EP 222: Classical Mechanics - Lecture 23 Dipan K. Ghosh Indian - - PowerPoint PPT Presentation

EP 222: Classical Mechanics - Lecture 23

Dipan K. Ghosh

Indian Institute of Technology Bombay dipan.ghosh@gmail.com

September 22, 2014

Dipan K. Ghosh EP 222: Classical Mechanics - Lecture 23

Euler Lagrange Equations with constraints

The Euler Lagrange equations were derived from the optimistion of the action integral X

j

Z dt  d dt ✓ ∂L ∂ ˙ qj ◆ − ∂L ∂qj

• δqj = 0

Using the fact that the generalized coordinates are independent, we derived the Euler Lagrange equation for each generalized coordinate d dt ✓ ∂L ∂ ˙ qj ◆ − ∂L ∂qj = 0 In case of non-holonomic constraints, the generalized coordinates are not independent (this can also happen if the set of generalized coordinates is not chosen to be independent).

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -23

October 1, 2014 1 / 19

Euler Lagrange Equations with constraints

Suppose there are m number of non-holonomic constraints of type involving generalized coordinates in differential form

n

X

j=1

ar,jdqj + br,jdt = 0 where ar,j and brj may depend on j and t. r is an index which runs from 1 to m. This is actually m equations, one equation for each value of r. Since virtual displacements take over constant time, we can vary the path from actual motion by replacing the constraints by Pn

j=1 ar,jdqj = 0. n

X

j=1

Z dt " d dt ✓ ∂L ∂ ˙ qj ◆ − ∂L ∂qj +

m

X

r=1

λrar,j # δqj = 0

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -23

October 1, 2014 2 / 19

Euler Lagrange Equations with constraints

The additional term is actually zero because for each r = 1, m, the sum over j is zero. δqj are not independent as they satisfy the constraint condition Pn

j=1 ar,jδqj = 0. However, since we have m of the λr s are yet

undetermined, we can choose them such that the integrands for j = 1 to m are zero. d dt ✓ ∂L ∂ ˙ qj ◆ − ∂L ∂qj +

m

X

r=1

λrar,j = 0 Note that the last term is not zero because the sum over j is missing.

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -23

October 1, 2014 3 / 19

Euler Lagrange Equations with constraints

This leaves us with

n

X

j=m+1

Z dt " d dt ✓ ∂L ∂ ˙ qj ◆ − ∂L ∂qj +

m

X

r=1

λrar,j # δqj = 0 However, these qj s are independent and, therefore for j = m + 1, n d dt ✓ ∂L ∂ ˙ qj ◆ − ∂L ∂qj +

m

X

r=1

λrar,j = 0

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -23

October 1, 2014 4 / 19

Euler Lagrange Equations with constraints

We thus have one single form of equation for each of the n generalized coordinates d dt ✓ ∂L ∂ ˙ qj ◆ − ∂L ∂qj +

m

X

r=1

λrar,jδqj = 0 Define Generalized Force corresponding to the constraint conditions Qj = − Pm

j=1 λrar,j.

The Euler Lagrange Equation is given by d dt ✓ ∂L ∂ ˙ qj ◆ − ∂L ∂qj = Qj

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -23

October 1, 2014 5 / 19

Euler Lagrange Equations with constraints- Example-1

Particle of mass m moving on the inner surface (smooth) of a paraboloid opt revolution x2 + y2 = az.

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -23

October 1, 2014 6 / 19

Euler Lagrange Equations with constraints- Example-1

The Lagrangian is L = T − V = 1 2m

• ˙

ρ2 + ρ2 ˙ ϕ2 + ˙ z2 − mgz where ρ = p x2 + y2 is the distance of the particle from the z-axis. The system is holonomic with two degrees of freedom. We can use ρ2 = az to eliminate one variable from the Lagrangian. The remaining two degrees of freedom will then be independent. Write the constraint as 2ρδρ − aδz = 0. Using the comparison with P3

j=1 ar,jδqj + br,jdt. we identify

aρ = 2ρ aϕ = 0 az = −a

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -23

October 1, 2014 7 / 19

Euler Lagrange Equations with constraints- Example-1

As there is only one constraint equation, there is only one Lagrange multiplier λ and the Euler Lagrange equation is d dt ✓ ∂L ∂ ˙ qj ◆ − ∂L ∂qj = Qj = λaj The Euler Lagrange equations for the three coordinates are d dt ✓∂L ∂ ˙ ρ ◆ − ∂L ∂ρ = λ × 2ρ d dt ✓∂L ∂ ˙ ϕ ◆ − ∂L ∂ϕ = 0 d dt ✓∂L ∂ ˙ z ◆ − ∂L ∂z = −λ × a

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -23

October 1, 2014 8 / 19

Euler Lagrange Equations with constraints- Example-1

We then have m(¨ ρ − ρ ˙ ϕ2) = 2λρ m d dt (ρ2 ˙ ϕ) = 0 = ⇒ mρ2ω0 = L m¨ z = −mg − λa These equations can be solved to determine ρ(t), z(t), ϕ(t) and λ Consider the case where the mass goes around in a horizontal circe at a height h from the bottom. z = h. This also implies ρ = constant = √

• ha. We then have (from the second equation)

λ = −mg a . ˙ ϕ = constant = ω0 = 2g a (from the first equation, using ¨ ρ = 0) λ = −mg a = −m 2 ω2

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -23

October 1, 2014 9 / 19

Euler Lagrange Equations with constraints- Example-1

Connection between constraint force and λ The force which is responsible for the circular motion is the component N sin ψ of the normal force. tan ψ = dz dρ = 2ρ a The component of the normal force towards the centre of the circle is −N sin ψ = −N 2ρ p 4ρ2 + a2 ≡ −mρω2 The vertical component of N must be equal to mg N cos ψ = N a p 4ρ2 + a2 = mg Thus λ = −mg a = − N p 4ρ2 + a2 = −1 2mω2

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -23

October 1, 2014 10 / 19

Euler Lagrange Equations with constraints- Example-2

A hoop rolling down the top of a cylinder without lipping. The constraint is holonomic till the mass slides off. Let r be the distance from O to C. θ is the polar coordinate of O and ϕ is the angle of rotation of the hoop about its own axis. R is the radius of the cylinder.

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -23

October 1, 2014 11 / 19

Euler Lagrange Equations with constraints- Example-2

The Lagrangian is L = 1 2m(˙ r2 + r2 ˙ θ2) + 1 2ma2 ˙ ϕ2 − mgr cos θ Constraints are : (i) r = R + a (ii) (R + a) ˙ θ = a ˙ ϕ The first is a holonomic constraint : f1 = r − (R + a) and the second is a non-holonomic (though integrable) constraint f2 = −(R + a) ˙ θ + a ˙ ϕ = 0. With the former we associate λ (associate with normal reaction) while with the latter we associate µ (associated with the tangential force required for rolling motion). The Euler Lagrange equations (for r, θ and ϕ) are d dt ✓ ∂L ∂qj ◆ − ∂L ∂qj = λ ∂f1 ∂qj + µ ∂f2 ∂ ˙ qj

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -23

October 1, 2014 12 / 19

Euler Lagrange Equations with constraints- Example-2

The equations give m¨ r − mr ˙ θ2 + mg cos θ = λ mr2¨ θ + 2mr ˙ r ˙ θ − mgr sin θ = −µ(R + a) ma2 ¨ ϕ = µa Hence µ = ma ¨ ϕ = m(R + a)¨ θ Substitute this in the second equation and use ˙ r = 0 ¨ θ = g sin θ 2(R + a) Integrate (first multiply with ˙ θ) ˙ θ2 = g R + a(1 − cos θ)

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -23

October 1, 2014 13 / 19

Euler Lagrange Equations with constraints- Example-2

Use this in the radial equation (put ˙ r = 0) λ = −mr ˙ θ2 + mg cos θ = −m(R + a) g R + a(1 − cos θ) + mg cos θ = mg(2 cos θ − 1) For θ = 0, λ = mg, the normal force. For θ > 60, λ becomes negative and contact with the cylinder is lost.

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -23

October 1, 2014 14 / 19

Euler Lagrange Equations with constraints- Example-3

A hoop rolling down an incline without sliding. It has only holonomic constraint and has one degree of freedom. We take the angle θ by which the hoop rolls and the distance by which the centre of mass moves x to be two generalized coordinates. For f (x, θ) = x − Rθ = 0 so that ax = ∂(x − Rθ) ∂x = 1 and aθ = ∂(x − Rθ) ∂θ = −R

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -23

October 1, 2014 15 / 19

Euler Lagrange Equations with constraints- Example-3

The Lagrangian is L = 1 2m ˙ x2 + 1 2mR2 ˙ θ2 − mg(L − x) sin α The Lagrangian equations for x and θ are m¨ x − mg sin α = λ mR ¨ θ = −λR The constraint is ˙ x − R ˙ θ = 0 which gives ¨ x = R ¨ θ = − λ m so that λ = −m¨ x. Substitute this in the x equation, ¨ x = g sin α + λ m = ⇒ ¨ x = g sin α 2 Thus λ = −mg sin α 2 which is the force of constraint for the x motion | λ∂f ∂x |

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -23

October 1, 2014 16 / 19

Euler Lagrange Equations with constraints- Example-4

In addition to rolling down the slope, let us allow the hoop to spin about a vertical axis. Take the direction along the slope as the x-axis.

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -23

October 1, 2014 17 / 19

Euler Lagrange Equations with constraints- Example-4

The Lagrangian is L = 1 2mv2 + 1 2mR2 ˙ θ2 + 1 2 mR2 2 ˙ ϕ2 − mg(L − x) sin α y is cyclic. The equation of motion for y will be ignored. Since the body rolls without slipping, the velocity of the rim is v = R ˙ θ. When the hoop has spun by an angle ϕ, we have, ˙ x = R ˙ θ cos ϕ ˙ y = R ˙ θ sin ϕ These are non-holonomic constraints as they cannot be used to connect x with θ and ϕ. The problem has two degrees of freedom

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -23

October 1, 2014 18 / 19

Euler Lagrange Equations with constraints- Example-4

Let us freeze the motion and make a virtual displacement δθ of the angle θ, the displacement of the tilt angle is of second order. we then have the constraint condition δx − R cos ϕδθ = 0 We then have ax = 1 and aθ = −R cos ϕ The Lagrange equations for the problem are −mg sin α − λ = 0 mR2¨ θ + λR cos ϕ = 0 mR2 ¨ ϕ = 0 These provide complete solution as x has been eliminated.

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -23

October 1, 2014 19 / 19