Classical Mechanics
Complete Solutions with Detailed Reasoning & Step-by-Step Explanations
Question 1
A double pendulum consists of two point masses $m$ attached by strings of length $l$. The strings make angles $\theta_1$ and $\theta_2$ with the vertical axis. Find the kinetic energy of the pendulum.
Reasoning:
In Lagrangian mechanics the kinetic energy $T$ must be expressed in terms of the generalised coordinates and their time derivatives. For a double pendulum the natural generalised coordinates are the angles $\theta_1$ and $\theta_2$. We first write the Cartesian coordinates of each mass, differentiate them with respect to time to obtain the velocity components, and finally form $\frac12 m v^2$ for each mass.
Step 1 – Cartesian coordinates
Place the fixed support at the origin.
Position of the first mass $m_1$:
$$
\begin{align*}
x_1 &= l\sin\theta_1, \\
y_1 &= -l\cos\theta_1.
\end{align*}
$$
Position of the second mass $m_2$:
$$
\begin{align*}
x_2 &= l\sin\theta_1 + l\sin\theta_2, \\
y_2 &= -l\cos\theta_1 - l\cos\theta_2.
\end{align*}
$$
Step 2 – Velocities
Differentiate with respect to time:
$$
\begin{align*}
\dot x_1 &= l\dot\theta_1\cos\theta_1, \\
\dot y_1 &= l\dot\theta_1\sin\theta_1.
\end{align*}
$$
$$
\begin{align*}
\dot x_2 &= l\dot\theta_1\cos\theta_1 + l\dot\theta_2\cos\theta_2, \\
\dot y_2 &= l\dot\theta_1\sin\theta_1 + l\dot\theta_2\sin\theta_2.
\end{align*}
$$
Step 3 – Squared speeds
$$
v_1^2 = \dot x_1^2 + \dot y_1^2 = l^2\dot\theta_1^2,
$$
$$
\begin{align*}
v_2^2 &= \dot x_2^2 + \dot y_2^2 \\
&= l^2\dot\theta_1^2 + l^2\dot\theta_2^2 + 2l^2\dot\theta_1\dot\theta_2\cos(\theta_1-\theta_2).
\end{align*}
$$
(The cross term arises from the product of the two velocity vectors and the angle between them.)
Step 4 – Total kinetic energy
$$
T = \frac12 m v_1^2 + \frac12 m v_2^2
= \frac12 m l^2\dot\theta_1^2 + \frac12 m\bigl(l^2\dot\theta_1^2 + l^2\dot\theta_2^2 + 2l^2\dot\theta_1\dot\theta_2\cos(\theta_1-\theta_2)\bigr).
$$
Simplifying:
$$
T = ml^2\dot\theta_1^2 + \frac12 ml^2\dot\theta_2^2 + ml^2\dot\theta_1\dot\theta_2\cos(\theta_1-\theta_2).
$$
Kinetic energy of the double pendulum:
$$
T = ml^{2}\dot{\theta}_{1}^{2}+\frac{1}{2}ml^{2}\dot{\theta}_{2}^{2}+ml^{2}\dot{\theta}_{1}\dot{\theta}_{2}\cos(\theta_{1}-\theta_{2}).
$$
Question 2
The Lagrangian of a particle is
$$
L=\frac{m}{2}(\dot{x}_{1}^{2}+\dot{x}_{2}^{2})-\frac{\lambda}{2}(x_{1}^{2}+x_{2}^{2}).
$$
Find the Hamiltonian.
Reasoning:
The Hamiltonian is obtained from the Legendre transform
$$
H=\sum_i p_i\dot q_i-L,
$$
where the generalised momenta are defined by $p_i=\partial L/\partial\dot q_i$. Because $L$ is quadratic in the velocities, $H$ will turn out to be the total energy.
Step 1 – Generalised momenta
$$
p_1=\frac{\partial L}{\partial\dot x_1}=m\dot x_1\qquad\Rightarrow\qquad\dot x_1=\frac{p_1}{m},
$$
$$
p_2=\frac{\partial L}{\partial\dot x_2}=m\dot x_2\qquad\Rightarrow\qquad\dot x_2=\frac{p_2}{m}.
$$
Step 2 – Construct $H$
$$
H=p_1\dot x_1+p_2\dot x_2-L
=p_1\Bigl(\frac{p_1}{m}\Bigr)+p_2\Bigl(\frac{p_2}{m}\Bigr)
-\frac{m}{2}\Bigl(\frac{p_1^2}{m^2}+\frac{p_2^2}{m^2}\Bigr)
+\frac{\lambda}{2}(x_1^2+x_2^2).
$$
$$
H=\frac{p_1^2}{m}+\frac{p_2^2}{m}-\frac{1}{2m}(p_1^2+p_2^2)+\frac{\lambda}{2}(x_1^2+x_2^2)
=\frac{1}{2m}(p_1^2+p_2^2)+\frac{\lambda}{2}(x_1^2+x_2^2).
$$
Hamiltonian:
$$
H=\frac{1}{2m}(p_1^2+p_2^2)+\frac{\lambda}{2}(x_1^2+x_2^2).
$$
Question 3
Show that the Hamiltonian of a system remains constant if the Lagrangian of the system does not depend upon time explicitly.
Reasoning:
We use the total time derivative of $H$ together with the Euler-Lagrange equations. When $\partial L/\partial t=0$, the energy function (which coincides with $H$ for standard kinetic-energy forms) is a constant of motion.
Step 1 – Definition of $H$
$$
H(q,p,t)=\sum_i p_i\dot q_i-L(q,\dot q,t).
$$
Step 2 – Total time derivative
$$
\frac{dH}{dt}=\sum_i\Bigl(\dot p_i\dot q_i+p_i\ddot q_i\Bigr)
-\sum_i\Bigl(\frac{\partial L}{\partial q_i}\dot q_i+\frac{\partial L}{\partial\dot q_i}\ddot q_i\Bigr)
-\frac{\partial L}{\partial t}.
$$
Because $p_i=\partial L/\partial\dot q_i$, the terms containing $\ddot q_i$ cancel:
$$
\frac{dH}{dt}=\sum_i\Bigl(\dot p_i\dot q_i-\frac{\partial L}{\partial q_i}\dot q_i\Bigr)-\frac{\partial L}{\partial t}.
$$
Step 3 – Use Euler-Lagrange equations
$$
\frac{d}{dt}\Bigl(\frac{\partial L}{\partial\dot q_i}\Bigr)=\frac{\partial L}{\partial q_i}\qquad\Rightarrow\qquad\dot p_i=\frac{\partial L}{\partial q_i}.
$$
Substitution yields
$$
\frac{dH}{dt}=-\frac{\partial L}{\partial t}.
$$
Step 4 – Conclusion
If the Lagrangian does not depend explicitly on time,
$$
\frac{\partial L}{\partial t}=0\qquad\Rightarrow\qquad\frac{dH}{dt}=0.
$$
Hence $H$ is a constant of the motion.
If $\partial L/\partial t=0$, then $H=\text{constant}$.
Question 4
Derive Hamilton’s equations of motion of a system. Show that the Hamiltonian of a conservative system is the total energy of the system.
Reasoning:
Hamilton’s equations follow directly from the differential of the Legendre transform that defines $H$. For a conservative system the potential depends only on coordinates, the kinetic energy is a homogeneous quadratic form in the velocities, and Euler’s theorem then shows that $H=T+V$.
Part A – Derivation of Hamilton’s equations
From the definition
$$
H(q,p,t)=\sum_i p_i\dot q_i-L(q,\dot q,t)
$$
we form the total differential:
$$
dH=\sum_i\bigl(\dot q_i\,dp_i+p_i\,d\dot q_i\bigr)
-\sum_i\Bigl(\frac{\partial L}{\partial q_i}dq_i+\frac{\partial L}{\partial\dot q_i}d\dot q_i\Bigr)
-\frac{\partial L}{\partial t}dt.
$$
The terms $p_i\,d\dot q_i$ and $\frac{\partial L}{\partial\dot q_i}d\dot q_i$ cancel because $p_i=\partial L/\partial\dot q_i$.
Using the Euler-Lagrange equation $\dot p_i=\partial L/\partial q_i$ we obtain
$$
dH=\sum_i\bigl(\dot q_i\,dp_i-\dot p_i\,dq_i\bigr)-\frac{\partial L}{\partial t}dt.
$$
On the other hand, treating $H$ as a function of the independent variables $q,p,t$,
$$
dH=\sum_i\Bigl(\frac{\partial H}{\partial q_i}dq_i+\frac{\partial H}{\partial p_i}dp_i\Bigr)+\frac{\partial H}{\partial t}dt.
$$
Comparing coefficients yields Hamilton’s canonical equations:
$$
\begin{align*}
\dot q_i&=\frac{\partial H}{\partial p_i},\\
\dot p_i&=-\frac{\partial H}{\partial q_i},\\
\frac{\partial H}{\partial t}&=-\frac{\partial L}{\partial t}.
\end{align*}
$$
Part B – $H$ is the total energy for a conservative system
For a standard mechanical system
$$
T=\frac12\sum_{i,j}m_{ij}(q)\dot q_i\dot q_j,\qquad V=V(q)
$$
(so $L=T-V$ does not depend explicitly on time).
$T$ is a homogeneous quadratic function of the velocities. Euler’s theorem on homogeneous functions gives
$$
\sum_i\dot q_i\frac{\partial T}{\partial\dot q_i}=2T.
$$
But $p_i=\partial L/\partial\dot q_i=\partial T/\partial\dot q_i$, therefore
$$
\sum_i p_i\dot q_i=2T.
$$
Hence
$$
H=\sum_i p_i\dot q_i-L=2T-(T-V)=T+V.
$$
Thus the Hamiltonian is identical with the total mechanical energy.
Hamilton’s equations:
$$
\dot q_i=\frac{\partial H}{\partial p_i},\qquad\dot p_i=-\frac{\partial H}{\partial q_i}.
$$
For a conservative system $H=T+V=$ total energy.
Question 5
(a) State the principle of Hamilton’s least action. Hence derive Euler-Lagrange’s equation of motion.
Reasoning:
Hamilton’s principle is a variational statement: the true path renders the action integral stationary. Applying the calculus of variations immediately produces the Euler-Lagrange equations.
Statement of Hamilton’s principle
The actual motion of a system between two fixed times $t_1$ and $t_2$ is such that the action integral
$$
S=\int_{t_1}^{t_2}L(q,\dot q,t)\,dt
$$
is stationary (usually a minimum) with respect to all varied paths that have the same end-points:
$$
\delta S=0.
$$
Derivation of the Euler-Lagrange equations
Consider a varied path $q_i(t)+\varepsilon\eta_i(t)$ where $\eta_i(t_1)=\eta_i(t_2)=0$.
The first variation of the action is
$$
\delta S=\varepsilon\int_{t_1}^{t_2}\sum_i\Bigl(\frac{\partial L}{\partial q_i}\eta_i+\frac{\partial L}{\partial\dot q_i}\dot\eta_i\Bigr)dt.
$$
Integrate the second term by parts:
$$
\int\frac{\partial L}{\partial\dot q_i}\dot\eta_i\,dt
=\Bigl[\frac{\partial L}{\partial\dot q_i}\eta_i\Bigr]_{t_1}^{t_2}
-\int\eta_i\frac{d}{dt}\Bigl(\frac{\partial L}{\partial\dot q_i}\Bigr)dt.
$$
The boundary term vanishes. Therefore
$$
\delta S=\varepsilon\int_{t_1}^{t_2}\sum_i\Biggl(\frac{\partial L}{\partial q_i}-\frac{d}{dt}\Bigl(\frac{\partial L}{\partial\dot q_i}\Bigr)\Biggr)\eta_i\,dt=0.
$$
Since the $\eta_i$ are arbitrary we must have
$$
\frac{d}{dt}\Bigl(\frac{\partial L}{\partial\dot q_i}\Bigr)-\frac{\partial L}{\partial q_i}=0
\qquad(i=1,2,\dots,n).
$$
These are the Euler-Lagrange equations of motion.
(b) A particle of mass $m$ is constrained to move on a circle under gravity. The circle is placed in a vertical plane and rests on the ground. Mention the generalised coordinate(s). Find Lagrange’s equation of motion.
Reasoning:
The constraint (motion on a fixed circle) reduces the degrees of freedom to one. The natural generalised coordinate is the polar angle measured from the downward vertical. Gravity supplies a potential linear in the height, which is easily written in terms of that angle.
Generalised coordinate
Let the radius of the circle be $R$.
Place the centre of the circle at height $R$ above the ground.
The single generalised coordinate is the angle $\theta$ that the radius vector makes with the downward vertical.
Lagrangian
Cartesian coordinates:
$$
x=R\sin\theta,\qquad y=R(1-\cos\theta)
$$
(so $y=0$ at the bottom).
Velocity:
$$
v^2=R^2\dot\theta^2.
$$
Kinetic energy:
$$
T=\frac12 m R^2\dot\theta^2.
$$
Potential energy (taking $y=0$ as zero):
$$
V=mgy=mgR(1-\cos\theta).
$$
Lagrangian:
$$
L=T-V=\frac12 m R^2\dot\theta^2-mgR(1-\cos\theta).
$$
Lagrange’s equation
$$
\frac{\partial L}{\partial\theta}=-mgR\sin\theta,\qquad
\frac{\partial L}{\partial\dot\theta}=m R^2\dot\theta.
$$
$$
\frac{d}{dt}\Bigl(\frac{\partial L}{\partial\dot\theta}\Bigr)=m R^2\ddot\theta.
$$
Euler-Lagrange equation:
$$
m R^2\ddot\theta+mgR\sin\theta=0
\qquad\Rightarrow\qquad
\ddot\theta+\frac{g}{R}\sin\theta=0.
$$
Generalised coordinate: $\theta$ (angle from downward vertical).
Equation of motion:
$$
\ddot{\theta}+\frac{g}{R}\sin\theta=0.
$$
Question 6
The Hamiltonian of a system is
$$
H=\frac{\alpha}{2}p^{2}q^{4}+\frac{\beta}{q^{2}},
$$
where $\alpha$ and $\beta$ are constants, and $q$ and $p$ are the generalised coordinate and momentum. Find the Lagrangian of the system.
Reasoning:
We recover the Lagrangian by the inverse Legendre transform. First express the velocity $\dot q$ in terms of $p$ from Hamilton’s equation $\dot q=\partial H/\partial p$, then substitute into $L=p\dot q-H$.
Step 1 – Velocity from Hamilton’s equation
$$
\dot q=\frac{\partial H}{\partial p}=\alpha p q^4
\qquad\Rightarrow\qquad
p=\frac{\dot q}{\alpha q^4}.
$$
Step 2 – Form the Lagrangian
$$
L=p\dot q-H
=\Bigl(\frac{\dot q}{\alpha q^4}\Bigr)\dot q
-\frac{\alpha}{2}\Bigl(\frac{\dot q}{\alpha q^4}\Bigr)^2 q^4
-\frac{\beta}{q^2}.
$$
$$
L=\frac{\dot q^2}{\alpha q^4}
-\frac{1}{2\alpha}\frac{\dot q^2}{q^4}
-\frac{\beta}{q^2}
=\frac{1}{2\alpha}\frac{\dot q^2}{q^4}-\frac{\beta}{q^2}.
$$
Lagrangian:
$$
L=\frac{1}{2\alpha}\frac{\dot{q}^{2}}{q^{4}}-\frac{\beta}{q^{2}}.
$$
Question 7
A particle of mass $m$ slides under gravity along the parabolic path $y=ax^{2}$ ($a=\text{constant}$). Find the Lagrangian and Lagrange’s equation of motion of the particle.
Reasoning:
The constraint $y=ax^{2}$ reduces the system to one degree of freedom. We choose $x$ as the generalised coordinate, express the kinetic energy in terms of $\dot x$ by differentiating the constraint, and write the gravitational potential in terms of $x$. The Euler-Lagrange equation then yields the equation of motion.
Step 1 – Constraint and velocity
$$
y=ax^{2}\qquad\Rightarrow\qquad\dot y=2ax\dot x.
$$
Speed squared:
$$
v^{2}=\dot x^{2}+\dot y^{2}=\dot x^{2}+4a^{2}x^{2}\dot x^{2}=\dot x^{2}(1+4a^{2}x^{2}).
$$
Step 2 – Lagrangian
Kinetic energy:
$$
T=\frac12 m\dot x^{2}(1+4a^{2}x^{2}).
$$
Potential energy (taking $y=0$ as reference):
$$
V=mgy=mg\,ax^{2}.
$$
Lagrangian:
$$
L=T-V=\frac12 m\dot x^{2}(1+4a^{2}x^{2})-mga x^{2}.
$$
Step 3 – Lagrange’s equation
$$
\frac{\partial L}{\partial\dot x}=m\dot x(1+4a^{2}x^{2}),
$$
$$
\frac{d}{dt}\Bigl(\frac{\partial L}{\partial\dot x}\Bigr)
=m\ddot x(1+4a^{2}x^{2})+m\dot x\cdot 8a^{2}x\dot x
=m\ddot x(1+4a^{2}x^{2})+8ma^{2}x\dot x^{2}.
$$
$$
\frac{\partial L}{\partial x}= \frac12 m\dot x^{2}\cdot 8a^{2}x -2mga x
=4ma^{2}x\dot x^{2}-2mga x.
$$
Euler-Lagrange equation:
$$
m\ddot x(1+4a^{2}x^{2})+8ma^{2}x\dot x^{2}
-4ma^{2}x\dot x^{2}+2mga x=0.
$$
Simplify:
$$
\ddot x(1+4a^{2}x^{2})+4a^{2}x\dot x^{2}+2ga x=0.
$$
Lagrangian:
$$
L=\frac12 m\dot{x}^{2}(1+4a^{2}x^{2})-mga x^{2}.
$$
Equation of motion:
$$
\ddot{x}(1+4a^{2}x^{2})+4a^{2}x\dot{x}^{2}+2ga x=0.
$$
All derivations follow the standard framework of analytical mechanics (Lagrange and Hamilton formalisms).
Symbols have their usual meanings: $m$ = mass, $g$ = acceleration due to gravity, etc.