EP 222: Classical Mechanics - Lecture 6 Dipan K. Ghosh Indian - - PowerPoint PPT Presentation

EP 222: Classical Mechanics - Lecture 6

Dipan K. Ghosh

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

August 1, 2014

Dipan K. Ghosh EP 222: Classical Mechanics - Lecture 6

Review of Lecture 5: Principle of Virtual Work

We discussed different types of constraints that limit the degrees of freedom of a many particle system. The Holonomic constraints are those which result in algebraic relations between the generalized coordinates which reduce the number of generalized coordinates required to describe the dynamic state of a system. Constraints, both holonomic and non-holonomic are either scleronomic or time independent or rheonomic or time dependent. We defined virtual displacements and enunciated the principle of virtual work. Today, we will illustrate the principle of virtual work and attempt to re-derive the Euler Lagrange Equations using this.

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -6

August 13, 2014 1 / 17

Principle of Virtual Work

The work done by applied forces in a virtual displacement of a body is zero.

• Fi = 0 =

⇒

i

Fi · δ ri = 0 Fi = F a

i + F c i but the constraint forces (reaction, tension etc.) do not

do any work. Thus

i

F a

i · δri = 0

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -6

August 13, 2014 2 / 17

Principle of Virtual Work- Illustration

The work done by applied forces in a virtual displacement of a body is zero.

• i
• F a

i · δ

ri = 0 Two particles joined by a taut string, constrained to move along a straight line. If F1 and F2 are applied forces acting on the two particles, F1δx1 + F2δx2 = 0. We have δx1 = δx2. Thus F1 + F2 = 0 (The individual applied forces need not be zero)

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -6

August 13, 2014 3 / 17

Principle of Virtual Work- Illustration

The work done by applied forces in a virtual displacement of a body is zero.

• i
• F a

i · δ

ri = 0 Consider a rigid body. It has six degrees of freedom, three rotations and two translations. Each of the six different types of displacements can be independently varied. for a rigid body, the magnitude of each ri must be the same. Thus each of the six types of terms Fx, Fy, Fz, Mx, My, Mz must vanish. No net force or torque on a rigid body in equilibrium.

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -6

August 13, 2014 4 / 17

Constraint Forces do not do work

Constraint forces do not do any work because they work to restrict motion along a surface which is perpendicular to the direction of such forces. Two bodies joined by a taut string: The constraint forces (tension) are equal and opposite and the displacement is the same. Hence no work. Rigid body: For any pair of particles j and k,

• f c

kj · δ

rj + f c

jk · δ

rk = ( f c

kj +

f c

jk) · δrj = 0

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -6

August 13, 2014 5 / 17

Virtual work - Examples

In the pulley system shown, when m1 moves by an amount δx to right, m2 moves by 2δx downward, so that the lengths of two ropes remain constant. The only external forces are gravity and friction. Virtual work = −µmgδx + mg(2δx) = 0 so that m2 = µm1/2

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -6

August 13, 2014 6 / 17

Virtual work - Examples

In the pulley system shown, y1 + y2 = constant is the constraint, so that δy1 = −δy2. The only external forces are m1g and m2g Virtual work = m1gδy1 + m2gδy2 = (m1 − m2)gδy1 = 0 so that m1 = m2.

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -6

August 13, 2014 7 / 17

Virtual work - Examples

The block on the vertical track has downward displacement of δx1, the block on the horizontal plane has a vertical displacement δx2 to the right. Constraint which keeps the length of the rod constant: δx1 sin θ = δx2 cos θ. Gravity does mgδx1, the applied force does Fδx2. Virtual work = −mgδx1 + Fδx2 = 0, so that F = −mg cot θ

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -6

August 13, 2014 8 / 17

Generalized Force

Express the expression for Virtual work in terms of the set of d generalized coordinates, δW =

• i
• F a

i · δ

ri =

• i

d

• α=1
• F a

i · ∂

ri ∂qα δqα ≡

d

• α=1

Qαδqα The Generalized force Qα is defined by Qα = F a

i · ∂

ri ∂qα

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -6

August 13, 2014 9 / 17

Generalized Force - Example

A bead sliding down an elliptical bead. Motion in x-z plane, but degrees of freedom is one as x = a cos α; y = b sin α; x2 a2 + y2 b2 = 1 Qα = −mgˆ z · ∂ ∂α (ˆ xa cos α + ˆ zb sin α) = −mgb cos α A disk rolling on a horizontal plane without slipping

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -6

August 13, 2014 10 / 17

Generalized Force - Example

A disk rolling down the incline without slipping, a force applied at C (which provides a torque). One degree of freedom : x the distance moves along the incline or the angle ϕ by which the disk rotates. Generalized force corresponding to x coordinate

• rc = xˆ

i, = ⇒ Qx = F Generalized force corresponding to ϕ

• rc = −Rϕˆ

i, = ⇒ Qϕ = −RF Generalized force need not have the dimension of force.

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -6

August 13, 2014 11 / 17

d’ Alembert’s Principle

Force results in rate of change of momentum Fi = ˙ pi The body can be brought to equilibrium by applying a pseudo force − ˙ pi

• Fi =

F c

i +

F a

i

Virtual work =

i(

F a

i +

F c

i −

˙ pi) · δ ri = 0 Constraint forces do not do any work

• I

( F a

i −

˙ pi) · δ ri = 0 (we henceforth drop the superscript a)

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -6

August 13, 2014 12 / 17

d’ Alembert’s Principle- consequence

In case the displacements are real, δ ri = d ri = ˙ ridt If, in addition, the force is conservative,

• i

( Fi − ˙ pi) · ˙ ri =

• i

[−∇iV − m ¨ ri] · ˙ ridt =

• i

[−∇iV · d ri − d dt (1 2(m˙ r2

i )dt]

= d(V +

• i

(Ti)) = d(T + V ) = 0

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -6

August 13, 2014 13 / 17

d’ Alembert’s Principle- Example

In case the displacements are real, δ ri = d ri = ˙ ridt If, in addition, the force is conservative,

• i

( Fi − ˙ pi) · ˙ ri =

• i

[−∇iV − m ¨ ri] · ˙ ridt =

• i

[−∇iV · d ri − d dt (1 2(m˙ r2

i )dt]

= d(V +

• i

(Ti)) = d(T + V ) = 0

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -6

August 13, 2014 14 / 17