Classical Solution bank

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Classical Mechanics Review

(Louisiana State University Qualifier Exam)

Jeff Kissel

May 18, 2007

A particle is dropped into a hole drilled straight through the center

of the earth. Neglecting rotational effects, show that the particle’s

motion is simple harmonic. Compute the period and give an estimate

in minutes. Compare your result with the period of a satellite orbiting

near the surface of the earth.

Figure 1: A particle dropped from R⊕.

Only the mass inside the shell of radius r contributes to the gravitationalforce. If we assume the Earth have uniform density, then the mass contributingto F g is

M

4

3π r3

= M ⊕

4

3π R3⊕

M = M ⊕

R3⊕r3 (1)

Thus, the force experienced by the particle m from mass M is

F g = GmM

r2

= Gm

r2

M ⊕R3⊕

r 3

F g = GM ⊕m

R

3

⊕ r (2)

= k r

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Jeff Kissel May 18, 2007 Classical Mechanics

where the last step absorbs all the constants G, M ⊕, R⊕, and m into the constant“k,” which shows that Eq. 2 is linearly proportional to the radius. Similarly,Hooke’s force law for simple harmonic motion states that the force of a simpleharmonic oscillator is proportional to the displacement from the equilibrium po-sition. Since the center of the earth would be the particles equilibrium position,r is the displacement, and thus by direct comparison, the particle obeys simple

harmonic motion, with “spring constant” k .To compute the period τ , we use what we know for a spring with resonant

frequency ω =

k/m,

τ = 1

f =

2π

ω = 2π

m

k

= 2π

m

R3

⊕

GM ⊕ m

τ = 2π

R3⊕GM ⊕

τ 2

=

4π2R3⊕GM ⊕ (3)

which is Kepler’s 3rd law for planetary motion (for a small particle), againindicating that motion is periodic.

Finally, one can find an estimate of this period by using Earth’s gravitationalacceleration of a small particle at the surface, g =

GM ⊕R2⊕

≈ 9.8 m/s ≈ π2.

τ 2 = 4π2R3⊕

GM ⊕

= 4 π2

R⊕

g

τ 2 ≈ 4R⊕ (4)

= 4(6.4× 106 m) [s]

τ 2 = 2.57× 107 s

τ = 5069 s = 84 min (5)

which is approximately equivalent to a satellite in low-Earth orbit. (At 185 km,i.e. when low-Earth orbit becomes stable, the correct period is 88.19 min.)

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Jeff Kissel

October 11, 2006

Two identical bodies of mass m are attached by identical springs of spring constant k as shown in the figure.

Figure 1: A system of two masses attached by springs (A) in equilibrium, and(B) after some oscillation.

a) Find the frequencies of free oscillation of this system.

The first goal will be to set up the system’s characteristic equation,

V− ω2T = 0 (1)where V is the potential energy matrix, T is the kinetic energy matrix, and ωare the desired eigenfrequencies (frequencies of free oscillation) of the system.The potential on the particles is the sum of each individual spring’s potentials,

V = 12

kd21 + 12

kd22 + 12

kd23 (2)

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Jeff Kissel October 11, 2006 Classical Mechanics

If we define the displacements between the masses in terms of generalized coor-dinates η ,

d1 = (x1 − x(0)1 ) + (0) = η1d2 = (x2 − x(0)2 ) + (x(0)1 − x1) = η2 − η1

d3 = (0) + (x(0)

2 − x2) = −η2we can then write the potential as,

V = 1

2k(η1)

2 + 1

2k(η2 − η1)2 + 1

2k(−η2)2

= 1

2kη21 +

1

2k(η22 + η

21 − η1η2 − η2η1) +

1

2kη22

= 1

2k

2η22 + 2η21 − η1η2 − η2η1

(3)

which in the desired matrix form is,

V

≡ 1

2

V ijηiηj = 2k −k−k 2k (4)The kinetic energy matrix is much easier to find because it is neatly diagonal:

T = 1

2mv21 +

1

2mv21

T ≡ 12

T ijηiηj =

m 0

0 m

(5)

Once we’ve got these matrices, we want to find the eigenfrequencies, ω of the charateristic equation

V− ω2T = 0 (6)Notice however, that Eq. 4 and Eq. 5 are symmetric, 2 × 2 matrices, so wecan use some trickery and write them as multiples of the identity matrix or theSU(2) spinor matrices:

V = 2k I − k σ̂xT = m I

Now we can just diagonalize Eq. 6 instead of the usual determinant method,since we know the eigenvalues of I are {1,1}, and for any σ̂i are {1,-1}. I’ll alsouse the fact that a diagonalized σ̂x is just σ̂z

0 = V− ω2T= 2k I − k σ̂x −mω2 I

D−1 (mω2 I)D = D−1 (2k I− k σ̂x)Dω2 (D−1ID) = 2k

m (D−1ID)− k

m(D−1σ̂xD)

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ω2 I = 2k

m I− k

m σ̂z (7)

ω2

1 00 1

=

2k

m

1 00 1

− k

m

1 00 −1

ω2 = 2k ± k

m (8)

So the eigen frequencies for this two mass system are

ω1 =

k

m or ω =

3k

m (9)

We’ll need the eigenvectors for part (b), so we’ll find them now. A bigadvantage to using this SU(2) group argument is that since Eq. 7 only involesI and σ̂z, we know immediately that the two (normalized) eigenvectors for thesystem are the (normalized) eigenvectors of those two matrices,

for I ⇒ û1 = 1√ 2

11

(10)

for σ̂z ⇒ û2 = 1√ 2

1−1

(11)

where û1 corresponds to when the two masses are sloshing back and forth inphase with each other, and û1 is when they oscillate out of phase, i.e. if M 1 ismoving to the left, M 2 is moving to the right, or vice versa.

b) M 1 is displaced from its position by a small distance A1 to theright while M 2 is not moved from its position. If the two masses arereleased with zero velocity, what is the subsequent motion of M 2?

With the (very safe) assumption that the system is periodic in time, we knowfrom Fourier’s theorem that the positions of each particle as a function of timemay be written as a sum of sines and cosines, whose phase is the eigenfrequen-

cies. So,

x1(t) = A sin ω1 t + B cos ω1 t + C sin ω2 t + D cos ω2 t (12)

x2(t) = E sin ω1 t + F cos ω1 t + G sin ω2 t + H cos ω2 t (13)

However, we can narrow this down at least a little because of the eigenvectors.

ω1 =

k

m ⇒ û1 = 1√ 2

11

⇒ A = E, B = F

ω2 =

3k

m ⇒ û2 = 1√ 2

1−1

⇒ C = −G, D = −H

which means Eqs. 12 and 13 (and there derivatives) simplify a lil’ bit to

x1(t) = A sin ω1t + B cos ω1t + C sin ω2t + D cos ω2t

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→ ẋ1(t) = Aω1 cos ω1t−Bω1 sin ω1t + Cω2 cos ω2t−Dω2 sin ω2tx2(t) = A sin ω1t + B cos ω1t− C sin ω2t−D cos ω2t

→ ẋ2(t) = Aω1 cos ω1t−Bω1 sin ω1t− Cω2 cos ω2t + Dω2 sin ω2t

Remember the problem statement said “M 1 is diplaced by a small distanceA1 [at t = 0]?” That means the initial conditions (where t = 0, so the sineterms are zero, and cosine terms are one) are

x1(0) = A1 = B + D

x2(0) = 0 = B −D⇒ B = D = 1

2 A1

ẋ1(0) = 0 = A + C √

3

ẋ2(0) = 0 = A− C √

3

⇒ A = C = 0

So then holy moly with extra cannoli, the positions of M 1 and M 2 at a funcionof time are

x1(t) = 1

2 A1 cos

k

m t

+

1

2 A1 cos

3k

m t

(14)

x2(t) = 1

2 A1 cos

k

m t

− 1

2 A1 cos

k

m t

(15)

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Jeff Kissel

May 18, 2007

Exam Iteration Question Number Notes/ChangesSpring 2006 X –

Fall 2006 X –Spring 2007 #3 Added to bank.

A planet is in circular motion about a much more massive star. The

star undergoes an explosion where three percent of its mass is ejected

far away, equally in all directions. Find the eccentricity of the new

orbit for the planet.

Initially the planet is in a circular Keplerian orbit, where we know severalthings. The force and potential are

F = GM m

r2 , V = −

GM m

r (1)

For any circular orbit, the radius is given by

F = mac

GM m

r 2c= m

v2t

rc For circular orbits, ṙ = 0

⇒ v2t = ṙ2 + r2 θ̇2 = r2c

θ̇2

GM

rc= θ̇2r2c

r3c = GM

θ̇2

rc =

GM

θ̇2

1/3(2)

Also, from Eq 3.57 on p94 of Goldstein, any mass m orbiting a r−2 central

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force will have eccentricity,

e =

1 +

2Eℓ2

mk2 (3)

with E as the total energy, ℓ = mr2 θ̇ as the angular momentum, and k is this

case as GM m. Since the orbit is circular, the eccentricity is zero, so the initialtotal energy is

0 =

1 +

2Eℓ2

mk2

0 = 1 + 2Eℓ2

mk2

−1 = 2Eℓ2

mk2

E = −mk2

2ℓ2 (4)

And finally, the total energy can also be written as the sum of the kinetic

an potential energies,

E = 1

2 m ṙ2

c + r2

c θ̇2−

GM m

rc

= 1

2 m

GM

θ̇

2/3θ̇ −

GM mGM θ̇

1/3=

1

2 m

GM θ̇2/3

−m

GM θ̇2/3

=

1

2 − 1

m

GM θ̇2/3

E = −1

2 mGM θ̇2/3 (5)

Phew! Now, just after the explosion, the star is at the same radius, rc.However the star has a new mass, M ′ = (1 − λ)M where I’ve denoted λ asthe percentage of mass loss. The only portion of the energy that is effects isthe potential, in which k′ = (1 − λ)k. The angular momentum ℓ = mr2c

θ̇ isindependent of M ′. So, the new energy is

E = 1

2 mr2c

θ̇2 −GM ′m

rc

= 1

2 m

GM

θ̇

2/3θ̇ −

G ((1 − λ)M ) m

GM θ̇

1/3

= 12

m

GM θ̇2/3

− (1 − λ) m

GM θ̇2/3

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=

1

2 − (1− λ)

m

GM θ̇2/3

E = (−1

2 + λ) m

GM θ̇

2/3(6)

The ratio of initial energy to final energy is then,

E ′

E =

(−12

+ λ) m

GM θ̇

2/3−

1

2 m

GM θ̇

2/3E ′

E = −2

−

1

2 + λ

E ′ = (1 − 2λ) E (7)

This in turn affects the eccentricity, which means Eq. 3 becomes,

e′ =

1 +

2E ′ℓ2

mk′2

=

1 + 2(1− 2λ) Eℓ2

m ((1 − λ) k)2

=

1 +

(1 − 2λ)

(1 − λ)22ℓ2

mk2 E

Eq. 4: E = −

mk2

2ℓ2

=

1 −

(1 − 2λ)

(1 − λ)2 2ℓ2

mk2 mk2

2ℓ

e′ =

1 −

1 − 2λ

(1 − λ)2 (8)

For λ = 0.03, the new eccentricity is

e′ = 0.030928 (9)

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Jeff Kissel

October 11, 2006

A homogeneous cube each edge of which has a length ℓ, is initiallyin a position of unstable equilibrium with one edge in contact witha horizontal plane. The cube is then given a small displacement andallowed to fall. Find the angular velocity of the cube when one facestrikes the plane if:

a) the edge cannot slip on the plane (friction).

b) sliding can occur (no friction).

Figure 1: A cube falling with friction holding one corner stationary, and thesame cube sliding on the corner without friction.

For both scenarios we can use a conservation of energy arguement to solvefor the angular velocity.

a) The initial kinetic energy, T i is zero, because the the cube is in (unstable)equilibrium. The potential we can say is simply that of a point mass at thecube’s center of mass, a height h above the ground,

E i = 0T i + V i = mgh

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V i = 1√

2mgℓ (1)

In this case, the edge that is stationary acts as a pivot around which thecube will rotate. Thus, the kinetic energy of the cube while in motion is onlyrotational. We can use the parallel axis theorem to find the moment of iner-

tia rotating about an edge of the cube, noting that the moment of inertia forspinning along the center of mass, I CoM = 1

6 mℓ2, and the displacement, d is

h = 1√ 2 ℓ ;

I || = I CoM + md2 =

1

6 mℓ2 + mh2

= 1

6 mℓ2 + m

ℓ√

2

2

= mℓ2

1

6 +

1

2

=

2

3 mℓ2 (2)

The final kinetic energy as the cube hits the plane is then

T f = 12 I || ω2f

= 1

2

2

3 mℓ2

ω2f

= 1

3 mℓ2ω2f (3)

The final potential is as a point mass at a height, 12 ℓ

V f = 1

2 mgℓ (4)

So conservation of energy dictates that

V i = T f + V f 1√

2mgℓ =

1

3 mℓ2ω2f +

1

2 mgℓ

1√

2g =

1

3 ℓω2f +

1

2 g

1

3 ℓω2f = g

1√

2−

1

2

ωf =

3g

2ℓ

√ 2 − 1

(5)

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b) In this case, the kinetic energy will be slightly different. Now, because theplane is frictionless, the cube rotates about its center of mass, which is moving(only) in the direction of gravity. The position and velocity of the center of mass become

rCoM = yCoM

= h sinφ + π

4

vCoM = ẏCoM

= h cosφ +

π

4

φ̇

= ωℓ√

2cos

φ +

π

4

taking φ̇ to be ω .Though the initial kinetic energy is still zero, the final kinetic energy changes

to

T f = 1

2

mv2f + 1

2

I CoM ω2

f

= 1

2 m

ωf ℓ√

2

2cos2

0φf +

π

4

+

1

2

1

6 mℓ2

ω2f

= 1

4 mℓ2ω2f

1

2

+

1

12 mℓ2ω2f

= 5

24 mℓ2ω2f (6)

Notice that the initial and final potential energies will remain the same,because the cube starts in the same position as in case a), and ends in the sameorientation, just displaced by 1

2 ℓ.

So, using Eq. 1 and 4, the conservation of energy equation is

V i = T f + V f 1√

2mgℓ =

5

24 mℓ2ω2f +

1

2 mgℓ

5

24 ℓω2f = g

1√

2−

1

2

ω2f = 24g

5ℓ

√ 2 − 1

2

ωf =

12g

5ℓ

√ 2 − 1

(7)

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Jeff Kissel

October 11, 2006

A chain of linear density (µ [g/cm]) is hanging vertically above atable. Its lowest point is at a height h above the table. The chain isreleased and allowed to fall. Calculate the force exerted on the tableby the chain when a length x of chain has fallen onto the surface.

Figure 1: A chain suspended a height h above the table at t0, and then x amountfallen onto the table after time t

The force exerted on the table will be equivalent to the sum of the gravita-tional (from whatever part of the chain has already landed on the table) andimpulse (from the part of the chain that has just hit the table) forces. Thegravitational force from the amount of mass m = µx already on the table is

simply F g = mg = µxg (1)

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During a time interval dt, the mass of the rope equal to µ(v dt) is hitting thetable. The change in momentum imparted onto the table is then

dp = dm v

= [µ(v dt)] v

= µv2 dt (2)

The impulse force is then

F impulse = dp

dt = µv2 (3)

However we would like to know how velocity v is related to x(t). Some onedimensional kinematics should shed light on the matter:

v2 − v2i = 2a∆x

⇒ v2 = 2g(x + h) (4)

So Eq. 3 becomes

F impulse = µv2

= 2µg(x + h) (5)Finally, the total force of the table from the falling chain is

F = F g + F impulse

= µxg + 2µ(x + h)g

= 3µxg + 2µhg

= µg(3x + 2h) (6)

which is equivalent to the weight of the length 3x of the rope, plus the correctionfor the initial height.

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Jeff Kissel

October 11, 2006

A thin circular ring of radius R and mass m is constrained to rotateabout a horizontal axis passing through two points on the circumfer-ence. The perpendicular distance from the axis to the center of thering is h.

Figure 1: A thin hoop confined to swing from an off-diameter axis of rotation.

a) Find a Lagrangian for this object.

Looking at the hoop from the side few, it’s apparent that this system is akinto a simple rigid pendulum of length h. Since we know the pendulum’s bob (i.e.the hoop’s center of mass) is confined the plane of the page (or in and out of the page for the front view) a convenient generalized coordinate for this systemis the angle φ as seen in Figure 1. The kinetic energy of the hoop is then

T = 1

2I φ̇2

= 1

2 mR2 + mh2

φ̇2

= 12 m

R2 + h2

φ̇2 (1)

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where the moment of inertia I was found using the parallel axis thereom,

I || = I CoM + md2 (2)

noting that d is the distance from the center of mass to the parallel axis, andI CoM =

1

2mr2 for a thin hoop.

The potential energy will be similar a simple pendulum:

V = mg · r

= −mgh cosφ (3)

So using Eq. 1 and 3 the Lagrangian for a thin hoop hung from an off diameteraxis of rotation is then

L ≡ T − V

= 1

2 m

R2 + h2

φ̇2 + mgh cosφ (4)

b) Find the period of small oscillations about this axis.

Retrieving the equation of motion from the usual Lagrangian formalism will

yield the frequency of small oscillations ω , from which we can find the period.

d

dt

∂L

∂ φ̇

−

∂L

∂φ ≡ 0

d

dt

mR2 + h2

φ̇

+ mgh sinφ = 0

mR2 + h2

φ̈ = − mgh sinφ

φ̈ = −gh

R2 + h2 sinφ (5)

From here, we make the approximation that sinφ ≈ φ so that Eq. 5 becomesa simplified 2nd order differential equation of the form ẍ = −φ̇2x, whose general

solution is x(t) = x0 cos ( φ̇t) in which φ̇ is the frequency. Hence,

φ̈ ≈ −gh

R2 + h2 φ

⇒ φ(t) = φ0 cos

gh

R2 + h2 t

(6)

The period of small oscillations is the time it takes Eq. 6 to repeat the initialconditions at t = 0, such that cos ( φ̇τ − 2π) = cos (0). From this we arrive atthe familiar expression for the period of a pendulum as expected:

τ = 2π

φ̇(7)

= 2π

ghR2 + h2

− 12(8)

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Hoop! There it is!

c) For what value of h is the period minimum?

We wish to minimize τ with respect to h, i.e. set the derivative of Eq. 8equal to zero,

∂τ ∂h

= 2π (gh)−1

2 (R2 + h2)1

2

0 = −1

2 (gh)−

3

2 (g) (R2 + h2)1

2 + 1

2 (gh)

1

2 (R2 + h2)−1

2 (2h)

= −g (R2 + h2)

1

2

2 (gh)3

2

+ 2h

2 (gh)1

2 (R2 + h2)1

2

= −g (R2 + h2)

1

2 (R2 + h2)1

2 + 2h (gh)

2 (gh)

3

2 (R2 + h2)1

2

= −g(R2 + h2) + 2gh2

= −gR2 − gh2 + 2gh2

= g (h2

− R2)

R2 = h2

⇒ h = ±R (9)

Only the non-trivial, positive value for h is physical, so the shortest periodwould result if the axis of rotation which has a distance

h = R (10)

from the center of mass of the hoop.Sanity check: if one wants to minimize the hoop’s period, Eq. 7 says that

one could maximize the frequency. The axis at which the hoop won’t oscillateat all is zero is through center, and an axis with the maximum frequency is thatwhich has the longest “pendulum arm,” i.e. at the edge of the hoop, or h = R.

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Jeff Kissel

May 18, 2007

An object is dropped from a tower of height h. The tower is locatedat the equator of the Earth. The rotational speed of the earth is Ω.

Figure 1: Da Erf. Note that +êx points North, and +êy points West, so Ω =Ω êx.

(a) If the Earth is treated as perfectly round and uniform and theacceleration due to gravity on a non-rotating earth is g, what is theacceleration of gravity at the tower?

With Earth rotating, we must not only consider the acceleration due togravity, but the centripetal and Coriolis acceleration. At the top of the tower,agrav = −g êz, the radial vector is r = +h êz, and the velocity is v = −vr êz.The effective acceleration (in the rotating frame) felt at the tower is then givenas

aeff = agrav + acentrifugal + aCoriolis

= agrav − Ω× Ω × r − 2 Ω ×v

= (−gêz) − (Ωêx) × (Ωêx) × (hêz) − 2(Ωêx) × (−vrêz)

= −g êz − hΩ2 (êx × êx × êz) − 2Ωvr (êx ×−êz)

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= −g êz − hΩ2 (êx ×−êy) − 2Ωvr(−êy)

= −g êz − hΩ2 (−êz) + 2Ωvrêy

aeff = 2Ωvr êy − (g − hΩ2) êz (1)

Yet, at the top of the tower, we assume the object to have no radial velocity,vr = 0. Thus, the effective acceleration is

geff = −(g − hΩ2) êz (2)

(b) Even though it is released from rest, this object will not landdirectly below the point from which it was dropped. Calculate theamount and direction (N, E, S, W, or elsewhere) of the horizontaldeflection of the object. You may assume the deflection is small.

The Coriolis effect is the only term that will have an effect in the horizontal(êy, East or West) direction, so pulling the first term from Eq. 1 if vr is nowgeff t,

F Coriolis = m aCoriolis

= 2mΩvr

êy

= 2mΩ(−(g − hΩ2) t) êy

m ÿ = −2mΩ(g − hΩ2) t

ÿ = −2Ω(g − hΩ2) t

⇒ ẏ = −2Ω(g − hΩ2)

1

2 t2

⇒ y = −2Ω(g − hΩ2)

1

6 t3

y = −Ω

3 (g − hΩ2) t3 (3)

This quantity y is the horizontal deflect, get we can be more explicit. The

time spent falling t can be found from the free-fall kinematics of a particlereleased from rest:

∆y = v0 t −1

2geff t

2⇒ h =

1

2 (g − hΩ2) t2

⇒ t2 = 2h

(g − hΩ2) ⇒ t =

2h

(g − hΩ2)

1/2(4)

which means the deflection becomes

y = −Ω

3 (g − hΩ2)

2h

(g − hΩ2)

3/2

y = −Ω (2h)3/2

3(g − hΩ2)1/2 (5)

The direction of this displacement is in the −êy direction, or to the East.

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Jeff Kissel

October 11, 2006

Consider a pendulum consisting of a uniform disk of radius, R andmass, m suspended from a massless rod that allows it to swing in theplane of the disk.

Figure 1: A circular disk, rotating in its plane on an axis a distance, d awayfrom the center of mass.

(a) Using the parallel axis theorem, or calculating it directly, findthe moment of inertia I for the pendulum about an axis a distance

d(0 < d < R) from the center of the disk

The moment of inertia through the center of a disk of radius R and mass mis

I CoM = 1

2mR2 (1)

Using the parallel axis theorem, which states that the moment of inertia arounda point parallel to the center of mass a distance d away is

I d = I CoM + md2 (2)

So, using Eq. 1 the moment of inertia displaced by |r| = d is

I d = 1

2mR2 + md2

I d = 12

m (R2 + 2d2) (3)

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(b) Find the gravitational torque on the pendulum when displacedby and angle φ.

From the definition of rotational torque from a given force:

τ = r × F g=

|r

| | F

| sin φ

= (d)(−mg) sin φ= −mgd sin φ (4)

(c) Find the equation of motion for small oscillations and give the fre-quency ω . Further, find the value of d corresponding to the maximumfrequency for fixed R and m.

If we note that τ = I α, then we know the equation of motion immediatelyfrom Eq. 4(using the small angle approximation that sin φ = φ),

I dα = −mgd φα = −mgd

I dφ

(5)

which is a 2nd order differential equation of the form φ̈ = −ω2φ from which wecan pull out ω . Subbing in Eq. 3,

ω2 = 2 mgd

m (R2 + 2d2)

ω =

2gd

R2 + 2d2

2(6)

To find the value of d for which ω is maximized, we shall differentiate withrespect to d and set to zero,

dωdd = ddd

(2gd)(R2 + 2d2)

−1

0 = 1

2

2gd

R2 + 2d2

− 12

2g (R2 + 2d2)−1 − (2gd)(R2 + 2d2)−2(4d)

0 = 1

2

R2 + 2d2

2gd

12

2g

R2 + 2d2 − 8gd

2

(R2 + 2d2)2

0 = 1

2

R2 + 2d2

2gd

12

2g(R2 + 2d2) − 8gd

(R2 + 2d2)2

0 =

R2 + 2d2 12

2gR2 + 4gd2 − 8gd2

0 = R2 + 2d21

2 ( 2g(R2

−2d2))

0 =

R2 + 2d2 12

R2 − 2d2

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R2 + 2d2 = 0 R2 − 2d2 = 0d =

iR√ 2

d = ± R√ 2

(7)

But we know the left set is unphysical and − R√ 2

is equivalent to R√ 2

, so

dmax = R√ 2

(8)

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Jeff Kissel

October 11, 2006

A heavy particle of mass m is placed close to the top of a frictionless

vertical hoop with radius R and allowed to slide down the hoop. Find

the angle at which the particle falls off.

We know the point at which the particle falls off the hoop is when the normalforce N acting on the particle goes to zero. So, we need an expression for N .

Figure 1: Ball of mass m on a hoop with radius R.

Before the particle begins to move, it has only potential energy V i. At theinstant it leaves the hoop, it has both potential V f and kinetic T f energies. Theparticle is allowed to slide, so there assumed to be no friction and therefore norolling, thus we need no worry about the moment of inertial of the ball. Theenergy conservation equation is then

E i = E f

V i = T f + V f

mgR = 1

2mv2 + mgR cos θ (1)

However, Eq. 1 does not contain the normal force N , which we need. It can be

found by assuming the hoop is circular, so that the normal force acting on the

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particle is just its centripetal acceleration subtracted from its weight, or

N = mg cos θ −mac

= mg cos θ −mv2

R (2)

Multiplying this by 1

2R will yield a more useful equation for the kinetic energy,

1

2R → (N = mg cos θ −mac)

1

2RN =

1

2mgR cos θ −

1

2mv2

1

2mv2 =

1

2mgR cos θ −

1

2RN (3)

which we can now plug into the conservation equation (Eq. 1) to arrive atexplicit equation for N ,

mgR = 1

2mgR cos θ −

1

2RN + mgR cos θ

12

RN = 32

mgR cos θ −mgR

N = 3mg cos θ − 2mg (4)

Since we’ve established that the particle falls off the hoop when N = 0, Eq. 4becomes,

0 = mg(3 cos θ − 2)

= 3 cos θ − 2

cos θ = 2

3 (5)

So the angle at which a heavy particle of mass M slides off a circular hoop or

radius R isθ ≈ 48.2◦ (6)

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Jeff Kissel

October 11, 2006

A solid sphere of radius R and mass M rolls without slipping down a

rough inclined plane of angle θ.

Figure 1: A sphere rolling down a rough ramp.

Take the coefficient of static friction to be µ and calculate the lin-

ear acceleration of the sphere down the plane, assuming that it rolls

without slipping.

In this problem, we can use conservation of energy between the top andbottom of the ramp. At the top of the ramp, the ball begins rolling from restand thus has no initial kinetic energy. This means the initial energy is

E i = 0

T i + V i

= mCoM g x sin θ (1)

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where the origin is set at the top of the ramp.The final energy has only kinetic terms, and is a sum of the motion of the

center of mass, and the rotation of the sphere,

E f = T f + 0

V f

= 12

mCoM v2 + 12

I CoM ω2 (2)

Setting Eqs. 1 and 2 equal to each other, plugging in for the moment of inertia of a sphere, I CoM =

25

mCoM R2 and noting that v = ẋ, and ω = ẋ

R,

E i = E f

mCoM g x sin θ = 5

10 mCoM ẋ

2 + 1

2

2

5 mCoM R

2

ẋ

R

2

ẋ2 = 10

7 g x sin θ

dx

dt = ξx

1

2 (3)

where I’ve defined a constant ξ =

107

g sin θ. At this point we integrate both

sides to find x as a function of time, which we can then differentiate twice withrespect to time to get the acceleration, ẍ = aCoM ,

x−

1

2 dxξ

= dt

2x1

2

ξ = t

x = 1

4 ξ 2 t2 ⇒ ẋ = 12 ξ

2 t ⇒ ẍ = 1

2 ξ 2

ẍ = 12

107 g sin θ

2

aCoM = 7

5 g sin θ (4)

Calculate the maximum value of θ for which the sphere will not slip.

The sphere will begin to slip once the x-component of the weight becomesgreater than the maximum frictional force providable by µ. The angle at which

this occurs, θc can be found by setting the normal force, F (x̂)g equal to the force

of friction F f = µN ,

mg sin θc = µN

mg sin θc = µ mg cos θc

µ = sin θc

cos θc= tan θc

⇒ θc = tan−1

(µ) (5)

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Jeff Kissel

October 11, 2006

A rocket is filled with fuel and is initially at rest. It starts moving by

burning fuel and expelling gases with the velocity u, constant relative

to the rocket. Determine the speed of the rocket at the moment when

its kinetic energy is largest.

Figure 1: A free rocket traveling in an inertial frame, without the force of gravity.

At some time t the rocket, traveling at a velocity v, has some mass m. Aninfinitesimal time dt later, the rocket has lost a mass dm, and gained a speeddv. At this time, an infinitessimal amount of fuel, dm′ is ejected at a velocityu with respect to the rocket. In an inertial frame however, that fuel’s velocityis v − u. Thus, using conservation of momentum, we can get an expression forthe mass of the rocket as a function of time,

procket = procket + pfuel

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m v = (m − dm)(v + dv) + dm′(v − u)

m v = m v + m dv − v dm′ − du dv + v dm′ − u dm′

m dv = u dm′ − du dv (1)

From here, we note that to first order, the term du dv is extremely small, so itis dropped, leaving

1

u dv =

1

m dm′ (2)

Noting that dm′ = −dm and integrating both sides give us an express forthe mass of the rocket as a function of time: v

0

1

u dv′ = −

mm0

1

m dm

−

v

u = ln

m

m0

m = m0e−

v

u (3)

Plugging into the expression of the rocket’s kinetic energy,

T = 12

mv2 = 1

2 m0v

2 e−v

u (4)

which we then maximize with respect to v to find the largest speed vmax,

∂T

∂v =

∂

∂v

1

2 m0v

2 e− v

u

0 = m0v e−

v

u−

1

2 m0v

2 1

u e−

v

u

m0v e−

v

u = v

2u

m0v e−

v

u

vmax = 2u (5)

Thus, the kinetic energy is maximum when the velocity of the rocket is twicethat of the ejected fuel.

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Jeff Kissel

May 19, 2007

Exam Iteration Question Number Notes/ChangesSpring 2006 #12 –

Fall 2006 #12 –Spring 2007 #12 Added damping.

A small lead ball of mass m is attached to one end of a vertical spring

with the spring constant k. The other end of the spring oscillates up

and down with amplitude A and frequency ω. Determine the motion

of the ball after a long period of time. You may assume that the

ball is subject to a small amount of damping, with the damping force

being given by F = −γv .

Figure 1: A driven harmonic oscillator sitting in a viscous damping fluid.

Hopefully, one can assume that all motion is in the ŷ direction.

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The kinetic energy of the system is

T = 1

2 mẏ2 (1)

and the potential, as long as we define the origin to be at the equilibrium pointsuch that the displacement ∆y ≡ y − ye = y, is as follows,

V = 1

2 ky2 + mgy (2)

This makes the Lagrangian

L ≡ T − V

L = 1

2 mẏ2 −

1

2 ky2 −mgy (3)

We can use the form of the Euler-Lagrange equations which has forces notderivable from a potential, Q, to determine the equation of motion. The forcesnot derivable from potential are the driving and damping forces,

F o

= A cos(ωt) (4)

F d = −γ ẏ (5)

So, the equation of motion is

d

dt

∂L

∂ ẏ

−

∂L

∂y = Qi

d

dt (mẏ)− (−ky −mg) = A cos(ωt)− γ ẏ

mÿ + ky + mg = A cos(ωt)− γ ẏ

mÿ + γ ẏ + ky = A cos(ωt)−mg

ÿ + γ

m ẏ +

k

m y =

A

m cos(ωt)− g

Let β = γ m

, ω20 = km

, and B = Am

ÿ + β ẏ + ω20 y = B cos(ωt)− g

Let x = − gω20

+ y ⇒ y = x + gω20

ẋ = ẏẍ = ÿ

ẍ + β ẋ + ω20

x +

g

ω20

= B cos (ωt) + g

ẍ + β ẋ + ω20x + g = B cos (ωt) + g

ẍ + β ẋ + ω20x = B cos (ωt) (6)

Yes folks, it’s the driven harmonic oscillator with damping. Your favoritedifferential equation to solve.

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We’ll start with the homogeneous solution.

0 = ẍH + β ẋH + ω20xH

The characteristic equation:

0 = am2± + bm± + c

⇒ 0 = (1)m2± + (β )m± + (ω20)

has roots

m = p± iq

with p = − b

2 and q =

c −

b2

4

⇒ p = − β

2 and q =

ω20 −

β 2

4

Such that the general solution is

e pt (C cos (q t) + D sin(q t))

xH (t) = e−

1

2βt

C cos

ω20 −

β 2

4 t

+ D cos

ω20 −

β 2

4 t

(7)

For the inhomogeneous solution, we can generalize the equation making itcomplex,

̈xI + β ̇xI + ω20xI = Be−iωtand then take the real part at the end. This way we can assume a complexexponential solution,

xI =

G ei(ωt), which knocks out the differentiation because

xI = G e−iωt ̇xI = − iω G e−iωt ̈xI = − ω2 G e−iωt (8)which means all we have to do is find the complex constant G,

Be−iωt = −ω2 G e−iωt +−iωβ G e−iωt + ω20 G e−iωtB = −ω2 G +−iωβ G + ω20 GB = G (ω20 − ω2)− iωβ G = B

(ω20 − ω2)− iωβ

= B

(ω20 − ω2)− iωβ

(ω20 − ω

2) + iωβ

(ω20 − ω2) + iωβ

G = B(ω20 − ω2) + iBωβ

((ω20 − ω2) + ω2β 2)

(9)

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So the inhomogeneous solution is

xI (t) = B(ω20 − ω2)((ω20 − ω

2) + ω2β 2) e−iωt +

Bωβ

((ω20 − ω2) + ω2β 2)

ie−iωt

⇒ xI (t) = ℜe

B(ω20 − ω

2)

((ω20 − ω2) + ω2β 2)

e−iωt + Bωβ

((ω20 − ω2) + ω2β 2)

ie−iωt

ℜe e−iωt = cos (ωt)ℜe

ie−iωt

= sin (ωt)

xI (t) =

B(ω20 − ω2)

((ω20 − ω2) + ω2β 2)

cos (ωt) + Bωβ

((ω20 − ω2) + ω2β 2)

sin (ωt) (10)

Yeah. The complete solution for x(t) is

x(t) = e−1

2βt

C cos

ω20 −

β 2

4 t

+ D cos

ω20 −

β 2

4 t

+ B(ω20 − ω

2)

((ω20 − ω2) + ω2β 2)

cos (ωt) + Bωβ

((ω20 − ω2) + ω2β 2)

sin (ωt)

(11)

Oh but wait, we have to plug back in for y(t) = x(t) + gω20

,

y(t) = e−

1

2βt

C cos

ω20 −

β 2

4 t

+ D cos

ω20 −

β 2

4 t

+ B(ω20 − ω

2)

((ω20 − ω2) + ω2β 2)

cos (ωt) + Bωβ

((ω20 − ω2) + ω2β 2)

sin (ωt)− g

ω20

And remember: β = γ

m, ω20 =

k

m, and B =

A

m

y(t) = e−

γt

2m

C cos

k

m −

γ 2

4m2 t

+ D cos

k

m −

γ 2

4m2 t

+ A(k −mω2)

(k −mω2)2 + ω2γ 2 cos(ωt) +

Aωγ

(k −mω2)2 + ω2γ 2 sin(ωt)−

mg

k

(12)

Which is the position as a function of time, where C and D are constantsdetermined by initial conditions.

OK so what happens after a long time? The exponential damping in thehomogeneous term kills off all resonances that would be found from ω0, and thespring just oscillates according to the inhomogeneous equation. In other words,the spring just sloshes along with the driving frequency, according to

limt→∞

y(t) = A(k −mω2)

(k −mω2)2 + ω2γ 2 cos(ωt) +

Aωγ

(k −mω2)2 + ω2γ 2 sin(ωt)−

mg

k

(13)

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Jeff Kissel

May 17, 2007

Exam Iteration Question Number Notes/ChangesSpring 2006 #13 –

Fall 2006 #13 –Spring 2007 #13 Last sentence changed

In movie cameras and projectors film speed is 24 frames/sec. Yousee on the screen a car that moves without skidding, and know that

the real-life diameter of its wheels is 1 m. The wheels on the screenmake 3 turns/sec. What is the speed of the car, assuming it is notmoving in excess of 200 mi/hr?

Figure 1: A wheel that appears to rotate one eighth of its circumference.

In one frame, it appears to the viewer that the wheel has turned and angularvelocity

ω =

3 [turns] [sec]

24 [frames]

[sec]

= 1

8

turns

frame (1)

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However, appearances can be deceiving. To the viewer, “measuring” the po-sition of the wheel discretely, the wheel has turned only 1/8 of its circumference.The car could be moving much faster, such that the wheel makes any integernumber N of extra turns before reaching the final 1/8 (for example, N=1 inFigure 1. The angular velocity should then technically be

ω =

N + 18

turnsframe

(2)

Finally, to find the (linear) speed v, we must convert Eq. 2 into the correctunits,

ω =

N +

1

8

[turns]

[frame]

× 24

[frames]

[sec] × 2π

[radians]

[turn]

ω = 48π

N +

1

8

rads/s (3)

and then convert angular to linear speed, noting that the wheel has a 1 mdiameter , and thus a 0.5 m radius.

v = ω r

=

48π

N +

1

8

rads/s

(0.5 m)

= 24π

N +

1

8

m/s (4)

Thus for N = 0, v = 3π m/s ≈ 6.7 mi/hr, and N = 1, v = 27π m/s ≈189.7 mi/hr. (1 m/s = 2.23693629 mi/hr).

Though this problem may seem pretty lame, it brings up an important con-cepts in signal processing: digital sampling of analog signals, which may resultin “aliasing.” Check out the Wikipedia article on it!

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Jeff Kissel

October 11, 2006

A ball of mass m collides with another ball, of mass M = am, initiallyat rest. The collision is elastic and central, i.e. the initial velocity of the ball is along the line connecting the centers of the two objects.

Figure 1: Two balls collide with mass ratio a.

(a) Determine how the energy lost by the moving ball depends onthe mass ratio a, and find the value of a for which the energy loss islargest. Describe what happens at that value of the mass ratio.

We can solve this problem using conservation of energy and momentum.The problem does not mention any potential, so we can find the velocity of thesecond ball in terms of the first starting with kinetic energy conservation.

E i = E f

1

2mv2i +

01

2M ui

2 = 1

2mv2f +

1

2M u2f

m v2

i = m v2

f + a m u2

f

v2i = v2

f + au2

f (1)

We can now bring in conservation of momentum to knock out one of theunknowns,

pi = pf

mvi + M ui = mvf + M uf

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m vi = m vf + a muf

vf = vi − auf (2)

Now, pluggind Eq. 1 into Eq. 2,

v2i = (vi − auf )2 + au2f

v2

i = v2

i + a2

u2

f − 2aviuf + av2

f

0 = a (a + 1) u 2

f − 2 a vi uf

−(a + 1)uf = −2 vi

uf = 2vi(a + 1)

(3)

Finally, the kinetic energy lost by the first ball is just that that is gained bythe second,

∆E = 1

2M u2

= 1

2 am

2via + 1

2

∆E = 2amv2i

(a + 1)2 (4)

To find the value for which the energy lost is largest, we can just maximizewith respect to a,

∂ ∆E

∂a =

∂

∂a

2mv2i a (a + 1)

−2

0 = 2mv2i (a + 1)−2− 4mv2i a (a + 1)

−3

2mv2i

1

(a + 1)2 =

2a

(a + 1)3 2mv2i

(a + 1) 3 = 2a

(a + 1)2

a + 1 = 2aa = 1 (5)

So, the change in energy is the greatest when the balls have the same exactmass: the first ball’s kinetic energy is split evenly between them during thecollision such that vf and uf are equal, but in opposite direcetions.

(b) Investigate the limiting cases of heavy and light balls and com-ment on your result.

In the limiting case that a ≪ 1, the first ball will be much more massivethan the second. In this case, the first ball will lose virtually none of its energy,but because the second ball is so less massive, it will propel off with uf = 2vi.

lima→0

∆E = 2amv2

i

( a + 1)2

= 2M v2i

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lima→0

vf = vi − auf = vi

lima→0

uf = 2vi( a + 1)

= 2vi

In the case where a ≫ 1, the energy imparted onto the second ball will beneglegible. In addition, since so little velocity is transferred to the second ball,

the first will be forced in the opposite direction with a final velocity equal to itsinitial.

lima→∞

∆E = 2amv2i(

∞a + 1)2

= 0

lima→∞

vf = vi − 2 avi

(a + 1) = −vi

lima→∞

uf = −2vi(

∞a + 1)

= 0

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Jeff Kissel

October 11, 2006

Two particles of mass m and m + 2 move in circular orbit around eachother under the influence of gravity. The period of the motion is T .

They are suddenly stopped and then released, after which they fall

towards each other. If the objects are treated as mass points, find

the time they collide in terms of T .

Figure 1: (Left) Two masses orbiting about each other suddenly stopped, as atwo body problem. (Right) The same system reduced to a one-body problem

using the reduced mass µ.

As noted in the picture, the problem can be simplified to a one body problemusing the reduced mass,

µ = m1m2

m1 + m2

= m2 + 2m

2m + 2

µ = 1

2

m + 2

(1 + m−1) (1)

Now a one body problem, we can find the Lagrangian of this more simple system,

V = − kr

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T = 1

2 µṙ2 +

1

2 µ r2 θ̇2

L = T − V

L = 1

2 µ (ṙ2 + r2 θ̇2) +

k

r (2)

where I’ve defined the gravitational constant of the system k = Gm1m2. Fromhere we crank the usual Lagrangian formalism,

d

dt

∂L

∂ ṙ

− ∂L

∂r = 0

d

dt (µṙ) − (µrθ̇ − k

r2) = 0

µr̈ − µrθ̇ + kr2

= 0 (3)

d

dt

∂L

∂ θ̇

− ∂L

∂θ = 0

d

dtµr2 θ̇− (0) = 0

µr2θ̈ = 0

θ̈ = 0 (4)

It appears as though the sooner will be a little more useful than the latter,so let’s roll with it. We know for circular motion, not only is Eq. 4 true, butr̈ = 0 is also true. Using this fact, we can distill Eq. 3 a bit to get a familiarlaw from my buddy Kepler:

µr̈ − µrθ̇ + k

r2 = 0

k

r2 = µrθ̇

r3 = kµθ̇2

(θ̇ = ω = 2πT

)

r3 = kT 2

4π2µ (5)

Now that we’ve naively re-derived Kepler’s 3rd, let’s stop the masses. TakingEq. 5 as our starting radius, r3

0, we know stop the angular motion of the masses,

i.e. set ˙theta, so the Eq. 3 instead becomes

0 = µr̈ − µrθ̇ + k

r2

r̈ = − kµr2

( r̈ = ∂ ṙ∂t

= ∂r∂t

∂ ṙ∂r

= ṙ ∂ ṙ∂r

)

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ṙ dṙ = − kµr2

dr

1

2 ṙ2 =

k

µr + C

ṙ2 = 2k

µr + 2C

We can use the initial condition that at r = r0, ṙ = 0, so that C = − kµr0 , andfinally we can solve for the linear merger time,

ṙ =

2k

µr − 2k

µr0

dr =

2k

µ

1

r − 1

r0

12

dt

dt = dr

2kµ

1

r − 1

r01

2

t =

µ2k

12 0r0

1r − 1

r0

− 12

dr (6)

which is a horribly disgusting integral, that can be solved analytically using afew “u substitutions,” (let u = 1

r, and then u = u0 sec

2 θ) but I’ll happily justleave it as Eq. 6. Numerically evaluated this turns out to be

t ≈ T 8√

2(7)

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Jeff Kissel

October 11, 2006

Three points masses of identical mass are located at (a, 0, 0), (0, a, 2a)and (0, 2a, a). Find the moment of inertia tensor around the origin,the principal moments of inertia, and a set of principal axes.

Figure 1: A system of three identical masses.

The moment of inertia tensor for a discrete mass system is given by

I = I jk =i

mi

δ jkr2

i − (xjxk)i

(1)

So for

r1 = a x̂ r2

1 = a2

r2 = a ŷ + 2a ẑ r2

2 = a2 + 4a2 = 5a2

r3 = 2a ŷ + a ẑ r2

3 = 4a2 + a2 = 5a2

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the diagonal elements of I are

I 11 =i

mi

r2i − xixi)

= m(r21 − x1x1) + m(r2

2 − x2x2) + m(r2

3 − x3x3)

= m(a2 − a2) + m(5a2 − 0) + m(5a2 − 0)

= 10ma2 (2)

I 22 =i

mi

r2i − yiyi)

= m(r21 − y1y1) + m(r

2

2 − y2y2) + m(r

2

3− y3y3)

= m(a2 − 0) + m(5a2 − a2) + m(5a2 − 4a2)

= 6ma2 (3)

I 33 =i

mi

r2i − zizi)

= m(r21 − z1z1) + m(r2

2 − z2z2) + m(r2

3 − z3z3)

= m(a2 − 0) + m(5a2 − 4a2) + m(5a2 − a2)

= 6ma2

(4)

And the off-diagonal elements are

I 12 = I 21 =i

mi(−xiyi)

= m(−x1y1) + m(−x2y2) + m(−x3y3)

= m(0) + m(0) + m(0)

= 0 (5)

I 13 = I 31 =i

mi(−xizi)

= m(−x1z1) + m(−x2z2) + m(−x3z3)

= m(0) + m(0) + m(0)

= 0 (6)

I 23 = I 32 =i

mi(−yizi)

= m(−y1z1) + m(−y2z2) + m(−y3z3)

= m(0) − m(2a2) − m(2a2)

= −4ma2 (7)

So the moment of inertia tensor for this system is

I =

10ma2 0 0

0 6ma2 −4ma2

0 −4ma2 6ma2

(8)

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The principle moments of inertia and principle axes are the eigenvalues andeigenvectors (respectively) of the moment of inertia tensor. So to find eigenval-ues, we find the characteristic equation of I

det

I − λ

1

=

10ma2 − λ 0 00 6ma2 − λ −4ma2

0 −

4ma2

6ma2−

λ

0 = (10ma2 − λ)

(6ma2 − λ)(6ma2 − λ) − (−4ma2)2− 0 + 0

= (10ma2 − λ)

36m2a4 − 12λma2 + λ2 − 16m2a4

= (10ma2 − λ)

λ2 − 12λma2 + 20m2a4

= (10ma2 − λ)(10ma2 − λ)(2ma2 − λ)

⇒ λ1 = 10ma2 λ2 = 10ma

2 λ3 = 2ma2 (9)

so there is a two-fold degeneracy in the moments of inertia which means thesystem is akin to a symmetrical top. To find the eigenvectors (principle axes),we demand that the principle moments of inertia obey the eigenvalue equation(I − λ

1) · x = 0. So for λ1 = λ2 = 10ma

2,

0 0 00 −4ma2 −4ma20 −4ma2 −4ma2

· xy

z

= 00

0

−4ma2 y − 4ma2 z = 0

⇒ y = − z

so, picking the simplest eigenvectors, the principle axes for λ1 & λ2 are

χ1 =

0−1

1

and χ2 =

10

0

(10)

And for λ3 = 2ma2,

8ma2 0 0

0 4ma2 −4ma2

0 −4ma2 4ma2

· xy

z

= 00

0

8ma2 x = 0

4ma2 y − 4ma2 z = 0

−4ma2 y + 4ma2 z = 0

⇒ x = 0, y = z

and again picking the simplest eigenvector, the principle axis for λ3 is

χ3 =

01

1

(11)

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Jeff Kissel

October 11, 2006

Consider a double pendulum consisting of a mass m suspended ona massless rod of length ℓ, to which is attached by a pivot anotheridentical rod with an identical mass m attached at the end, as shownin the Figure 1.

Figure 1: A double pendulum, with identical masses and string lengths.

Using the angles θ1 and θ2 as generalized coordinates,

a) Find a Lagrangian for the system.

Using θ1 and θ2 as generalized coordinates, the position (r1, r2) and velocity(v1, v2) vectors become

r1 = (ℓ sin θ1) x̂ + (ℓ cos θ1) ŷ (1)

v1 = ̇r1 = ℓθ̇1 cos θ1 x̂− ℓθ̇1 sin θ1 ŷ (2)r2 = (ℓ sin θ1 + ℓ sin θ2) x̂ + (ℓ cos θ1 + ℓ cos θ2) ŷ

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= ℓ (sin θ1 + sin θ2) x̂ + ℓ (cos θ1 + cos θ2) ŷ (3)

v2 = ̇r2 =

ℓθ̇1 cos θ1 + ℓθ̇2 cos θ2

x̂−

ℓθ̇1 sin θ1 + ℓθ̇2 sin θ2

ŷ

= ℓ

θ̇1 cos θ1 + θ̇2 cos θ2

x̂− ℓ

θ̇1 sin θ1 + θ̇2 sin θ2

ŷ (4)

In the kinetic energy term of the Lagrangian, we’ll need the squares of thevelocities, which are

v21 = v1 · v1 =

ℓθ̇1 cos θ1

2+

ℓθ̇1 sin θ1

2= ℓ2 θ̇2

1

cos2 θ1 + sin

2 θ1

= ℓ2 θ̇21 (5)

v22

= v2 · v2 =

ℓθ̇1 cos θ1 + ℓθ̇2 cos θ22

+

ℓθ̇1 sin θ1 + ℓθ̇2 sin θ22

= ℓ2 θ̇21 cos2 θ1 + ℓ

2 θ̇22 cos2 θ2 + 2ℓ

2 θ̇1 θ̇2 cos θ1 cos θ2

+ ℓ2 θ̇21

sin2 θ1 + ℓ2 θ̇2

2 sin2 θ2 + 2ℓ

2 θ̇1 θ̇2 sin θ1 sin θ2

= ℓ2 θ̇21 cos2 θ1 + sin

2 θ1 + ℓ2 θ̇22 cos

2 θ2 + sin2 θ2

+ 2ℓ2 θ̇1 θ̇2 (cos θ1 cos θ2 + sin θ1 sin θ2)= ℓ2 θ̇2

1 + ℓ2 θ̇2

2 + 2ℓ2 θ̇1 θ̇2 cos (θ2 − θ1) (6)

Finally, noting that g = −g ŷ and m1 = m2 = m we can combine the resultsof Eqs. 1, 3, 5, and 6 to find the kinetic and potential energies:

T = 1

2mv2

1 +

1

2mv2

1

= 1

2 mℓ2 θ̇2

1 +

1

2 m

ℓ2 θ̇2

1 + ℓ2 θ̇2

2 + 2ℓ2 θ̇1 θ̇2 cos (θ2 − θ1)

= mℓ2 θ̇2

1 +

1

2 mℓ2 θ̇2

2 + mℓ2 θ̇1 θ̇2 cos (θ2 − θ1) (7)

V = mg · r1 + mg · r2

= −mgℓ cos θ1 −mgℓ (cos θ1 + cos θ2)

= −2mgℓ cos θ1 −mgℓ cos θ2 (8)

which we combine to form a Lagrangian for a double pendulum with identicalmasses and string lengths:

L ≡ T − V

= mℓ2 θ̇21 + 1

2mℓ2 θ̇22 + mℓ

2 θ̇1 θ̇2 cos (θ2 − θ1) + 2mgℓ cos θ1 + mgℓ cos θ2

= 1

2mℓ2

2θ̇21 +

θ̇22 + 2θ̇1 θ̇2 cos (θ2 − θ1)

+ mgℓ (2cos θ1 + cos θ2) (9)

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b) Find an approximate Lagrangian that is appropriate for small oscil-lations and obtain from it the equations of motions when θ1 & θ2 ≪ 1.

For small oscillations, we can approximate cos x ≈ 1 − 12

x2 + O(x4), and

drop those terms which are of higher order than quadratic in θi and θ̇i so that

Eq. 9 is

L ≈ mℓ2 θ̇21

+ 1

2mℓ2 θ̇2

2 + mℓ2 θ̇1 θ̇2

1−

1

2

(θ2 − θ1)2

+ 2mgℓ

1−

1

2θ21

+ mgℓ

1−

1

2θ22

where what is canceled above will turn out to be quartic in θi and θ̇i so it isdropped. Simplified, the approximate Lagrangian is then

L ≈ mℓ2 θ̇21 + 1

2mℓ2 θ̇22 + mℓ

2 θ̇1 θ̇2 + 2mgℓ −mgℓθ2

1 + mgℓ − 1

2mgℓθ22

≈ 1

2

mℓ2 2θ̇21 + θ̇22 + 2θ̇1 θ̇2− 12

mgℓ 2θ21 + θ22 + 3mgℓ (10)

c) Assuming that each angle varies as θ1,2 = A1,2 eiωt, find the frequen-

cies for small oscillations.

Performing the usual Lagrange processes using Eq. 10 to find the equationsof motion, we can then use these to solve for the frequencies of small oscillations.In general, Lagrange’s equations of motion are defined by

d

dt

∂L

∂ q̇ i

−

∂L

∂q i= 0 (11)

where q i are the generalized coordinates of the system, and q̇ i are their respective

time derivatives.So the equations of motion for θ1 and θ2 are

∂L

∂ θ̇1= 2mℓ2 θ̇1 + mℓ

2 θ̇2∂L

∂θ1= −2mgℓ θ1

⇒ 2mℓ2θ̈1 + mℓ2θ̈2 + 2mgℓ θ1 = 0

ℓ

2θ̈1 + θ̈2

+ 2gθ1 = 0 (12)

∂L

∂ θ̇2= mℓ2 θ̇2 + mℓ

2 θ̇1∂L

∂θ2= −mgℓ θ2

⇒ mℓ2θ̈2 + mℓ2θ̈1 + mgℓ θ2 = 0

ℓθ̈1 + θ̈2 + g θ2 = 0 (13)

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If we assume θj = Aj eiωt, then θ̈j → −ω

2θj , and the equations of motion yieldfour oscillation frequencies:

ℓ−2ω2θ1 − ω

2θ2

+ 2g θ1 = 0

−ω2ℓ

2A1

eiωt + A2 eiωt

+ 2gA1

eiωt = 0

ω

2

ℓ (2A1 + A2) = 2gA1

ω2 = 2gA1ℓ(2A1 + A2)

ω± = ±

2gA1

ℓ(2A1 + A2) (14)

ℓ−ω′2θ1 − ω

′2θ2

+ g θ2 = 0

−ω′2ℓ

A1

eiω′t + A2

eiω

′t

+ gA2

eiω′t = 0

ω′2ℓ (A1 + A2) = gA2

ω′2 = gA2

ℓ(A1 + A2)

ω′± = ±

gA2

ℓ(A1 + A2) (15)

So ’der you go.

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Jeff Kissel

October 22, 2006

A particle of mass m. at rest initially, slides without friction on awedge of angle θ and and mass M that can move without friction ona smooth horizontal surface.

Figure 1: A mass m slides in the +x̂ direction down a frictionless wedge M ,which can slide across a frictionless plane in the x̂ direction. Note, this picturedoes not represent the actual motion; realistically Ẋ is in the −x̂ direction aswill be shown later.

a) What is the Hamiltonian?

First we must find the Lagrangian (L = T − V ), to determine whether it isexplicitly independent of time. If this is so, then the Hamiltonian is simply thetotal energy (H = T + V ).

We can define a coordinate system where the origin is the top of the wedgeat some initial time as in Figure 1, such that the position vector for the topcorner of the wedge at time, t will be

R = X x̂ + (0) ŷ (1)

and particle’s position vector will be

r = R + (x x̂ + y ŷ)

r = (X + ℓ cos θ) x̂ + (ℓ sin θ) ŷ (2)

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where ℓ is the distance the mass has traveled down the ramp. While the particleis on the ramp, it is constrained to move along ℓ, or

tan θ = ∆y

∆xsin θ

cos θ

= y − Y

x − X (x − X )sin θ = (y − (0)) cos θ

f : (x − X )sin θ − y cos θ = 0 (3)

which is an equation of constraint related to the normal force.The kinetic energy of the system is that of two free particles, since there is

no friction:

T = 1

2 m(ẋ2 + ẏ2) +

1

2 M Ẋ 2 (4)

The only potential in the system is gravitational, with g = g ŷ. We can ignorethe potential on the wedge because it is constant, so the system’s potential is

just the gravitational potential of the particle,

V g ≡ − mg · r

V = −mgy (5)

The Lagrangian is then potential subtracted from the kinetic energy,

L ≡ T − V

= 1

2 m(ẋ2 + ẏ2) +

1

2 M Ẋ 2 + mgy

L = 1

2 mẋ2 +

1

2 mẏ2 +

1

2 M Ẋ 2 + mgy (6)

which does not explicitly depend on time, so the Hamiltonian is the sum of thekinetic and potential energies,

H = 1

2 mẋ2 +

1

2 mẏ2 +

1

2 M Ẋ 2 − mgy (7)

b) Derive the equation of motion from the Lagrangian.

With Lagrange multipliers, The Euler-Lagrange equation in general is

d

dt

∂L

∂ q̇ i

−

∂L

∂q i= λ

∂f

∂q i+ µ

∂f

∂ q̇ i(8)

yet the second term on the right-hand side is zero because Eq. 3 is not a functionof any q̇ i. From Eq. 8, the equations of motion for each coordinate are,

ddt

(mẋ) = λ (sin θ)

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mẍ = λ sin θ (9)

d

dt (mẏ) − mg = λ (− cos θ)

mÿ − mg = −λ cos θ (10)

d

dt

m Ẋ

= λ (− sin θ)

m Ẍ = −λ sin θ (11)

Immediately, if we add Eq. 11 to Eq. 9, we see why Figure 1 has the directionof motion for the wedge incorrect,

mẍ + M Ẍ = 0

ẍ = −M

mẌ (12)

Also, this tells us that the x̂ position of the system’s center of mass stays sta-tionary because the only external force on the system is gravity, which is in the

ŷ direction. We’ve assumed that the particle’s initial position is at the origin,but if we also assume that the initial velocity of the center of mass is also zerothen from Eq. 12,

∂

∂t

∂x

∂t = −

M

m

∂

∂t

∂X

∂t∂x

∂t ∂t = −

M

m

∂X

∂t ∂t

x0

∂x = −M

m

X0

∂X

x = −M

m X or X = −

m

M x (13)

giving us an equation relating X in terms of x. We can use the equation of constraint (Eq. 3) to get y in terms of x,

(x − X )sin θ = y cos θ

(x + m

M x)sin θ = y cos θ

y = x(1 + m

M ) tan θ (14)

from which only y and x are a function of time, so

ẏ = ẋ(1 + m

M ) tan θ

ÿ = ẍ(1 + m

M ) tan θ (15)

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OK, OK, let’s finally get the equations of motion. Starting with Eq. 9,

mẍ = λ sin θEq. 10: mÿ − mg = −λ cos θ ⇒ λ = −

m(ÿ − g)

cos θ

mẍ = m(g − ÿ)tan θ (16)

Eq. 15: ÿ = ẍ(1 + mM

) tan θ

mẍ = mg tan θ − mẍ(1 + m

M ) tan2 θ

mẍ + mẍ tan2 θ + mẍ m

M tan2 θ = mg tan2 θ

m(1 + tan2 θ +

m

M tan2 θ)ẍ = mg tan θ

(cos2 θ

cos2 θ +

sin2 θ

cos2 θ +

m

M

sin2 θ

cos2 θ)ẍ = g

sin θ

cos θ

(

1cos2 θ + sin2 θ) + (m/M )sin2 θ

ẍ = g sin θ cos θ

ẍ =

g sin θ cos θ

1 + (m/M )sin2 θ

ẍ = g M sin θ cos θ

M + m sin2 θ(17)

Since the acceleration in the x̂ direction is a constant in time, ax Eq. 17 hasthe usual 1-D kinematic solution,

x(t) = x0 + v0,xt + 1

2axt

2

set x0 → 0, and v0,x → 0

x(t) = 1

2 g

M sin θ cos θ

M + m sin2 θ

t2 (18)

For the ŷ equation of motion, we plug Eq. 17 back into Eq. 15,

ÿ = ẍ(1 + m

M ) tan θ

= g

M sin θ

cos θ

M + m sin2 θ

1

M (M + m)

sin θ

cos θ

ÿ = g (M + m)sin2 θ

M + m sin2 θ(19)

which is also constant, so with the same initial conditions (y0 → 0, y0,y → 0)we get a similar free-fall equation,

y = 12 g

(M + m)sin

2

θM + m sin2 θ

t2 (20)

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Finally we get the solution for X plugging Eq. 17 into Eq. 12,

Ẍ = −m

M ẍ

= −m

M g M sin θ cos θ

M + m sin2 θ

Ẍ = −g m sin θ cos θM + m sin2 θ

X 0 → 0, and v0,X → 0

⇒ X = −1

2g

m sin θ cos θ

M + m sin2 θ

t2 (21)

One could also solve for the normal force λ via Eq. 9, but the problem doesnot ask for it, so it is left as an exercise to the reader. (Ew. I feel so dirty sayingthat.)

c) What are the constants of motion?

To find the constants of motion, i.e. conserved quantities, we need to express

the Lagrangian (Eq. 6), in terms of independent coordinates. We can use theconstraint equation (Eq. 3) to solve for X ,

(x − X )sin θ − y cos θ = 0

x − X −y

tan θ = 0

X = x −y

tan θ

⇒ Ẋ = ẋ −ẏ

tan θ

⇒ L = 1

2 mẋ2 +

1

2 mẏ2 +

1

2 M Ẋ 2 + mgy

L = 1

2

mẋ2 + 1

2

mẏ2 + 1

2

M ẋ −ẏ

tan θ2

+ mgy (22)

Because this Lagrangian is explicitly independent of x, then the ∂L∂x

term inthe Euler-Lagrange equation will be zero. Thus, the canonical momentum inthe x̂ direction is a conserved quantity.

px = ∂L

∂ ẋ

= mẋ + M

ẋ −

ẏ

tan θ

(1)

px = (m + M )ẋ −M ẏ

tan θ (23)

General Rule: If the Lagrangian is independent of any generalized coordi-

nate, that coordinate is “cyclic,” and therefore it’s respective canonical momen-tum ( pi ≡ ∂L/∂ q̇ i) is conserved.

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Also, the Lagrangian does not explicitly depend on time, so as previouslystated, the total energy is conserved (which is why we could write the Hamilto-nian as the sum of kinetic and potential energies).

d) Describe the motion of the system.

Figure 2: An alternate view of the particle-wedge system at times t = 1, 2, 3and 4.

Since friction is not involved in the system, the x̂ coordinate of the center of mass between the two objects will remain stationary as has been shown in part

b. The ŷ coordinate will accelerate as though in slow free-fall (a la Eq. 20), untilthe particle reaches the surface on which the wedge slides. From then on, it willsimply travel according to Eq. 18. Thus, (as in Figure 2) from a reference pointfollowing the center of mass, the wedge and particle will fly apart from eachother. If M ≫ m, it will not move much ( Ẍ ≈ 0), and the system approximatesthat of a mass sliding down a fixed incline.

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Jeff Kissel

October 26, 2006

A uniform rod slides with its ends inside a smooth (frictionless) verti-cal circle of radius a. The rod of uniform density and mass m subtendsan angle of 120◦ at the center of the circle.

Figure 1: A rod, with length L (which is the length of a chord subtending 120◦)is restricted to slide inside a frictionless circular pipe.

(a) Compute the center of mass moment of inertia of the rod I CoM interms of m and a (not the length of the rod).

In most general terms, the moment of inertia tensor is calculated as

I CoM =

ρ(r)(δ ijr

2 − xixj) d3r (1)

However, we assume the rod is only has one dimension so we need onlyintegrate along dℓ instead of d3r; there are no off-diagonal terms (i.e. xixj = 0);and the rod has constant density, so ρ(r) = m/L, which means Eq. 1 becomesconsiderable less intimidating,

I CoM = mL

+L/2−L/2

ℓ2 dℓ = 2 mL

L/20

ℓ2 dℓ

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= 2 m

L

ℓ3

3

L/20

= 2 m

L

1

3

L

2

3

= 2

24 m

L3

L

I CoM = 1

12

mL2 (2)

Finally, because we know angle which the rod subtends is 120◦, it forms anisosceles triangle whose sides are a : a : L, and interior angles are 120 : 30 : 30,which means L/2 = a cos (30◦), or L =

√ 3 a. Thus, in terms of m and a, the

center of mass moment of inertia is

I CoM = 1

4 ma2 (3)

(b) Obtain the potential energy, the kinetic energy, and the La-grangian L(θ(t), θ̇(t);m,a,g) for this system, where θ(t) the dynamicalvariable, is the instantaneous angular position relative to its equilib-rium position and m,a, and g are constant parameters of the system.

As the question implies later, once we have the center of mass moment of inertia, we can treat the rod like a pendulum, as long as we displace I CoM tothe center of the circle, a distance d = a sin(30◦). So, following the parallelaxis theorem,

I d = I CoM + md2

= 1

4 ma2 + m(a sin (30◦))2

= 1

4 ma2 + m

1

2 a

2

= 1

4 ma2 +

1

4 ma2

I d = 1

2

ma2 (4)

Now the kinetic energy T is only rotational, the potential V is only gravita-tional since there is no friction, and the Lagrangian is L = T − V ,

T = 1

2I d θ̇

2

= 1

2

1

2 ma2

θ̇2

T = 1

4ma2θ̇2 (5)

V = mga cos θ (6)

L = 14ma2 θ̇2 − mga cos θ (7)

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(c) Compare the dynamics of this system with that of a simple pendu-lum with mass M and length L (exactly, without any approximationssuch as the small oscillation approximation). Compare the parame-ters L and M to a and m.

With out making any approximations (I promise!) we can find the equa-

tion of motion, and compare those to that of a simple pendulum. Let’s getLagrangimiphysical!

d

dt

∂L

∂ θ̇

− ∂L

∂θ = 0

d

dt

1

2ma2 θ̇

+ mga sin θ = 0

1

2 ma

2θ̈ = − mg a sin θ

θ̈ = 2g

a sin θ (8)

which is indeed comparable to a simple pendulum’s equation of motion whichis different only by the factor of 2 if L

→ a and M

→ m. For a pendulum of

length L, and mass M ,

T = 1

2ML2θ̇2 V = MgL cos θ

L = 1

2ML2θ̇2 − MgL cosθ

d

dt

∂L

∂ θ̇

− ∂L

∂θ = 0

d

dt

ML2 θ̇

+ MgL sinθ = 0

ML 2θ̈ = − Mg L sin θθ̈ =

g

L

sin θ (9)

(c) Find the frequency of small oscillations of the system.

Oh, NOW it’s OK to make approximations, huh? Fine. Gosh.

In using the small angle approximation, sinθ ≈ θ, which means Eq. 8becomes a second-order differential equation of the form ẍ = −ω2x which hasthe general solution x(t) = x0 cos (ωt), where ω is the frequency (small) of oscillation(s).

θ̈ ≈ − 2ga θ

⇒ ω = 2g

a (10)

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Jeff Kissel

May 18, 2007


Fall 2006 X –Spring 2007 #22 Added to bank.

A mass m is attached to a spring of spring constant k that can slidevertically on a pole without friction, and moves along a frictionless

inclined plane as shown in Fig 1. After an initial displacement along

the plane, the mass is released. Find an expression for the x and yposition of the mass as a function of time. The initial displacement

of the mass is x0. You may assume that the object never slides down

the ramp so far that it strikes the floor.

Figure 1: A box of undisclosed materials oscillating on a ramp.

If one sets up the coordinates as in Fig 1, where the equilibrium is at theorigin, then two important simplifications occur:

1) y = ax + b → y = ax (1)

2) ∆x = x− xe → ∆x = x (2)

With these simplifications, the kinetic energy is

T = 12 mẋ2 + ẏ2

(3)

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and the potential energy is the sum of that from the spring and that fromgravity,

V = 1

2 k∆x2 + mgy

(Because of Eq. 2, ∆x = x)

V = 12 kx2 + mgy (4)

Making the Lagrangian for the system in terms of x and y ,

L = T − V

= 1

2 mẋ2 + ẏ2

−

1

2 kx2 + mgy

L = 1

2 mẋ2 +

1

2 mẏ2 −

1

2 kx2 −mgy (5)

However, there is a fixed relationship between x and y , namely the Eq. 1, so

y = ax → ẏ = aẋ (6)

which makes Eq. 5 a function only of x and ẋ,

L = 1

2 mẋ2 +

1

2 m(aẋ)2 −

1

2 kx2 −mg(ax)

= 1

2 mẋ2 +

1

2 ma2ẋ2 −

1

2 kx2 −mgax

L = 1

2 m(1 + a2)ẋ2 −

1

2 kx2 −mgax (7)

Using the Euler-Lagrange equation, we can then find an expression for x(t),

0 = d

dt ∂L

∂ ẋ−∂L

∂x

= d

dt

m(1 + a2)ẋ

− (−kx−mga)

0 = m(1 + a2)ẍ + kx + mga

ẍ = −k

m(1 + a2) x−

ga

m(1 + a2)

Let ω2 = k

m(1 + a2), and R = −

ga

m(1 + a2)

ẍ = −ω2 + R

Let x′ = −ω2x + Rẋ′ = −ω2ẋẍ′ = −ω2ẍ

ẍ′

−ω2 = −ω2

x

′

−R−ω2

+ R

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ẍ′

−ω2 = x′

ẍ′ = −ω2x′

⇒ x′(t) = A cos(ωt) (8)

Where the last step we know comes from the typical harmonic oscillator solution,

with A as some constant. Plugging back in for x, and using the initial conditionsthat x(0) = x0,

−ω2x(t) + R = A cos(ωt)

x(t) = −A

ω2 cos(ωt)−

R

ω2 x0 = x(0) = − Aω2 (1)− Rω2x0 + Rω2 = − Aω2

A = −ω2x0 −R

= −(−ω2x0 −R)

ω2 cos(ωt)−

R

ω2

=x0

+

R

ω2

cos(ωt)−

R

ω2R

ω2 = −

ga

m(1 + a2)

m(1 + a2)

k

= −

ga

k

ω =

k

m(1 + a2)

x(t) =x0 −

ga

k

cos

k

m(1 + a2) t

+ ga

k (9)

and using Eq. 1, we can get y(t),

y(t) = ax(t)

y(t) = ax0 −

ga

k

cos

k

m(1 + a2) t

+ ga

k (10)

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Jeff Kissel

May 17, 2007


Fall 2006 X –Spring 2007 #22 –

An astronaut is on the surface of a spherical asteroid of radius r andmean density similar to that of the earth. On the earth this astronautcan jump about a height h of 0.5 m. If he jumps on this asteroid, hecan permanently leave the surface.

(a) Taking the radius of the earth as 6.4×106 m, find the largest radiusthe asteroid can have.

With the initial velocity on earth v⊕, the astronaut jumps a height h. If hecan leave the asteroid with this same velocity, we can treat v⊕ as the escapeveloctiy for the asteroid vesc. We can determine his velocity on Earth from 1-Dkinematics (using the usual trick that his final velocity on the way down is equalto his initial velocity on the the way up),

v2f −

0v2i = 2a∆x

v2⊕ = 2g⊕h

v⊕ =

2g⊕h

The escape velocity on the asteroid occurs when the astronaut’s total kineticenergy equals the potential felt from the asteroid, i.e. his total energy is zero,

E = 1

2 mav

2−

GM AmarA

0 = 1

2 mav

2

esc −GM Ama

rA

12 mav

2

esc = GM A ma

rA

1


61/205


v2esc = 2GM A

rA

vesc =

2GM A

rA

ρA =

M AV A

⇒

M A = ρAV AM A = 4π

3 r3AρA

=

2G

rA

4π

3 r3AρA

= rA

2G

4πρA3

ρA = ρ⊕ρA =

M ⊕4π

3 R3⊕

= rA

2G 4π

3

M ⊕

4π3

R3⊕

vesc = rA

2GM ⊕

R3⊕g⊕ ≡

GM

R2⊕

, v

Date post:	07-Jul-2018
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