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Classical Mechanics Solutions of Assignment -1 August 23, 2015 Prob.1 Given that z = 4ay 2 Let us take z = 4cy 2 We can write the Lagrangian Equations for this motion 1 T = m(r˙ 2 + r2 θ˙2 + z˙ 2 ) 2 U = mgz In our case r = y and z = cy 2 so we can say that z˙ = 2ycy˙ and we know that θ. = ωt and θ˙ = ω Now we can write the Lagrangian as L = T − U 1 L = m(y˙2 + y 2 ω 2 + 4y 2 c2 y˙ 2 ) − mgcy 2 2 Now d ∂L = my˙˙ + 4my 2 c2 y˙˙ dt ∂ y˙ and ∂L = myω 2 − 2mgcy + 4mc2 y y˙ ∂ y˙ putting these two in Lagrangian equation we get m¨ y (1 + 4y 2 c2 ) = myω 2 − 2mgcy + 4mc2 y y˙

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Prob.2 2

r Given that F = r12 (1 − r˙ c−2¨ 2
) + dtd ∂V F is defined as ∂V ∂r ∂ r˙ So comparing with the respective factors we have     1 d ∂V r˙ 2 ∂V = ; = 2 ∂r r dt ∂ r˙ cr 2

So, V (r) = 1r + cr˙2 r Hence the Lagrangian would be : 1 2 1 r˙ 2 L = mr˙ − + 2 2 r cr

Prob.3 Writing the Lagrangian Equation : 1 T = m(r˙ 2 + r2 θ˙2 + z˙ 2 ) 2 and V = mgz Writing L 1 L = m(r˙ 2 + r2 θ˙2 + z˙ 2 ) − mgz 2 As we know θ = ωt and θ˙ = ω Constraint : r = R d ∂L dt partialθ˙

= m¨ r;

d ∂L dt ∂ θ˙

= mr2 θ˙ ;

d ∂L dt ∂ z˙

V = −mgr cos φ Now put

2

= m¨ z

x = r cos θ sin φ ; y = r sin θ sin φ ; z = r cos φ ; ˙ φ=ω Finding x˙ ; y˙ and z˙ x˙ = r˙ cos φ cos θφ˙ + r˙ cos θ sin φ − r sin θsinφθ˙ y˙ = r˙ sin θ sin φ + rcosθ sin φθ˙ + r sin θ cos φφ˙ z˙ = r˙ cos φ − r sin φφ˙ Now writing Lagrangian 1 L = m(r˙ 2 + r2 θ˙2 + r2 sin2 θφ˙ 2 ) − mgr cos θ 2 = mrθ¨ + mr sin2 θφ˙ 2 − mgr cos θ

d ∂L dt ∂ r˙

= m¨ r and

d ∂L dt ∂ θ˙

= mr2 θ¨ and

d ∂L dt ∂ φ˙

= mr2 sin2 φ¨ and

∂L ∂r

∂L ∂θ

= ∂L ∂φ

=

Prob.4 We know that : ∂L d ∂L = dt ∂ q˙m ∂qm So substituting it in Lagrange equations of motion, We obtain   d ∂ d ∂ ∂α 0 L + F (q1 , q2 ...qm ) = dt ∂ q˙m dt ∂t ∂ q˙m Now we can write LHS as :      ∂ ∂F ∂L d ∂F d ∂L + = + dt ∂ q˙ ∂ q˙ ∂t ∂q dt ∂q 3

Now : ∂ L˙ ∂ q˙

d dt Hence :

! =

∂ 2F ∂ 2F + q˙ ∂t∂q ∂q 2

d ∂ F˙ ∂ F˙ = dt ∂ q˙ ∂q

Prob.5 L=

k m (ax˙ 2 + 2bx˙ y˙ + cy˙ 2 ) − (ax2 + 2bxy + cy 2 ) 2 2 d dt



∂L ∂ x˙

 = ma¨ x + mb¨ y

d ∂L = mb¨ x + mc¨ y dt ∂ y˙ and ∂L = −(Kax + Kby) ∂x and

∂L = −K(bx + cy) ∂y

Now write equations of motion by using Lagrangian

Prob.6 x1 = l1 sin θ1 and y1 = l1 cos θ1 x2 = l1 sin θ1 − l2 sin θ2 and y2 = l1 cos θ1 + l2 cos θ2 therefore : ˙ 1 and y˙1 = −l1 sin θ1 θ˙1 x˙1 = l1 cos θ1 θ − 4

x˙2 l1 cos θ1 θ˙1 − l2 sin θ2 and y˙2 = l1 (− sin θ1 )θ˙1 − l2 sin θ2 θ˙2 Now writing Lagrangian : So, V = −m1 gl1 cos θ1 − m2 gl2 cos θ2 − m2 gl1 cos θ1 1 1 T = m1 (x˙1 2 + y˙1 2 ) + m2 (x˙1 2 + y˙ 2 ) 2 2

So 1 1 2 T = m1 l1 θ˙1 + m2 (l12 cos2 θ1 θ˙2 + l22 θ2 − 2l1 l2 cos(theta1 + θ2 ) 2 2 Now writing Lagrangian : L=T −V

1 2 1 = m1 l1 θ˙1 + m2 (l12 cos2 θ1 θ˙2 +l22 θ2 −2l1 l2 cos(theta1 +θ2 )−(−m1 gl1 cos θ1 −m2 gl2 cos θ2 −m2 gl1 co 2 2 d dt



∂L ∂ θ˙1



= m1 l12 θ¨1 + m2 l1 θ¨1 − m2 gl1 l2 cos(θ1 + θ2 )θ¨2

∂L = −m1 gl1 sin θ1 − m2 gl1 sin θ1 ∂θ1 Same way find d ∂L dt ∂ θ˙2 and

∂L ∂θ2 and use Lagrangian equations to find the final relations of motions.

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Prob.7 Given that

 2  d ∂L q + mq˙ = eγt m¨ dt ∂ q˙ ∂L = −eγt kq ∂q

Therefore, m¨ q + γmq˙ + kq = 0 .....(1) { equation of damped harmonic oscillator } γt

For s = e 2 q q˙ = − se ˙ −γt/2

q¨ = e

−γt/2


1 γse−γt/2 2

  1 γ 4s 1 ˙ + s¨ − s˙ γ˙ − sγ 2 2 4

put the value of q˙ q¨ in eq (1) We will get  s¨ +

γ4 k − m 4/



Prob.8 x = l cos θ sin φ ; z = cos φ ; y = l sin θ sin φ Now find x, ˙ y˙ and z˙ 1 T = m(x˙ 2 + y˙ 2 + x˙ 2 ) 2 m T = (l2 sin2 φθ˙2 + l2 φ˙ 2 ) 2 Writing Lagrangian L=T −V =

m 2 2 ˙2 (l sin φθ + l2 φ˙ 2 ) − mgl cos φ 2

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Prob.9 Given that F = −kr cos θ~r

since there no θ component in this force ∂V = (kr2 /2) cos θ + F (θ) ∂θ ˆ is cyclic Since no θˆ component, hence angular momentum is conserved ( θ)

Prob.10 Equation of paraboloid x2 + y 2 = az Potential Energy V = −mgz Let us say : x = uv cos θ ; y = uv sin θ ; z = 0.5(u2 − v 2 ) Therefore V =

mg (u2 2

− v2)

Prob.11 For a charged particle in electromagnetic field the force experinced is given by Lorentz force ~ + V~ xB) ~ F = q(E now on moving through field let it be accelerated by a velocity v such that :1 T = mv 2 2

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F = −δU −

d ∂U dt ∂ r˙

~ = −δφ − ∂A E ∂t ~ is the potential and where A φ is the electric potential and B = δXA So

1 ~ r˙ L = mr˙ 2 − qφ + q A. 2

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