4.1: Non-Conservative Forces

4.2 Forces of Constraints and Generalized Coordinates

Suppose that a particle is free to move in all directions. Three coordinates are needed to specify its location.

• Cartesian: $(x,y,z)$

• Cylinderical: $(\rho, \psi, z)$

• Spherical: $(r, \theta,\phi)$

The presence of constrains mean that some coordinates might be less than three positions. A constraint that reduces the number of position of a particle is called a holonomic constraint. The minimal set of required independent coordinates are called generalized coordinates, denoted by $q_k$.

4.3: Hamilton’s Mechanics

The Lagrangian is defined as

where $T$ is kinetic energy and $U$ is potential energy. Having chosen a set of generalized coordinates, the Lagrangian can be written as:

We define its action $S[q_k(t)]$ as the functional of the time integrand over the Lagrange $L$, from a starting time $t_a$ to an ending time $t_b$.

When $S$ is stationary, $\delta S$(the change in position) is zero, giving us

Using knowledge from the previous chapter, the Lagrange is equal to (The integral represents the area under curve, where the curve/function is defined by the integrand. In this case, the integrand is $L$. When $S$ is stationary, it has reached a minimum or a maximum. As we derived in chapter, when the value of a function is extremeized, by Euler’s equation $\color{red}0={\partial L}/{\partial q_k}-({d}/{dt})({\partial L}/{\partial \dot{q_k}})$:

4.32: Newton’s Law and Hamilton Mechanics

Hamilton Mechanics is consistent of Newton’s law. The second term of the Lagrange equation essentially represents $F=ma$, while the first term sums of force components. This is illustrated by below examples 4., 4.2. and 4.3.

Example 4.1: A Simple Pendulum

An inertial observer sees that a small plumb bob of mass m is free to swing back and forth in a vertical x–z plane at the end of a string of length R. The position of the bob can be specified uniquely by its angle $\theta$ measured up from its equilibrium position at the bottom, so we choose $\theta$ as the generalized coordinate, as illustrated in Figure 4.2. Find the equation of motion and the second order differential equation.

• Solution

  The origin of the system is defined at the bottom. Kinetic energy is $K=1/2mv^2$, where velocity $v$ requires three coordinates to specify its direction. By cylindrical coordinates, we know that $(x,y,z)=(r, r\theta, z)$. Speed $v$ is equal to:

  $$v=\dot{x}^2+\dot{y}^2+\dot{z}^2=r^2+r^2\dot{\theta}^2+\dot{z}^2$$

  Kinetic energy is equal to:

  $$K=\frac{1}{2}mv^2=\frac{1}{2}m(r^2+r^2\dot{\theta}^2+\dot{z}^2)$$

  Given the constraint that the string length $R$ is fixed, $\dot{r}=0$ and initial speed $\dot{z}=0$, the equation becomes:

  $$K=\frac{1}{2}mv^2=\frac{1}{2}m(R^2\dot{\theta}^2)$$

  The simple pendulum, with $\theta$ as the generalized coordinate.

  $$L=T-U=\frac{1}{2}m(R^2\dot{\theta}^2)-mgR(1-\cos\theta)$$

  The Euler’s equation for $L=L(t, \theta, \dot{\theta})$ is equal to:

  $$0=\frac{\partial L}{\partial \theta}-\frac{d}{dt}\frac{\partial L}{\partial \dot\theta}=-mgR\sin\theta-\frac{d}{dt}(\frac{1}{2}mR^22\dot\theta\ddot\theta)$$

  Simplifying the expression, we get

  $$0=\ddot\theta+\frac{g}{R}\sin\theta$$

  The second order differential equation shows the pendulum is in simple harmonic motion with an angular frequency of $\omega=\sqrt{g/R}$. Torque is equal to $\tau=I\omega=mR^2g/R=mgR$.

Example 4.2: A Bead Sliding on a Vertical Helix

A bead of mass m is slipped onto a frictionless wire wound in the shape of a helix of radius $R$, whose symmetry axis is oriented vertically in a uniform gravitational field, as shown in Figure 4.3. As always, we assume the description is from an inertial frame’s perspective (unless explicitly stated otherwise). Using cylindrical coordinates to find equations of motion.

Example 4.3: Block on an Inclined Plane

A bead of mass m is slipped onto a frictionless wire wound in the shape of a helix of radius R, whose symmetry axis is oriented vertically in a uniform gravitational field, as shown in Figure 4.3

• Solution

  The Lagrange $L$ is equal to:

  $$L=T-U=\frac{1}{2}mv^2-mgX\sin\alpha=\frac{1}{2}m\dot{X}^2-mgX\sin\alpha$$

  Using Euler’s equation, we let $L=L(t, \dot{X}, X)$, giving us:

  $$0=-mg\sin\alpha-\frac{d}{dt}\frac{1}{2}m(2\dot{X})=-mg\sin\alpha-\frac{d}{dt}m\dot{X}$$

  The equation is consistent with Newton’s law, where $ma=-mg\sin\alpha$ in the horizontal (inclined) direction.

Summary for finding $L$ and equations of motion

1. Use cartesian/polar/spherical/cylindrical coordinate to find positions of any arbitrary position $\vec s$.

2. Differentiate position $\vec s$ to find velocity $v$

3. Find the Lagrange $L=T-U$
   1. Find the expression for kinetic energy $T$
   2. Find the expression for potential energy $U$

4. Use Euler’s equation to find equations of motion

4.4: Generalized Momenta and Cyclic Coordinates

In the last section, we has shown that Lagrange equation is consistent with Newton’s second law. The first term represents force components, and the second term is $ma$. Because $F=dp/dt$, the term $\partial L/\partial \dot{x}$ can be seen as a component of momentum. Thus, it is natural to define the generalized momenta using generalized coordinates $q_k$

The generalized momenta is defined as:

By Newton’s second law, $dp/dt=ma$. In terms of $p_k$, the Lagrange equation becomes:

Lagrange is expressed in the form of $L=L(t, q_k, \dot{q_k})$. Sometimes, the coordinate $q_k$ is missing while $\dot{q_k}$ is presents. A missing coordinate is said to be a cyclic coordinate. For such coordinate, the Lagrange equation tells us that the time derivative of corresponding generalized momentum is zero, so that the particular momentum is conserved.

Explanation:

The explanation seems confusing. All it is trying to say is that: if there is a cyclic coordinate, there must be a conserved quantity. This is better explained by the below example. Consider the Lagrange $L=L(t, q_k, \dot{q_k})$, where $q_k$ is missing. Using Euler’s equation, we get:

Because $q_k$ is missing, the equation is independent of $q_k$ and $\partial L/\partial q_k=0$. Then,

so $\partial L/\partial \dot{q_k}=k$ is a constant. Thus, $\partial L/\partial \dot{q_k}$ is a conserved quantity.

Example: Particle on a Tabletop

Particle moving on a tabletop. There is a central for $F(r)=kr\hat{r}$ directed towards the origin. Derive the Lagrange equation and expressions for effective potential, and time.

• Solution

  1. Lagrange and Euler’s Equations:

  For a particle moving in a plane (two dimensions) and subject to a central force, it is better to use polar coordinates $(r, \theta)$ about the origin. The kinetic energy is given by:

  $$T=\frac{1}{2}mv^2=\frac{1}{2}m(\dot{r}^2+(r\dot{\theta})^2)$$

  The Lagrange $L=T-U$ is equal to:

  $$L=T-U(r)=\frac{1}{2}m(\dot{r}^2+(r\dot{\theta})^2)-U(r)$$

  The generalized coordinates are $r$ and $\theta$. Notice that $\theta$ is the cyclic coordinate since it is missing and only $\dot{\theta}$ is present. Thus, there must be a conserved quantity. Using Euler’s equation, we know that:

  $$0=\frac{d}{dt}\frac{\partial L}{\partial \dot{\theta}}$$

  such that $\partial L/\partial \dot{\theta}=k$ is equal to a constant. Using chain rule for multivaraible functions, the conserved is:

  $$p_\theta=\frac{\partial L}{\partial \dot\theta}=mr^2\dot\theta$$

  We recognize this as the angular momentum (The makes sense inutitively and theoretically. As shown in the previous section, the second term in Lagrange equation is equal to $d/dt(\partial L/\partial \dot\theta)=ma$ and $F=dp/dt$, so it is natural that $\partial L/\partial \dot\theta$ is some form of momentum).

  The Lagrange equations are (Notice that there are two independent variables/generalized coordinates, so we use variational calculus for several variables in chapter three The Variational Principle):

  $$\frac{\partial L}{\partial r}=\frac{d}{dt}\frac{\partial L}{\partial \dot{r}}\quad \frac{\partial L}{\partial \theta}=\frac{d}{dt}\frac{\partial L}{\partial \dot{\theta}}$$

  which becomes:

  $$m(r\dot\theta^2)-\frac{\partial U(r)}{\partial r} =\frac{d}{dt}(m\dot{r}), \quad 0=\frac{d}{dt}mr^2\dot\theta$$

  $$-\frac{\partial U(r)}{\partial r}=m\ddot{r}-mr\dot\theta^2, \quad p_\theta=mr^2\dot\theta$$

  This is equal to:

  $$F(r)=ma_r=m(\ddot{r}-r\dot\theta^2), \quad F_\theta=m(r\ddot\theta+2\dot{r}\dot\theta)$$

  2. Newton’s Law and Lagrange Equations: 2nd Order DE Derivations

  Notice this is consistent with Newton’s second law, where the acceleration for an object in circular motion is equal to $a=a_r+a_\theta=(\ddot r-r\dot\theta)\hat r+(r+\ddot\theta+2\dot{r}\dot{\theta})\hat{\theta}$. Because $p_\theta=mr\dot\theta^2$ is a conserved quantity, we can solved for $\dot\theta$, giving us:

  $$\dot\theta=\frac{p_\theta}{mr^2}$$

  Substituting the expression for $\dot\theta$ into the Lagrange equation for $-\partial U(r)/\partial r$, we get:

  $$-\frac{\partial U(r)}{\partial r}=m\ddot{r}-m{r}(\frac{p_\theta}{mr^2})^2$$

  where the central force $F(r)=\partial U(r)/\partial r=-kr$ (This is shown in example 1.6 in chapter one Newtonian Mechanics). Then, the expression becomes:

  $$-kr=m\ddot{r}-\frac{mp_\theta^2}{m^2r^3}$$

  Rearranging the equation, we get:

  $$0=m\ddot{r}-\frac{p_\theta^2}{mr^3}+kr$$

  $$0=\ddot{r}-\frac{p_\theta^2}{m^2r^3}+\frac{k}{m}r$$

  $$\boxed{0=\ddot{r}-\frac{p_\theta^2}{m^2r^3}+\omega^2 r}$$

  3. Effective Potential/Energy Conservation Derivation

  Rearranging the second order differential equation, $m\ddot r$ is equal to:

  $$m\ddot{r}=\frac{p_\theta^2}{mr^3}-m\omega^2 r=F(r)$$

  where $F(r)=m\ddot r$ is the central force directed towards the origin. Recall that effective potential $U_{eff}$ is defined as (Note that effective potential is NOT same as potential energy. It is defined to simplifies two dimensional motion into one dimension. For more details, please refer to the :

  $$U_{eff}=-\int F(r) dr$$

  Integrate $F(r)$ alone the path for $r$, $U_{eff}$ is equal to:

  $$U_{eff}=-\int (\frac{p_\theta^2}{mr^3}-m\frac{k}{m} r) dr=\frac{p_\theta^2}{2mr^2}+\frac{1}{2}kr^2+C$$

  where $C$ is a constant. Then, the second first integral of motion is given by energy conservation in this one-dimensional system.

  $$\boxed{E=K+U_{eff}(r)=\frac{1}{2}m\dot r^2+\frac{p_\theta^2}{2mr^2}+\frac{1}{2}kr^2}$$

  Note that $K\ne T$ in the Lagrange equation. The kinetic energy $K$ includes the radial component only. This is because kinetic energy in tangential component in $\hat\theta$ direction is a part of the effective potential $U_{eff}$, so that the motion is simplified from two dimensional into one dimensional.

  

  4. Time

     Rearranging the equation for energy $E$, the expression becomes:

     $$E=\frac{1}{2}m\dot r^2+U_{eff}(r)$$

     Solving for $\dot r$, we get:

     $$\dot{r}=\sqrt{\frac{2E}{m}-U_{eff}(r)}$$

     Expressing $\dot r$ as a differential, we get:

     $$\dot r=\frac{dr}{dt}$$

     Using separable integration, time $t$ is equal to:

     $$\boxed{t=\int dt=\int \sqrt{\frac{2E}{m}-U_{eff}(r)} dr}$$

Example 4.5: Spherical Pendulum

A ball of mass m swings on the end of an unstretchable string of length R in the presence of a uniform gravitational field g. This is often called the “spherical pendulum,” because the ball moves as though it were sliding on the frictionless surface of a spherical bowl. We aim to find its equations of motion.

• Solution

  Before solving the problem, we recognize the constraint that the pendulum has a fixed radius of R. Then, the ball has two degrees of freedom:

  1. It can move horizontally around a vertical axis, corresponding to the azimuthal angle $\phi$.

  2. It can also move in the polar direction, described by the angle $\theta$.

  When finding equations of motion, we follow the below steps:

  1. Specify the position of an arbitrary point using Cartesian/Spherical/Cylindrical coordinates.

  2. Differentiate position $\vec{s}$ to find velocity $\vec{v}$.

  3. Find kinetic energy $T$ and potential energy $U$ to get the Lagrange $L=T-U$.

  4. Apply Euler’s equation to $L$ to find equations of motion.

  5. First integrals of motion.

  STEP 1 Find $\vec{s}$:

  Using spherical coordinates, the position of a point is specified by:

  $$\vec{s}=(r,r\theta, r\sin\theta\phi)$$

  Given that radius $r=R$, the expression becomes:

  $$\vec{s}=(R,R\theta, R\sin\theta\phi)$$

  STEP 2: Velocity

  Differentiating $\vec{s}$ with respect to time, velocity $\vec{v}$ is equal to:

  $$\vec{v}=(0, R\dot\theta, R\sin\theta\dot\phi)$$

  where $\dot R=0$ since $R$ is fixed (a constant).

  STEP 3: Lagrangian

  Using $\vec{v}$, kinetic energy $T$ is equal to:

  $$T=\frac{1}{2}mv^2=\frac{1}{2}mR^2(\dot\theta^2+\sin^2\theta\dot\phi^2)$$

  Potential energy $U$ is equal to:

  $$U=mgR\cos\theta$$

  The Lagrange $L=T-U$ is equal to:

  $$L=T-U=\frac{1}{2}mR^2(\dot\theta^2+\sin^2\theta\dot\phi^2)=mgR\cos\theta$$

  STEP 4: Euler’s equation to find equations of motion

  Using Euler’s equation, we get:

  $$0=\frac{\partial L}{\partial \theta}-\frac{d}{dt}\frac{\partial L}{\partial \dot\theta},\quad 0=\frac{\partial L}{\partial \phi}-\frac{d}{dt}\frac{\partial L}{\partial \dot\phi}$$

  STEP 5: First Integral of motion

  Because $\phi$ is cyclic, we know that $p_\phi$, the angular momentum, is conserved. Note that the expression for angular momentum $p_\phi$ is a first integral of motion..

  $$p_\phi=mR^2\sin^2\theta\dot\phi=\text{constant}$$

  Rearranging the Lagrange equation for $\theta$, we get:

  $$\ddot\theta-\sin\theta\cos\theta\dot\phi^2+\frac{g}{R}\sin\theta=0$$

  Solving for $\dot\phi$ using $p_\phi$, the expression becomes:

  $$\ddot\theta-(\frac{p_\theta}{mR^2})^2\sin\theta\cos\theta+\frac{g}{R}\sin\theta=0$$

  The second first integral of motion is the expression for conservation of energy, where $E=T+U$.

4.5: Systems of Particles

The last section considers the motion for a single particle only. The section studies the motion of a system of particles

Recall that the action for the system is defined as:

Consider a system of two particles confined in the horizontal direction and is moving alone a frictionless rail.

The Lagrange of the system is equal to:

where $U(x_2-x_1)$ is the potential energy between masses, describing the interaction between two particles. The action of the system is equal to:

Notice that there are two generalized coordinates, $x_1$ and $x_2$, for the system. Using Euler’s equation, Lagrange equations are:

Using chain rule for multivariable functions, we get:

giving us:

The expression for $m_2$ is:

giving us:

According to Newton’s third law, the force exerted by $m_1$ on $m_2$ is equal to the magnitude of the force exerted by $m_2$ on $m_1$.

Because the total momentum of the system is conserved, the change in momentum $\Delta p=dp/dt=0$, giving us:

The new coordinates are center of mass, giving by:

Central Force, Reduced Mass, Energy

Let $x=x_2-x_1$. The Lagrange of the system is:

where $\mu$ is the reduced mass

The corresponding momentum $P$ is equal to:

and is conserved.

Example: Pulley

A contraption of pulleys. We want to find the accelerations of all three weights. We assume that the three pulleys have negligible mass so they have negligible kinetic and potential energies.

• Solution

  1. Lagrangian

  In order to write the Lagrange, we need to know the system’s kinetic and potential energies. Kinetic energy of the system is equal to:

  $$T=\frac{1}{2}m_1\dot y_1^2+\frac{1}{2}m_2\dot y_2^2+\frac{1}{2}M\dot Y^2$$

  2. Length Constraint

  Because there is only two degrees of freedom, we need to eliminate one variable in $\dot y_1, \dot y_2$ and $\dot Y$. When solving problems involving pulley, one common technique is to use length constraint to relate length with acceleration to eliminate a variable. Because the total length of the string is fixed, we know that:

  $$(H-y_1)+2(H-Y)+(H-y_2)=4H-2Y-y_2-y_1=\text{constant}=L$$

  Differentiating the equation with respect to time $t$, the expression is equal to:

  $$-2\dot Y-\dot y_2-\dot y_1=0$$

  giving us:

  $$\dot Y=-\frac{1}{2}(\dot y_1+\dot y_2)$$

  3. Substitution

  Substituting the expression for $\dot Y$ into $T$, we get:

  $$T=\frac{1}{2}m_1\dot y_1^2+\frac{1}{2}m_2\dot y_2^2+\frac{1}{2}M\dot (\frac{1}{2}(\dot y_1+\dot y_2))2$$

  Knowing the kinetic energy, we need to find potential energy $U$, giving us:

  $$U=m_1gy_1+m_2gy_2+MgY$$

  $$U=m_1gy_1+m_2gy_2-Mg(\frac{y_1+y_2}{2})+\text{constant}$$

  4. Euler’s equation

  The Lagrange is equal to:

  $$L=T-U=\frac{1}{2}m_1\dot y_1^2+\frac{1}{2}m_2\dot y_2^2+\frac{1}{2}M\dot (\frac{1}{2}(\dot y_1+\dot y_2))2-m_1gy_1-m_2gy_2+Mg(\frac{y_1+y_2}{2})+\text{constant}$$

  Using Euler’s equation, we get:

  $$0=\frac{\partial L}{\partial y_1}-\frac{d}{dt}\frac{\partial L}{\partial \dot y_1}, \quad 0=\frac{\partial L}{\partial y_2}-\frac{d}{dt}\frac{\partial L}{\partial \dot y_2}$$

  Using chain rule, we get:

  $$m_1\ddot y_1+\frac{M}{4}(\ddot y_1+\ddot y_2)=-m_1g+\frac{Mg}{2}$$

  $$m_2\ddot y_2+\frac{M}{4}(\ddot y_1+\ddot y_2)=-m_2g+\frac{Mg}{2}$$

  5. Solving the system of equations

  Solving the system of equations, we get:

  $$\dot y_1=-g+\frac{-4m_2g}{m_1+m_4m_1m_2/M}$$

  $$\dot y_2=-g+\frac{-4m_1g}{m_1+m_4m_1m_2/M}$$

  $$\dot Y=g-\frac{2(m_1+m_2)g}{m_1+m_4m_1m_2/M}$$

Example: A block on a Inclined Plane

Let us return to the classic problem of a block sliding down a frictionless inclined plane, as in Example 4.2, except that we will make things a bit more interesting: now the inclined plane itself is allowed to move. Figure 4.11(a) shows the system. A block of mass $m$ rests on an inclined plane of mass $M$: both the block and the inclined plane are free to move without friction. The plane’s angle is denoted by $\alpha$. The problem is to find the acceleration of the block

• Solution

  This is a classic problem. In the approach of Newton’s law, we would draw force diagrams and use the constraint of a fixed angle to solve the system of equations to find acceleration. In Hamilton’s mechanics, we will find the Lagrange $L=T-U$ first.

  The first observation we need to make is that the plane is not fixed. As the block slides down, it exerts a force on the plane, causing the object to move leftward (the ground is frictionless or the force exerted by the block exceeds stative friction). Kinetic energy of the system the includes two components: the block and the plane.

  Define the coordinate system: Another thing we need to do before solving the problem is to define the coordinate system. Let $X$ be the horizontal distance slided by the plane, and $x$ and $y$ be the horizontal and vertical distances slide by block. The origin is shown in the figure.

  

  Kinetic energy of the system is:

  $$T=\frac{1}{2}M(\dot X^2+\dot Y^2)+\frac{1}{2}m(\dot x^2+\dot y^2)$$

  Because $\dot Y$ is fixed, $\dot Y=0$. Then, $T$ us equal to (This makes sense and can be seen esaily as the plane moves in the horizontal only. The problem uses $v^2=\dot X^2+\dot Y^2$ to show a more comprehensive approach):

  $$T=\frac{1}{2}M\dot X^2+\frac{1}{2}m(\dot x^2+\dot y^2)$$

  Potential energy is equal to:

  $$U=U_{plane}-mgy$$

  where $U_{plane}$ is the plane’s gravitational potential energy, and the quantity remains unchanged.

  There are three variables $\dot X, \dot x$ and $\dot y$ in $T$. Because the degree of freedom is two (The degree of freedom of two refer to $X$ and $Y$. We start with four coordinates, $\dot X, \dot Y, \dot x$, and $\dot y$. After identifying two constraints: $Y$ is fixed and $\dot x$ and $\dot y$ can be represented using $D$ and $\alpha$, the degree of freedom is 2), we need to replace one of the variables. Using trigonometry

  $$x=\dot X+D\cos\alpha, \quad y=-D\sin\alpha$$

  Differentiating expressions with respect to $t$, we get

  $$\dot x=\dot X+\dot D\cos\alpha, \quad \dot y=-\dot D\sin\alpha$$

  Substituting, we get:

  $$T=\frac{1}{2}M\dot X^2+\frac{1}{2}m((\dot X+\dot D\cos\alpha)^2+(-\dot D\sin\alpha)^2)$$

  $$T=\frac{1}{2}M\dot X^2+\frac{1}{2}m\dot X^2+m\dot X\dot D\cos\alpha +\frac{1}{2}m\dot D^2\cos^2\alpha +\frac{1}{2}m\dot D^2\sin^2\alpha$$

  $$T=\frac{1}{2}M\dot X^2+\frac{1}{2}m\dot X^2+m\dot X\dot D\cos\alpha +\frac{1}{2}m\dot D^2$$

  The Lagrangian is equal to:

  $$L=T-U=\frac{1}{2}M\dot X^2+\frac{1}{2}m\dot X^2+m\dot X\dot D\cos\alpha +\frac{1}{2}m\dot D^2-U_{plane}+mgD\sin\alpha$$

  Using Euler’s equation, we get:

  $$0=\frac{\partial L}{\partial X}-\frac{d}{dt}\frac{\partial L}{\partial \dot X}, \quad 0=\frac{\partial L}{\partial D}-\frac{d}{dt}\frac{\partial L}{\partial \dot D}$$

  Using Chain Rule, we get

  $$0=\frac{d}{dt}(M\dot X+m\dot X+m\dot D\cos\alpha)=M\ddot X+m\ddot X+m\ddot D\cos\alpha$$

  $$0=mg\sin\alpha-\frac{d}{dt}(m\dot X\cos\alpha+m\dot D)=mg\sin\alpha-m\ddot X\cos\alpha+m\ddot D$$

  This gives us a system of equations

  $$0=(M+m)\ddot X+m\ddot D\cos\alpha$$

  $$mg\sin\alpha=m\ddot X\cos\alpha+m\ddot D$$

  Solving the system of equations, we get:

  $$\boxed{\ddot X=\frac{-mg\cos\alpha\sin\alpha}{M+m\sin^2\alpha}}$$

  $$\boxed{\ddot D=\frac{(M+m)g\sin\alpha}{M+m\sin^2\alpha}}$$

  Then, $\ddot x$ and $\ddot y$ are equal to:

  $$\ddot x=\frac{-mg\cos\alpha\sin\alpha}{M+m\sin^2\alpha}+\frac{(M+m)g\sin\alpha}{M+m\sin^2\alpha}\cos\alpha$$

  $$\boxed{\ddot x=\frac{Mg\cos\alpha\sin\alpha}{M+m\sin^2\alpha}}$$

  $$\ddot y=-\ddot D\cos\alpha=-\frac{(M+m)g\sin\alpha}{M+m\sin^2\alpha}\sin\alpha$$

  $$\boxed{\ddot y=-\frac{(M+m)g\sin^2\alpha}{M+m\sin^2\alpha}}$$

4.6: Hamiltonian

Let the Lagrange be $L=L(t, q_k, \dot q_k)$. Using chain rule to multi-variable functions, we differentiate $L$ with respect to time, giving us:

Using product rule for multi-variable functions, we recognize that:

We notice that:

The Hamiltonian $H$ of a particle is defined as:

where $p_k=\partial L/\partial \dot q_k$. From the equation, we can see that:

The result is interesting if $L$ is not an explicit function of time. For example, if $\partial L/\partial t=0$, $H$ is equal to a constant and thus $H$ is conserved.

Though the Hamiltonian seems to be an abstract quantity, it is closely related to energy. Suppose that a particle is free to move in three dimensions in a potential $U(x, y, z)$ without constraints, and that we are using Cartesian coordinates. Momentums are $p_x=m\dot x, p_y=m\dot y$, and $p_z=m\dot z$. Then, $\sum \dot q_kp_k=m(\dot x^2+\dot y^2+\dot z^2)$. The Hamiltonian is equal to:

4.6.1: Interpretation of the Hamiltonian