PHYS 251 HONOURS CLASSICAL MECHANICS 2014

Homework Set 6 Solutions

6.25. Spring on a T

Let the `-rod be horizontal (parallel to the x axis) at t = 0. Then, as it rotates with constant angular frequencyω, its angle with respect to the x axis will be ωt. Thus, we can write the coordinates of the mass as follows:

x = ` cos(ωt)− r sin(ωt) , (1)

y = ` sin(ωt) + r cos(ωt) , (2)

and so the velocity of the mass is given by

x = ω[−` sin(ωt)− r cos(ωt)]− r sin(ωt) , (3)

y = ω[` cos(ωt)− r sin(ωt)] + r cos(ωt) . (4)

So after simplification,v2 = x2 + y2 = ω2(`2 + r2) + r2 + 2ω`r . (5)

We can then write the Lagrangian:

L = T − V =1

2mv2 − 1

2kr2 =

1

2m[ω2(`2 + r2) + r2 + 2ω`r

]− 1

2kr2 . (6)

Applying the Euler-Lagrange equation, we find

∂L

∂r− d

dt

(∂L

∂r

)= 0 (7)

=⇒ mr = mω2r − kr (8)

=⇒ r +

(k

m− ω2

)r = 0 . (9)

We recognize this last equation as the equation of motion of a simple harmonic oscillator. There are threedifferent cases, which yield different solutions:

1. If ω <√k/m, then the solution is r(t) = A cos(ωt) + B sin(ωt), where A and B are constants which

depend on the initial conditions. Note that we defined ω =√

km − ω2.

2. If ω >√k/m, then the solution is r(t) = Aeωt +Be−ωt, where this time we define ω =

√ω2 − k

m . A and

B are still constants of integration.

3. If ω =√k/m, then the solution is simply r(t) = At+B.

The special value of ω is√k/m. When it is larger than this value, the centrifugal force in the rotating frame

is dominant over the spring force so r grows exponentially. For an angular velocity smaller than√k/m, then

the usual oscillator motion is dominant.

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Figure 1: Maple code for Problem 6.27. Note that e is some small time interval and that i is some iterationvariable.

6.27. Coffee cup and mass

The kinetic energy of the coffee cup is simply 12Mr2. Since the mass m can also rotate, its kinetic energy is

12mr

2 + 12mr

2θ2. Their potential energy is then Mgr and −mgr sin θ, respectively. Therefore, the Lagrangianis

L = T − V =1

2Mr2 +

1

2mr2 +

1

2mr2θ2 −Mgr +mgr sin θ , (10)

and the equations of motion are

∂L

∂r− d

dt

(∂L

∂r

)= 0 (11)

=⇒ (M +m)r = mrθ2 −Mg +mg sin θ (12)

and

∂L

∂θ− d

dt

(∂L

∂θ

)= 0 (13)

=⇒ rθ = −2rθ + g cos θ . (14)

For the second part, see Figure 1 for an example of code written in Maple. We find the ratio of the r atthe lowest point to the r at the start to be ≈ 0.2076 for m/M = 1/10. We also find that this lowest point isattained after a time of ≈ 0.4831 s, and that the angle θ at that moment is ≈ 5.3488 rad.

6.34. x dependence

With a x dependent term in the Lagrangian, Eq. (6.19) becomes

∂

∂aS[xa(t)] =

∫ t2

t1

(∂L

∂xaβ +

∂L

∂xaβ +

∂L

∂xaβ

)dt (15)

We then integrate the last term by parts twice:∫∂L

∂xaβ dt =

∂L

∂xaβ −

∫d

dt

(∂L

∂xa

)β dt (16)

=∂L

∂xaβ −

[d

dt

(∂L

∂xa

)β −

∫d2

dt2

(∂L

∂xa

)β dt

]. (17)

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Therefore, we obtain

∂S[xa(t)]

∂a=

∫ t2

t1

β

[∂L

∂xa− d

dt

(∂L

∂xa

)+d2

dt2

(∂L

∂xa

)]dt+ β

[∂L

∂xa− d

dt

(∂L

∂xa

)]t2ti

+∂L

∂xaβ

∣∣∣∣t2t1

. (18)

The boundary term proportional to β is equal to 0, since we assume that β vanishes at the endpoints. However,the boundary proportional to β may not be zero, since we do not assume that the derivative is zero at theendpoints. Therefore, the proposed result is not valid.

6.38. Constraint on a curve

The Lagrangian is L = 12m(x2 + y2)− V (r) where we denote r to be the distance of the bead from the curve.

Then,

∂L

∂x− d

dt

(∂L

∂x

)= 0 =⇒ mx = −∂V (r(x, y))

∂x= −∂V

∂r

∂r

∂x, (19)

∂L

∂y− d

dt

(∂L

∂y

)= 0 =⇒ my = −∂V (r(x, y))

∂y= −∂V

∂r

∂r

∂y, (20)

where we note that F (r) = −∂V∂r , so

mx = F (r)∂r

∂xand my = F (r)

∂r

∂y. (21)

We are told that the curve is described by a function y = f(x), so we have

y =df(x(t))

dt= f ′x =⇒ y = f ′x+ f ′′x2 , (22)

where ′ denotes a derivative with respect to x. Substituting (21) into (22), we obtain

F

m

∂r

∂y= f ′

F

m

∂r

∂x+ f ′′x2 (23)

=⇒ F =mf ′′x2

∂r∂y − f ′

∂r∂x

. (24)

Now, to determine the partial derivatives of r with respect to x and y, let θ be the angle between the curve andthe x axis at any time. Then, the tangent to the curve at any point on the x axis is simply

f ′(x) = tan θ . (25)

With the same geometrical description, a bead a distance r from the curve will have

∂r

∂x= − sin θ and

∂r

∂y= cos θ . (26)

Using (25), sin and cos can be re-written in terms of f ′, so one obtains

∂r

∂x= − f ′√

1 + f ′2and

∂r

∂y=

1√1 + f ′2

. (27)

Therefore, the expression for the force found in (24) becomes

F =mf ′′x2√1 + f ′2

, (28)

but we note that since v = xx + yy, we have x = v cos θ = v√1+f ′2

, so in the end

F =mf ′′v2

(1 + f ′2)3/2. (29)

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6.40. Atwood’s machine

By conservation of string, the height of the 4m mass is −x+y2 , so the Lagrangian is simply

L =T − V =1

2(4m)

(x+ y

2

)2

+1

2(5m)x2 +

1

2(3m)y2 −

[−(4m)g

(x+ y

2

)+ (5m)gx+ (3m)gy

](30)

=m(3x2 +mxy + 2my2)−mg(3x+ y) . (31)

We note that under the transformation

x→ x+ ε , (32)

y → y − 3ε , (33)

the Lagrangian is left invariant (L→ L). Therefore, we can use Noether’s theorem with

Kx = 1 and Ky = −3 , (34)

and the conserved momentum is thus

P =∂L

∂xKx +

∂L

∂yKy = m(6x+ y)− 3m(x+ yy) = m(3x− 11y) , (35)

which tells us that m(3x− 11y) = constant all the times in this system.

6.42. Spring on a spoke

Let (x, y) be the coordinates of the mass relative to the center of the wheel at t = 0. Also, let r be the lengthof the spring and θ be the angle through which the wheel has rolled relative to the position where the spring isvertical. Then,

x = Rθ − r sin θ and y = R− r cos θ , (36)

so

x = Rθ − rθ cos θ − r sin θ and y = rθ sin θ − r cos θ (37)

=⇒ v2 = x2 + y2 = R2θ2 + r2θ2 + r2 − 2rRθ2 cos θ − 2rRθ sin θ . (38)

We assume that θ, θ, r � 1, so the velocity becomes

v2 = R2θ2 + r2θ2 + r2 − 2rRθ2(

1− θ2

2+O(θ4)

)− 2rRθ

(θ − θ3

6+O(θ5)

)(39)

= R2θ2 + r2θ2 + r2 − 2rRθ2 , (40)

where the second equality is valid up to second order in small quantities. The Lagrangian is then

L = T − V =1

2m(R2θ2 + r2θ2 + r2 − 2rRθ2)− 1

2kr2 +mgr cos θ . (41)

So,

∂L

∂θ− d

dt

(∂L

∂θ

)= 0 (42)

=⇒ md

dt

[(R− r)2θ

]= −mgr sin θ (43)

=⇒ (R− r)2θ − 2(R− r)rθ = −gr sin θ . (44)

To first order in small quantities, this EOM simplifies to become

(R− r)2θ = −grθ . (45)

For small oscillations, r � 1 =⇒ r ≈ 0 =⇒ r ≈ constant. Thus, we have a simple harmonic oscillator,

θ + ω2θθ = 0 , (46)

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with ω2θ ≡

gr(R−r)2 . Also, we have

∂L

∂r− d

dt

(∂L

∂r

)= 0 (47)

=⇒ d

dt(mr) = m(r −R)θ2 − kr +mg cos θ (48)

=⇒ r = −krm

+ g , (49)

up to first order in small quantities. This EOM can be re-written as follows:

ρ+ ω2rρ = 0 , (50)

with ρ ≡ r − mgk and ω2

r ≡ km . From this, we read the equilibrium value of r to be when ρ = 0, i.e. when

r =mg

k≡ req . (51)

Thus, we can re-write ωr =√g/req. Finally, the two frequencies are equal when

ωθ = ωr =⇒√greq

R− req=

√g

req=⇒ req =

R

2. (52)

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