of the skier. Taking

= 1kg/m3,A = 1 m2 , = 45 deg, and m = 100 kg, estimate the length l neededto achieve 90% of the maximum velocity. How does it compare with the length ofa typical ski jump track of about 100 m?

mmmkg

kgl 166833.0

2/1/1

1002390

==

So, the track is nearly as long as it takes to reach the maximum velocity. The

coefficient of air resistance is probably actually smaller because of the

aerodynamic suits.

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2. (30 points) Falling Ladder

2l

x

y

CM

A ladder of length 2l is standing up against a vertical wall with

initial angle relative to the horizontal. There is no frictionbetween the ladder and the wall or the floor. The ladder begins

to slide down with zero initial velocity. Denote by (t) theangle the ladder makes with the horizontal after it starts to

slide.

a) Write down Lagrangian equations of motion with twoconstraints describing the contact of the ladder with thevertical wall and the floor.

0cossin31)3(

0)2(

0)1(

0sin

0cos

3

1)2(

12

1;

2

1)(

2

1

2122

21

1

22

2

11

1

2

1

22222

=+=++

==

+

==

+

==

==

==

==++=

lmlggLdtdL

ymmgy

g

y

L

dt

d

y

L

xmx

g

x

L

dt

d

x

L

ylg

xlg

UTL

mgyU

mllmIIyxmT

&&&

&&&

&&&

&&&

b) Find the expressions for and as a function of2& &&

Differentiating the constraint equations and plugging them into the equation for

kinetic energy:

cos4

3

03

4cos

sin

3

2

6

1

)cos(sin2

1

cos

sin

2

22222222

l

g

mlmglL

dt

dL

mglU

mlmlmlT

ly

lx

=

==

==++=

=

=

&&

&&

&

&&&

&&

&&

From energy conservation Uinit=T+U

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)sin(sin2

3

3

2sinsin

2

22

=

+=

l

g

mlmglmgl

&

&

c) Find the angle c when the ladder losses contact with the vertical wall.The constraint force given by 1 is equal to the normal force from the vertical wall.Find when it goes to zero.

sin3

2sin

0sincos4

9sincos

2

3)sin(sincos

2

3cossin

4

3cossin

0cossin

2

2

1

=

===+=+

=+==

l

g

l

g

l

g

l

g

mlmlxm

&&&

&&&&&

3. (40 points) Conical surfaceA small mass m can slide without friction on the inside of a

conical surface with opening angle . Use cylindricalcoordinates , , andz to describe the position of the masswithz = 0 at the apex of the cone.

a) Find the Lagrangian for the mass m in terms ofcoordinates and only by eliminating z dependenceusing the surface constraint.

z

m

cot))cot1((2

1

cot

)(2

1

2222

2222

mgmL

z

mgzzmL

++=

=

++=

&&

&&&

b) Is there a conserved momentum in the problem? If so, use it to reduce the problemto only a single dynamical variable.

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cot2)cot1(2

)cot1(

;0

2

2

2

2

2

2

2

mgm

p

m

pH

mL

p

m

p

constmL

pL

+++

=

+=

=

=

=

==

&&

&

&&

Since there is no dependence and p is a constant, we have only one dynamical

variable,

c) Show that the solution can be described by motion of a particle with a certaineffective mass in a one-dimensional effective potential. What are the effective

potential and the effective mass?

This Hamiltonian corresponds to motion of a single particle with mass

in effective potential 22 sin/)cot1( mm =+=

cot2 2

2

mgm

pVeff +=

Note that the effective potential has to be identified from the Hamiltonian, not

Lagrangian, because the Hamiltonian is the function of the momentum and

coordinate.

d) Use the effective potential to calculate the radius of a circular orbit. If you arestuck on other parts, use freshman physics techniques to find the radius for acircular orbit.

32

2

3

2

tan

cot

gm

p

mgm

p

d

dVeff

=

+=