Homework 4 Mechanics 701, Fall 2009 Due: Friday, riday, September 11 1
Pro Problem 2.12 The term “generalized mechanics” came to designate a variety of classical mechanics in which the Lagrangian contains the derivatives of qi higher higher than the first. Problems Problems for which x... = f (¨ (¨x, x, x,x,t) x,x,t ˙ ) have been referred to as “jerky” mechanics. Such equations of motion have interesting applications applications in chaos theory theory (cf. Chapter Chapter 11). By applying the methods of calculus of variations, variations, show that if there is a Lagrangian of the form L(¨qi , q˙i , qi , t) and Hamilton’s principle holds with zero variation of both qi and q˙i at the end points, then the Euler-Lagrange equations are d2 dt2
∂L ∂ ¨ ∂ q¨i
−
d dt
∂L ∂ ˙ ∂ q˙i
+
∂L = 0 i = 1,...,n. 1 ,...,n. ∂q i
(1)
Apply this result to the Lagrangian L=−
m k q q¨ − q 2 2 2
Do you recognize the equation of motion? Solution. The variation of action equals
δS =
∂L ∂L ∂L δq¨i + δq˙i + δq i dt ∂ ¨ ∂ q¨i ∂ ˙ ∂ q˙i ∂q i
Equati Equation on (1) is prove proven n by integra integratin tingg by parts parts twice twice in the first term and once once in the second second term. term. The substitution terms are equal to zero in both cases because of zero variations of q and q˙ at the ends of the interval. Applying it to the given Lagrangian we get d2 (−mq/2) mq/2) + (−mq/2 q¨/2 − kq) kq ) = 0 dt2 or mq¨ = −kq The latter is the harmonic oscillator oscillator equation. equation.
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Pro Problem 2.13 A heavy particle is placed on top of a fixed vertical hoop. Calculate the reaction of the hoop on the particle by means means of Lagrange Lagrange undetermined undetermined multipliers multipliers and Lagrange Lagrange equations. equations. Find the height at which the particle falls off. 1
Note: This problem is almost solved in Sec. 2.4 (page 47). The undetermined multiplier is related to the
reaction force. Since mass is sliding on top of the hoop, the reaction force can only repel from the hoop, i.e. can only have one sign. Thus the particle will fall off when that force well become equal to zero. Solution: We determine the position of the mass by coordinates of its center ( x, y) that satisfy the constraint
x2 + y2 = R2 . The kinetic energy can be expressed as T =
m 2 (x˙ + y˙ 2 ) 2
and the potential energy is given by V = mgy. We now have to perform the conditional minimization of the action with a constraint, i.e. minimize the time integral
S = where
L dt
L = T − V + λ(x2 + y 2 − R2 )
with respect to functions x(t), y(t) and λ(t). We obtain the equations 2m¨ x = 2λx 2m¨ y = −mg + 2λy 2 x + y 2 − R2 = 0
(2)
To solve this system we parameterize x = R cos φ, y = R sin φ. This automatically satisfies the third equation of the system (2). With this parametriazation ¨ x ¨ = −R(cos φ φ˙ 2 + sin φ φ) ¨ y¨ = −R(sin φ φ˙ 2 − cos φ φ) and we can rewrite the first two equations of (2) as 2λ cos φ φ˙ 2 + sin φ φ¨ = − cos φ m g 2λ sin φ φ˙ 2 − cos φ φ¨ = − sin φ R m Now multiply the first equation by sin φ and the second by cos φ and subtract the second from the first: g φ¨ = − cos φ R
(3)
Next, multiply the first equation by cos φ and the second by sin φ and add the two: g 2λ φ˙ 2 = sin φ − R m Equation (3) is equivalent to d dt or
φ˙ 2 g + sin φ 2 R
=0
φ˙ 2 g + sin φ = const 2 R 2
(4)
This is of course the energy conservation condition, i.e, the constant on the right hand side is E/mR2 , and ˙ has to be determined from the initial conditions φ(0) = 0, φ(0) = π/2. This gives E/mR2 = g/R, thus φ˙ =
2
g (1 − sin φ) R
We now substitute this expression for φ˙ into equation (4) and get 2λ g g g = sin φ − φ˙ 2 = sin φ − 2 (1 − sin φ) m R R R
(5)
Recall that the Lagrange multiplier λ represents the reaction force. At the initial stage of the motion this is the normal force of the cylinder directed outwards from its center. From (5) we see that λ goes through zero at the angle φ0 = arcsin(2/3) This means that for the mass to continue to move along the surface of the cylinder the reaction force has to change the sign. We know, however, that physically this is impossible – the cylinder can repel the mass outwards from the center, but cannot attract it to the center. Thus we conclude that at φ0 the mass will detach from the surface of the cylinder and start a free motion in the air.
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Problem 2.22 Suppose a particle moves in space subject to a conservative potential V (r) but is constrained to always move on a surface whose equations is σ(r, t) = 0. (The explicit dependence on t indicates that the surface may be moving.) The instantaneous force of constraint is taken as always perpendicular to the surface. Show analytically that the energy of the particle is not conserved if the surface moves in time. What physically is the reason for non-conservation of the energy under this circumstance?
Solution: The Lagrangian of the free particle in external potential equals
L=
m˙r2 − V (r) . 2
The modified Lagrangian L = L + λ(t)σ(r, t) with yet unknown Lagrange multiplier λ(t) will describe the particle constrained to the surface. From L we get the following system of differential equations
m¨ ri
= −
∂V ∂σ +λ , ∂r i ∂r i
i = x,y,z
σ(r, t) = 0 for unknown functions ri (t) and λ(t). Energy non-conservation. Suppose we found a trajectory r(t). The change of the particle energy E = m˙r2 /2 + V (r) on the trajectory will be given by dE ∂V = (m¨ ri )r˙i + r˙i . dt ∂r i
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Substituting (m¨ ri ) from the first Lagrange equation we get dE ∂σ =λ r˙i . dt ∂r i Vector ∂σ/∂ri is directed perpendicular to the σ = 0 surface. The velocity r˙i is not completely arbitrary but has to be compatible with the second Lagrange equation. From σ(ri (t), t) = 0 we get ∂σ ∂σ r˙i + =0 . ∂r i ∂t
(6)
Using the above we can rewrite
dE ∂σ = −λ , dt ∂t which shows that E is not conserved when the surface is moving. Physical reason for energy non-conservation. For the moving surface the velocity of the particle is not tangential to the surface (see Eq. (6). Therefore the reaction force can have a non-zero scalar product with the velocity, N · r˙ = 0. That means that the reaction force performs some work which changes the total energy of the particle. For example, in the case of σ = z − vt, V = mgz one simply lifts the particle in the gravitational field. The energy is obviously increased.
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Problem 2.23 Consider two particles of masses m1 and m2 . Let m1 be confined to move on a circle of radius a in the z = 0 plane, centered at x = y = 0. Let m2 be confined to move on a circle of radius b in the z = c plane centered at x = y = 0. A light (massless) spring of spring constant k is attached between the particles. (a) Find the Lagrangian of the system (b) Solve the problem using the Lagrange multipliers and give physical interpretation for each multiplier.
Note: Here you are supposed to use coordinates x1 , y1 for the first particle and x2 , y2 for the second one.
These coordinates are not independent, thus Lagrange multipliers are necessary to write down the equations. Solution: The modified Lagrangian equals
L =
m1 (x˙ 21 + y˙ 12 ) m2 (x˙ 22 + y˙ 22 ) k(l − l0 )2 + − 2 2 2
where l is the current distance between the particles and l0 is the unstrained length of the spring. The distance between the particles is given by the expression l=
(x1 − x2 )2 + (y1 − y2 )2 + c2 .
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The Lagrange equations read l − l0 (x1 − x2 ) − 2λ1 x1 = 0 l l − l0 m1 y¨1 + k (y1 − y2 ) − 2λ1 y1 = 0 l l − l0 m1 x ¨2 + k (x2 − x1 ) − 2λ1 x2 = 0 l l − l0 m1 y¨2 + k (y2 − y1 ) − 2λ2 y2 = 0 l x21 + y12 = a2 m1 x ¨1 + k
x22 + y22 = b2 Consider two first Lagrange equations. They reproduce Newton’s equations of motion for the first particle with two forces acting on it. The first one is the spring force. The second force has components {−2λx1 , −2λy1 }. It is directed along the radius vector of the particle and is nothing else then the projection of the normal reaction force on the horizontal plane: N1,plane = {−2λ1 x1 , −2λ1 y1 } = −2λ1 an1 ,
where n1 is the unit vector along r1 . The reaction force N1 keeps the particle on the circle. In principle, it has all three projections with N z = 0. We could find only the horizontal projection of this force because we didn’t explicitly introduce coordinate z1 with a constraint z1 = 0. The same argument is true for the second particle. It experiences the reaction force with a projection N2,plane = {−2λ2 x2 , −2λ2 y2 } = −2λ2 bn2
Now we are going to exclude the unknown functions λ1,2 from the system. From the constraints we get x ¨1 x1 + x˙ 21 + y¨1 y1 + y˙ 12 = 0 , x ¨2 x2 + x˙ 22 + y¨2 y2 + y˙ 22 = 0 .
(7) (8)
To find the Lagrange multiplier λ1 , substitute expressions for x ¨ 1 , y¨1 from the first two Lagrange equations into Eq. (7). You get (l − l0 ) a2 − x1 x2 − y1 y2 x˙ 21 + y˙ 12 2λ1 = k (9) − l a2 a2 Analogous calculation for λ2 gives 2λ2 = k
(l − l0 ) b2 − x1 x2 − y1 y2 x˙ 22 + y˙ 22 − l b2 b2
(10)
Expressions (9, 10) for the Lagrange multipliers should be now substituted in the Lagrange equations. We then obtain a system of differential equations on x1 (t), y1 (t), x2 (t), y2 (t) such that its solution will automatically satisfy the constraints (provided the initial conditions satisfy them). To illustrate the last point, consider the simple case without the spring, k = 0. Equations on x1 and y1 assume the form m1 v12 x1 a2 m1 v12 y1 m1 y¨1 = − a2
m1 x ¨1 = −
where v12 = x˙ 22 + y˙ 22 is the square of the particle’s velocity. We recognize that the right hand side of this equation is exactly the centripetal force required to keep the particle on the circular orbit of radius a. 5
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Problem 2.26
Small oscillations
Slack of the string
Full rotation
Figure 1: Motion of the mass on a string for small, medium, and large initial velocity.
A particle of mass m is suspended by a massless string of length L. It hangs without initial motion in a gravitational field of strength g. The it is struck by an impulsive horizontal blow which introduces an angular velocity ω. If ω is sufficiently small, it is obvious that the mass moves as a simple pendulum. If ω is sufficiently large, the mass will rotate about the support. Use Lagrange multiplier to determine the conditions under which the string becomes slack at some point of the motion Note: This problem is an inverse of Problem 2.13. The force of the string can be directed only towards the
suspension point. Slack of the string happens when this force is equal to zero. It is probably easier to solve this problem by the “undergraduate” use of Newton’s equations, but the idea is to practice the Lagrange multipliers method. Solution: We describe the mass by the ( x, y) coordinates. The modified Lagrangian is
m(x˙ 2 + y˙ 2 ) L = + mgy + λ(x2 + y 2 − L2 ) 2
and the Lagrange equations read m¨ x = 2λx m¨ y = 2λy − mg x2 + y2 − L2 = 0 We will use the same method as in Problem 2.23 to find λ. Differentiating twice the last equation and substituting the second derivatives ¨x and y¨ from the first two equations we get λ=
m m (gy − x˙ 2 − y˙ 2 ) = (gy − v2 ) 2 2 2L 2L
The energy is conserved in this problem. To prove it formally, we multiply the first Lagrange equations by x, ˙ the second by y˙ and add them. m¨ xx˙ + m¨ y y˙ = 2λ(xx˙ + y y) ˙ − mgy˙ 6
or
d dt
mv2 + mgy 2
=λ
d 2 dL2 (x + y2 ) = λ =0 dt dt
(here L is the length of the string). Using energy conservation we write mv 2 mω2 L2 + mgy = E 0 = − mgL 2 2 or v2 = −2gy − 2gL + ω2 L2 For a given initial angular velocity ω the mass rises from y = −L to ymax = min[−L + ω2 L2 /2g, L], with ymax = L being realized for full rotations. If on this interval λ turns to be zero, that means that the reaction force had to change its sign. Such a sign change is impossible for a mass suspended on a string, and means that slack of the string happened. Mathematically, slack happens if we can find y0 , such that λ(y0 ) =
m (3gy 0 + 2gL − ω2 L2 ) = 0 2L2
and
−L < y0 < min[−L + ω2 L2 /2g, L] One finds that this can be satisfied for
2g 5g < ω2 < L L
That inequality can be rewritten as mgL <
mω2 L2 5 < mgL 2 2
Its meaning is as follows. First, the initial kinetic energy T 0 = mω2 L2 /2 has to be large enough so raise the mass form y = −L up to y > 0. This is because for y < 0 the projection of mg on the string always points outwards and thus slacking is impossible. Second, if there is enough energy to raise the mass to y = L, the centripetal acceleration at the top has to be smaller than g. Otherwise the particle will make a full circle maintaining tension in the string. Note that in the interval of kinetic energies 2mgL < T 0 < (5/2)mgL the particle would have had enough energy to raise to y = L, but this does not happen because slacking occurs before the particle reaches the top. A particle held by a flexible string cannot reach the top with too small velocity without inducing a slack.
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