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 Dipan K. Ghosh EP 222: Classical Mechanics - Lecture 23 September 22, 2014

Euler Lagrange Equations with constraints The Euler Lagrange equations were derived from the optimistion of the action integral  d ✓ ∂ L ◆ � Z − ∂ L X dt δ q j = 0 dt ∂ ˙ q j ∂ q j j Using the fact that the generalized coordinates are independent, we derived the Euler Lagrange equation for each generalized coordinate ✓ ∂ L ◆ d − ∂ L = 0 ∂ ˙ dt q j ∂ q j 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 di ff erential form n X a r , j dq j + b r , j dt = 0 j =1 where a r , j and b rj 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 P n j =1 a r , j dq j = 0. ✓ ∂ L n " m # ◆ Z d − ∂ L X X dt + λ r a r , j δ q j = 0 ∂ ˙ dt q j ∂ q j j =1 r =1 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. δ q j are not independent as they satisfy the constraint condition P n j =1 a r , j δ q j = 0. However, since we have m of the λ r s are yet undetermined, we can choose them such that the integrands for j = 1 to m are zero. ✓ ∂ L m ◆ d − ∂ L X + λ r a r , j = 0 ∂ ˙ dt q j ∂ q j r =1 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 ✓ ∂ L n " m # ◆ Z d − ∂ L X X dt + λ r a r , j δ q j = 0 dt ∂ ˙ q j ∂ q j r =1 j = m +1 However, these q j s are independent and, therefore for j = m + 1 , n ✓ ∂ L m ◆ d − ∂ L X + λ r a r , j = 0 ∂ ˙ dt q j ∂ q j r =1 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 ✓ ∂ L m d ◆ − ∂ L X + λ r a r , j δ q j = 0 dt ∂ ˙ q j ∂ q j r =1 Define Generalized Force corresponding to the constraint conditions Q j = − P m j =1 λ r a r , j . The Euler Lagrange Equation is given by ✓ ∂ L ◆ − ∂ L d = Q j ∂ ˙ dt q j ∂ q j 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 x 2 + y 2 = 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 ρ 2 + ρ 2 ˙ ϕ 2 + ˙ z 2 � � 2 m ˙ − mgz x 2 + y 2 is the distance of the particle from the z-axis. p where ρ = The system is holonomic with two degrees of freedom. We can use ρ 2 = az to eliminate one variable from the Lagrangian. The remaining two degrees of freedom will then be independent. Write the constraint as 2 ρδρ − a δ z = 0. Using the comparison with P 3 j =1 a r , j δ q j + b r , j dt . we identify a ρ = 2 ρ a ϕ = 0 a z = − 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 ✓ ∂ L d ◆ − ∂ L = Q j = λ a j dt ∂ ˙ q j ∂ q j The Euler Lagrange equations for the three coordinates are ✓ ∂ L ◆ − ∂ L d ∂ρ = λ × 2 ρ dt ∂ ˙ ρ ✓ ∂ L ◆ − ∂ L d ∂ϕ = 0 ∂ ˙ dt ϕ ✓ ∂ L ◆ − ∂ L d ∂ z = − λ × a ∂ ˙ dt z Dipan Ghosh (I.I.T. Bombay) Class. Mech. -23 October 1, 2014 8 / 19

Euler Lagrange Equations with constraints- Example-1 We then have ϕ 2 ) = 2 λρ m (¨ ρ − ρ ˙ m d dt ( ρ 2 ˙ ⇒ m ρ 2 ω 0 = L ϕ ) = 0 = m ¨ z = − mg − λ a 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 = ω 0 = 2 g ˙ a (from the first equation, using ¨ ρ = 0) λ = − mg = − m 2 ω 2 0 a 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 d ρ = 2 ρ a The component of the normal force towards the centre of the circle is 2 ρ 4 ρ 2 + a 2 ≡ − m ρω 2 − N sin ψ = − N 0 p The vertical component of N must be equal to mg a N cos ψ = N 4 ρ 2 + a 2 = mg p Thus 4 ρ 2 + a 2 = − 1 λ = − mg N 2 m ω 2 = − 0 p a 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 o ff . 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 θ 2 ) + 1 r 2 + r 2 ˙ 2 ma 2 ˙ ϕ 2 − mgr cos θ 2 m (˙ Constraints are : (i) r = R + a (ii) ( R + a ) ˙ θ = a ˙ ϕ The first is a holonomic constraint : f 1 = r − ( R + a ) and the second is a non-holonomic (though integrable) constraint f 2 = − ( 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 ✓ ∂ L ◆ d − ∂ L = λ ∂ f 1 + µ ∂ f 2 ∂ ˙ dt ∂ q j ∂ q j ∂ q j q j Dipan Ghosh (I.I.T. Bombay) Class. Mech. -23 October 1, 2014 12 / 19

Euler Lagrange Equations with constraints- Example-2 The equations give θ 2 + mg cos θ = λ r − mr ˙ m ¨ mr 2 ¨ r ˙ θ + 2 mr ˙ θ − mgr sin θ = − µ ( R + a ) ma 2 ¨ ϕ = µ a ϕ = m ( R + a )¨ Hence µ = ma ¨ θ Substitute this in the second equation and use ˙ r = 0 g sin θ ¨ θ = 2( R + a ) Integrate (first multiply with ˙ θ ) g θ 2 = ˙ 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) θ 2 + mg cos θ λ = − mr ˙ g = − m ( R + a ) 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 − R θ = 0 so that a x = ∂ ( x − R θ ) = 1 and ∂ x a θ = ∂ ( x − R θ ) = − R ∂θ 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 x 2 + 1 2 mR 2 ˙ θ 2 − mg ( L − x ) sin α 2 m ˙ The Lagrangian equations for x and θ are m ¨ x − mg sin α = λ mR ¨ θ = − λ R θ = − λ x − R ˙ x = R ¨ The constraint is ˙ θ = 0 which gives ¨ m so that λ = − m ¨ x . Substitute this in the x equation, x = g sin α + λ x = g sin α ¨ m = ⇒ ¨ 2 Thus λ = − mg sin α which is the force of constraint for the x motion 2 | 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 mR 2 L = 1 2 mv 2 + 1 θ 2 + 1 2 mR 2 ˙ ϕ 2 − mg ( L − x ) sin α ˙ 2 2 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