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 + r 2 ˙θ2 + z ˙ 2 ) 2 U = mgz mg z In our case r = y = y and z = cy 2 so we can say that z ˙ = 2yc y˙ and we know that θ. = ω = ωtt and a nd θ˙ = ω = ω Now we can write the Lagrangian as L = T = T − U 1 L = m(y ˙2 + y 2 ω 2 + 4y 4y 2 c2 y˙ 2 ) − mgcy 2 2 Now d ∂L = m y˙˙ + 4my 4 my 2 c2 y˙˙ dt ∂ ˙ y˙ and ∂L = myω my ω 2 − 2mgcy 2mgcy + + 4mc 4 mc2 y ˙y ∂ ˙y ˙ putting these two in Lagrangian equation we get my¨(1 + 4y 4y2 c2 ) = myω my ω2 − 2mgcy 2mgcy + + 4mc 4 mc2 y ˙y
1
Prob.2 2
Given that F = r1 (1 − r˙ c 2¨r ) F is defined as ∂V + dtd ∂V ∂r ∂ r ˙ So comparing with the respective factors we have 2
−
2
∂V 1 d ∂V r˙ 2 = ; = 2 ∂r r dt ∂ r˙ cr
2
So, V (r) = 1r + cr˙ r Hence the Lagrangian would be : 2
1 2 1 r˙ 2 − L = mr˙ + 2 r c2 r
Prob.3
Writing the Lagrangian Equation : 1 T = m(r˙ 2 + r 2 ˙θ2 + z ˙ 2 ) 2 and V = mgz Writing L
L =
1 m(r˙ 2 + r 2 ˙θ2 + 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 ˙
= mr 2 ˙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 ˙ ˙ r˙ cos φ cos θ ˙φ + r˙ cos θ sin φ − r sin θsinφθ˙ x = ˙ ˙ 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 + r 2 ˙θ2 + r 2 sin2 θ ˙φ2 ) − mgr cos θ 2 ¨ ˙ 2 − mgr cos θ = mr θ + mr sin2 θφ
d ∂L dt ∂ r˙
= m¨ r and
∂L ∂r
d ∂L dt ∂ ˙θ
= mr 2 θ¨ and
d ∂L ˙ dt ∂ φ
= mr 2 sin2 φ¨ and
∂L ∂θ
= ∂L ∂φ
=
Prob.4
We know that : d ∂L ∂L = dt ∂ q ˙m ∂q m So substituting it in Lagrange equations of motion, We obtain d ∂ d ∂ ∂α L + F (q 1 , q 2 ...q m ) = dt ∂ q ˙m dt ∂t ∂ q ˙m
Now we can write LHS as : d dt
∂L ∂ ∂F + ∂ q ˙ ∂ q ˙ ∂t
3
∂L d = + ∂q dt
∂F ∂q
Now : d dt
∂ ˙L ∂ 2 F ∂ 2 F = + q ˙ ∂ q ˙ ∂t∂q ∂q 2
Hence :
˙ ˙ d ∂ F ∂ F = dt ∂ q ˙ ∂q
Prob.5
L =
m k (ax˙ 2 + 2bx˙ y + c ˙ y˙ 2 ) − (ax2 + 2bxy + cy 2 ) 2 2 d dt
∂L = ma¨ x + mb¨ y ∂ x˙
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 = l 1 sin θ1 and y 1 = l 1 cos θ1 x2 = l 1 sin θ1 − l2 sin θ2 and y2 = l 1 cos θ1 + l2 cos θ2 therefore : ˙ 1 and y˙1 = − l1 sin θ1 θ˙1 x˙1 = l 1 cos θ1 θ − 4
x˙2 l1 cos θ1 θ˙1 − l2 sin θ2 and y˙2 = l 1 (− sin θ1 )θ ˙1 − l2 sin θ2 θ˙2 Now writing Lagrangian : So, V = − m1 gl 1 cos θ1 − m2 gl 2 cos θ2 − m2 gl 1 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 ˙θ2 + l22 θ2 − 2l1 l2 cos(theta1 + θ2 ) 2 2 Now writing Lagrangian : L = T − V 1 1 2 = m1 l1 θ˙1 + m2 (l12 cos2 θ1 ˙θ2 +l22 θ2 −2l1 l2 cos(theta1 +θ2 )−(−m1 gl 1 cos θ1 −m2 gl 2 cos θ2 −m2 gl 1 c 2 2 d dt
∂L ∂θ 2 and use Lagrangian equations to find the final relations of motions.
5
Prob.7
Given that
d ∂L = e γt m¨ q 2 + mq ˙ dt ∂ q ˙
∂L = − eγt kq ∂q Therefore, m¨ q + γmq ˙ + kq = 0
.....(1) { equation of damped harmonic oscillator } γt
For s = e q 2
γt/ 2
−
q ˙ = − se ˙
q¨ = e
γt/ 2
−
put the value of q ˙ q¨ in eq (1) We will get
1 γse 2
γt/ 2
−
1 1 γ 4 s s¨ − ˙sγ ˙ − ˙sγ + 2 2 4
k γ 4 s¨ + − 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 ˙φ2 ) 2 Writing Lagrangian L = T − V =
m 2 2 ˙2 (l sin φθ + l2 ˙φ2 ) − mgl cos φ 2
6
Prob.9
Given that F = − kr cos θr
since there no θ component in this force ∂V = (kr 2 /2)cos θ + F (θ) ∂θ ˆ ˆ is cyclic Since no θ 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 2
(u2 − v 2 )
Prob.11
For a charged particle in electromagnetic field the force experinced is given by Lorentz force + F = q (E V xB) now on moving through field let it be accelerated by a velocity v such that :1 T = mv 2 2
7
F = − δU −
d ∂U dt ∂ r˙
= − δφ − ∂ A E ∂t the potential and where A is φ is the electric potential and B = δXA So
