EP 222: Classical Mechanics
Tutorial Sheet 1
This tutorial sheet contains problems on the Newton’s laws of motion and Lagrangian
formalism.
1. Show that for a single particle with a constant mass the equation of motion implies
the following differential equation for the kinetic energy:
dT
= F · v,
dt
while if the mass varies with time the corresponding equation is
d(mT )
= F · p.
dt
Soln: (a) We know T = 21 mv 2 = 21 mv · v. Thus, if m is constant
dT
dv
=m
· v = ma · v = F · v
dt
dt
(b) For a variable mass particle, let us consider
d(mT )
d 1 2
dm
dm
2
=
m v·v =m
v·v+m v·a=
v + ma · (mv) = F · p,
dt
dt 2
dt
dt
dp
dt
because for a variable mass particle
=
dm
v
dt
+ ma.
2. Prove that the magnitude R of the position vector for the center of mass from an
arbitrary origin is given by the equation
M 2 R2 = M
X
mi ri2 −
i
1X
2
mi mj rij
2 i,j
Soln: R is defined as
P
mi ri
XM
=⇒ M R =
mi ri
i
R=
i
X
X
=⇒ M R = (
mi ri ) · (
mj rj )
2
2
2
2
i
=⇒ M R =
X
j
mi mj ri · rj =
i,j
X
i
1
m2i ri2 +
X
i6=j
mi mj ri · rj
Using
2
rij
= (ri − rj )2 = ri2 + rj2 − 2ri · rj
1
2
),
=⇒ ri · rj = (ri2 + rj2 − rij
2
we obtain
1X
2
mi mj (ri2 + rj2 − rij
)
2
i
i6=j
X
X
1
1X
1X
2
=
m2i ri2 +
mi mj ri2 +
mi mj rj2 + −
mi mj rij
2 i6=j
2 i6=j
2 i6=j
i
X
X
1X
2
mi mj rij
=
mi mj ri2 +
m2i ri2 −
2 i6=j
i
i6=j
X
X
1X
2
mi mj rij
=(
mj )(
mi ri2 ) −
2
j
i
i6=j
X
X
1
2
mi mj rij
= M(
mi ri2 ) −
2
i
i6=j
X
X
1
2
= M(
mi ri2 ) −
mi mj rij
2
i
i,j
M 2 R2 =
X
m2i ri2 +
In the second sum on the RHS, we can perform unrestricted sum over i and j, because
for i = j, rij = 0.
3. Show that the Lagrange equations
∂T
d ∂T
−
= Qj ,
dt ∂ q˙j
∂qj
can also be written as
∂T
∂ Ṫ
−2
= Qj .
∂ q˙j
∂qj
These are sometimes called the Nielsen form of Lagrange equations.
Soln: Assuming that
T = T (qi, q˙i , t),
we have
Ṫ =
so that
X ∂T
X ∂T
dT
∂T
=
q̈i +
q̇i +
dt
∂ q̇i
∂qi
∂t
i
i
X ∂ 2T
X ∂ 2T
∂ Ṫ
∂T
∂ 2T
=
q̈i +
q̇i +
+
∂ q̇j
∂ q̇j ∂ q̇i
∂ q̇j ∂qi
∂qj ∂ q̇j ∂t
i
i
2
leading to
X ∂ 2T
X ∂ 2T
∂T
∂ 2T
∂ Ṫ
−
=
q¨i +
q̇i +
∂ q̇j
∂qj
∂ q˙j ∂ q˙i
∂ q̇j ∂qi
∂ q̇j ∂t
i
i
X ∂
∂T
∂
∂T
∂ ∂T
q̈i +
q̇i +
=
∂
q̇
∂
q̇
∂q
∂
q̇
∂t ∂ q̇j
i
j
i
j
i
d ∂T
=
dt ∂ q̇j
Substituting this on the LHS of Lagrange equation, we obtain
∂T
d ∂T
−
= Qj
dt ∂ q˙j
∂qj
=⇒
∂ Ṫ
∂T
∂T
−
−
= Qj
∂ q̇j
∂qj
∂qj
=⇒
∂ Ṫ
∂T
−2
= Qj
∂ q̇j
∂qj
4. If L is a Lagrangian for a system of n degrees of freedom satisfying the Lagrange
equations, show by direct substitution that
0
L =L+
dF (q1 , . . . , qn , t)
dt
also satisfies Lagrange’s equations where F is any arbitrary, but differentiable, function of its arguments.
Soln: We have
X ∂F
∂F
dF
=
q̇i +
.
dt
∂qi
∂t
i
Therefore
0
L =L+
X ∂F
dF (q1 , . . . , qn , t)
∂F
=L+
q̇i +
,
dt
∂q
∂t
i
i
so that
0
∂L
∂L ∂F
=
+
,
∂ q̇j
∂ q̇j ∂qj
leading to
d
dt
0
∂L
∂ q̇j
d ∂L
d ∂F
=
+
dt ∂ q̇j
dt ∂qj
X 2
d ∂L
∂ F
=
+
q̇i +
dt ∂ q̇j
∂q
∂q
i
j
i
X
d ∂L
∂ 2F
=
+
q̇i +
dt ∂ q̇j
∂q
∂q
j
i
i
3
∂ 2F
∂t∂qj
∂ 2F
∂qj ∂t
and
0
∂L X ∂ 2 F
∂ 2F
∂L
=
+
q̇i +
.
∂qj
∂qj
∂q
∂q
∂q
∂t
j
i
j
i
Thus
d
dt
d
=⇒
dt
0
∂L
∂ q̇j
0
∂L
∂ q̇j
0
∂L
d
−
=
∂qj
dt
0
∂L
d
−
=
∂qj
dt
∂L
∂ q̇j
∂L
∂ q̇j
+
X ∂ 2F
∂ 2F
∂L X ∂ 2 F
∂ 2F
q̇i +
−
−
q̇i −
∂qj ∂qi
∂qj ∂t ∂qj
∂qj ∂qi
∂qj ∂t
i
i
−
∂L
=0
∂qj
QED.
5. Obtain the Lagrange equations of motion for a spherical pendulum, i.e., a point mass
suspended by a rigid weightless rod.
Soln: It is best to use spherical polar coordinates here. In the lectures we showed
that the kinetic energy of a particle in spherical polar coordinates is
1 2
2 2
2
2
2
T = m ṙ + r θ̇ + r sin θφ̇ .
2
If length of the rod is l, then r = l, and ṙ = 0, we are left with two generalized
coordinates (θ, φ), with
1 2 2
2
2
2
T = m l θ̇ + l sin θφ̇ ,
2
and using the point of suspension as the reference for potential energy, we have
V = −mgl cos θ.
4
Thus
1 L = T − V = T = m l2 θ̇2 + l2 sin2 θφ̇2 + mgl cos θ.
2
Now the two Lagrange equations are
d ∂L
∂L
−
=0
dt ∂ θ̇
∂θ
1
=⇒ ml2 θ̈ − ml2 sin 2θφ̇2 + mgl sin θ = 0
2
and
∂L
∂L
−
=0
∂φ
∂ φ̇
d(ml2 sin2 θφ̇)
=⇒
= 0.
dt
d
dt
6. Obtain the Lagrangian and equations of motion for a double pendulum, where the
lengths of the pendula are l1 and l2 with corresponding masses m1 and m2 , confined
to move in a plane.
Soln: As discussed in the lectures, this system has two generalized coordinates θ1
and θ2 , the angles which the upper and the lower pendula make with respect to the
vertical.
With the motion of the pendula confined in a plane (say, xy plane), then the Cartesian
coordinates of the two particles can be written as
x1 = l1 sin θ1
y1 = −l1 cos θ1
and
x2 = x1 + l2 sin θ2 = l1 sin θ1 + l2 sin θ2
y2 = y1 − l2 cos θ2 = −l1 cos θ1 − l2 cos θ2
5
So that
1
1
T = m1 (ẋ21 + ẏ12 ) + m2 (ẋ22 + ẏ22 ).
2
2
Now
ẋ1 = l1 cos θ1 θ̇1
ẏ1 = l1 sin θ1 θ̇1
ẋ2 = l1 cos θ1 θ̇1 + l2 cos θ2 θ̇2
ẏ2 = l1 sin θ1 θ̇1 + l2 sin θ2 θ̇2
Easy to verify
ẋ21 + ẏ12 = l12 θ̇12 ,
ẋ2 + ẏ 2 = l2 θ˙2 + l2 θ̇2 + 2l1 l2 θ̇1 θ̇2 cos(θ1 − θ2 ).
2
2
1 1
2 2
With this
1
1
T = (m1 + m2 )l12 θ̇12 + m2 l22 θ̇22 + m2 l1 l2 θ̇1 θ̇2 cos(θ1 − θ2 ),
2
2
and
V = m1 gy1 + m2 gy2
= −(m1 + m2 )gl1 cos θ1 − m2 gl2 cos θ2 ,
so that
L=T −V
1
1
= (m1 + m2 )l12 θ̇12 + m2 l22 θ̇22 + m2 l1 l2 θ̇1 θ̇2 cos(θ1 − θ2 )
2
2
+ (m1 + m2 )gl1 cos θ1 + m2 gl2 cos θ2 .
θ1 equation of motion
d
dt
∂L
∂ θ̇1
−
∂L
= 0,
∂θ1
leads to
d (m1 + m2 )l12 θ̇1 + m2 l1 l2 θ̇2 cos(θ1 − θ2 ) + m2 l1 l2 θ̇1 θ̇2 sin(θ1 − θ2 ) + (m1 + m2 )gl1 sin θ1 = 0.
dt
Upon taking the time derivative, we obtain the final form
(m1 + m2 )l12 θ̈1 + m2 l1 l2 cos(θ1 − θ2 )θ̈2 + m2 l1 l2 sin(θ1 − θ2 )θ̇22 + (m1 + m2 )gl1 sin θ1 = 0
For θ2
d
dt
∂L
∂ θ̇2
6
−
∂L
= 0,
∂θ2
which yields
d m2 l22 θ̇2 + m2 l1 l2 θ̇1 cos(θ1 − θ2 ) − m2 l1 l2 θ̇1 θ̇2 sin(θ1 − θ2 ) + m2 gl2 sin θ2 = 0.
dt
Upon taking the time derivative, we obtain the final form
m2 l1 l2 cos(θ1 − θ2 )θ̈1 + m2 l22 θ̈2 − m2 l1 l2 sin(θ1 − θ2 )θ̇12 + m2 gl2 sin θ2 = 0
7. If we want to obtain the equations of motion for a charged particle of mass m, moving
in an electromagnetic field (E, B), the potential in the Lagrangian has to be velocity
dependent U = qφ − qA · v, where q is the charge of the particle, and φ, and A,
respectively, are the scalar and vector potentials of the electromagnetic field so that
E = −∇φ −
∂A
∂t
B = ∇ × A.
Show that using this Lagrangian, we obtain the correct equations of motion for the
particle.
Soln: Using Cartesian coordinates and the fact that v = ẋî + ẏ ĵ + ż k̂, and A =
Ax î + Ay ĵ + Az k̂, we have
1
L = m(ẋ2 + ẏ 2 + ż 2 ) − qφ + q(Ax ẋ + Ay ẏ + Az ż).
2
Lagrange equation for x component
d ∂L
∂L
=0
−
dt ∂ ẋ
∂x
∂φ
∂Ax
∂Ay
∂Az
d
(mẋ + qAx ) + q
− q ẋ
− q ẏ
− q ż
=0
=⇒
dt
∂x
∂x
∂x
∂x
dAx
∂φ
∂Ax
∂Ay
∂Az
=⇒ mẍ = −q
−q
+ q ẋ
+ q ẏ
+ q ż
dt
∂x
∂x
∂x
∂x
But
dAx
∂Ax
∂Ax
∂Ax
∂Ax
=
ẋ +
ẏ +
ż +
,
dt
∂x
∂y
∂z
∂t
so that
∂φ ∂Ax
∂Ax
∂Ax
∂Ax
∂Ax
∂Ay
∂Az
mẍ = q −
−
−q
ẋ − q
ẏ − q
ż + q ẋ
+ q ẏ
+ q ż
∂x
∂t
∂x
∂y
∂z
∂x
∂x
∂x
∂φ ∂Ax
∂Ay
∂Az
∂Ax
∂Ax
=q −
−
+ q ẏ
+ q ż
−q
ẏ − q
ż
∂x
∂t
∂x
∂x
∂y
∂z
Using the fact that
E = −∇φ −
B = ∇ × A,
7
∂A
∂t
we obtain above
mẍ = qEx + q(v × B)x ,
which is the x component of the Lorentz force equation. Using the same procedure for
y and z components, we obtain
mr̈ = qE + q(v × B).
8. The electromagnetic field is invariant under a gauge transformation of the scalar and
vector potential given by
A → A + ∇ψ(r, t),
∂ψ
,
φ → φ−
∂t
where ψ is arbitrary (but differentiable). What effect does this gauge transformation
have on the Lagrangian of a moving particle in the electromagnetic field? Is the equation of motion affected?
Soln: On performing the gauge transformations, we have
0
A → A = A + ∇ψ(r, t),
∂ψ
0
φ → φ =φ−
,
∂t
1
∂ψ
0
0
0
L → L = m(ẋ2 + ẏ 2 + ż 2 ) − qφ + qA · v = L − qv · ∇ψ − q
2
∂t
Or
∂ψ
∂ψ
∂ψ
∂ψ
L =L−q
ẋ +
ẏ +
ż +
∂x
∂y
∂z
∂t
dψ
=L−q
dt
0
Because L0 differs from L by the total time derivative of a differentiable function
ψ = ψ(r, t), hence, from the result of Prob 4, it will lead to the same Lagrange
equations as L.
8