Classical Mechanics in HKDSE context

Note: Some of the content are similar to lecture notes from my Dynamics courses, Introductory Dynamics (PHYS08052) and Lagrangian Dynamics (PHYS10015). I tried my best to avoid copying, but some of the flow may + similar.

Intended Audience: Year 1 Math/Physics major.

Motivation

HKDSE is a public exam taken by Hong Kong high school students. It covers a wide range of physics (Dynamics, Thermal Physics, Electromagnetism, etc.)

I am here using knowledge I gained in university to explain some advanced topics strongly related to the HKDSE Physics syllabus.

Calculus skills are expected.

Part I:

The first part covers orbits and gravitational forces.

Terminologies

\(\omega\) refers to angular velocity. We have \(v= r \omega\). \(\dot{r}\) is the derivative of \(r\) (with respect to time). We have \(F=m \ddot{r}\) (Make sure you can do this!)

Here we have \(\omega= \dot{theta}\)

\(\hat{r}\) is a unit-normal vector pointing at the direction of \(r\).

Centripetal Force

• The section below is quite hard. You need to know vector calculus operations.

From HKDSE Physics, we have Centripetal force as \(mr\omega^2\), or equivalently \(mr \dot{\theta}^2\). This is a very special form and only works for circular orbit, as radius is a constant.

Going back to Planar Polar Coordinates. \(r\) and \(\theta\) are the two degrees of freedom.

\[\hat{r}=\cos (\theta) \hat{x}+\sin (\theta) \hat{y}, \quad \hat{\theta}=-\sin (\theta) \hat{x}+\cos (\theta) \hat{y}\]

By Chain Rule,

\[\dot{\hat{r}}=\dot{\theta}[-\sin (\theta) \hat{x}+\cos (\theta) \hat{y}]=\dot{\theta} \hat{\theta}, \quad \dot{\hat{\theta}}=\dot{\theta}[-\cos (\theta) \hat{x}-\sin (\theta) \hat{y}]=-\dot{\theta} \hat{r} .\]

And

\[\underline{\dot{r}}=\frac{d}{d t}(r \hat{r})=\dot{r} \hat{r}+r \dot{\hat{r}}=\dot{r} \hat{r}+r \dot{\theta} \hat{\theta}\]

Further Differentiation yields:

\[\begin{aligned} \underline{\ddot{r}}=\frac{d}{d t}(\dot{r} \hat{r}+r \dot{\theta} \hat{\theta}) & =\ddot{r} \hat{r}+\dot{r} \dot{\hat{r}}+\dot{r} \dot{\theta} \hat{\theta}+r \ddot{\theta} \hat{\theta}+r \dot{\theta} \dot{\hat{\theta}} \\ & =\ddot{r} \hat{r}+\dot{r} \dot{\theta} \hat{\theta}+\dot{r} \dot{\theta} \hat{\theta}+r \ddot{\theta} \hat{\theta}-r \dot{\theta} \dot{\theta} \hat{r} \\ & =\left(\ddot{r}-r \dot{\theta}^2\right) \hat{r}+(2 \dot{r} \dot{\theta}+r \ddot{\theta}) \hat{\theta} \end{aligned}\] \[\begin{aligned} \hat{F}=\left(\ddot{r}-r \dot{\theta}^2\right) \hat{r}+(2 \dot{r} \dot{\theta}+r \ddot{\theta}) \hat{\theta} \end{aligned}\]

Notice that \(\dot{r}=0\) for a circle (What can we deduce for \(\ddot{r}\)?)

Kepler’s Third Law

• This section is the easiest and all HKDSE students should know by heart.

We can easily deduce Kepler’s Third Law below:

\[\begin{align} F=mr \omega^2 &= \frac{GMm}{r^2} \\ \omega^2 &= \frac{GM}{r^3} \\ \left (\frac{2 \pi}{T} \right)^2 &=\frac{GM}{r^3} \\ \left (\frac{4 {\pi}^2}{T^2} \right) &=\frac{GM}{r^3} \\ T^2&=\frac{4\pi^2}{GM}r^3 \space \space \blacksquare \end{align}\]

Which is Kepler’s third law (in circular motion).

Example 1

Consider Geostationary Satellites. It takes \(24 \text{ hours} =1440\text{ minutes}=86400 \text{ seconds}\) for a complete revolution, hence \(T=86400 \text{s}\). Applying the formula we have \(r^3=\frac{GMT^2} {4\pi^2}\). Hence \(r=(4.22 \times 10^7) \text{ m}\), meaning radius of orbit of geostationary satellite is around \(42200\text{ km}\) from center of earth!

Its velocity is: \(v= r \omega= 3070 ms^{-1}\)

Example 2

Estimate Peter Griffin’s weight:

With \(T=2.75s\) and \(r=1m\), we have \(M=7.82 \times 10^{10} \text{kg}\). Hmmmm

Generalized Case

• This section requires knowledge of ellipses. Standard integration skills (Up to M2) is expected.

Unfortunately the case above only works for circles. Below I will show it works for all ellispes.

Proposition: \(T^2 \propto a^3 \space \forall\) ellipses, where a is the semi-major axis

Proof:

First Note Kepler’s Second Law, the rate at which the area is swept by the Radius Vector (surprising a constant!)

\[\dot{A}=\frac{d A}{d t}=\frac{r(r d \theta) / 2}{d t}=\frac{1}{2} r^2 \dot{\theta}=\frac{h}{2}\]

Next, by substitution and using the results from Area of Ellipse:

\[\Delta t=\frac{A}{\dot{A}}=\frac{\pi a b}{h / 2}=\frac{2 \pi a b}{h}\]

Next, I’ll state the formulae below without proof. Remember that \(\epsilon\) is defined by us so the below are almost definitions

\[a=\frac{l}{1-\epsilon^2}, \quad b=a \sqrt{1-\epsilon^2}=\frac{l}{\sqrt{1-\epsilon^2}}\]

Using \(l=\frac{mh^2}{k}\), we have:

\[\Delta t=\frac{2 \pi a b}{h}=\frac{2 \pi}{h} a^2 \sqrt{1-\epsilon^2} \quad \Longrightarrow \quad(\Delta t)^2=\frac{(2 \pi)^2 a^3}{h^2} l=(2 \pi)^2 a^3 \frac{m}{k}\]

We deduce \((\Delta t)^2 / a^3=(2 \pi)^2 / G M\), which is consistent with the analysis from HKDSE

Orbit Equation:

• More Vector Calculus

We are trying to find what happens if we disturb a orbiting physical system:

From above, we can quickly identify two equations:

\[\begin{aligned} \ddot{r}-r \dot{\theta}^2 & =F_r(r) / m \\ 2 \dot{r} \dot{\theta}+r \ddot{\theta} & =0 . \end{aligned}\]

From the second equation:

\[\frac{1}{r} \frac{d}{d t}\left(r^2 \dot{\theta}\right)=0 \quad \Longrightarrow \quad r^2 \dot{\theta} \equiv h=\text { const }\] \[\underline{L}=\underline{r} \times \underline{p}=m r \hat{r} \times \underline{\dot{r}}=m r \hat{r} \times(\dot{r} \hat{r}+r \dot{\theta} \hat{\theta})=m r \hat{r} \times r \dot{\theta} \hat{\theta}=m r^2 \dot{\theta} \hat{n} \equiv m h \hat{n}\]

Hence:

\[r^2 \dot{\theta} \equiv h=\frac{|\underline{L}|}{m}\]

The smart reader (see if it’s you!) will realize that this means \(L=m r^2 \dot{\theta}\). This is just angular momentum!

Exercise 1

From the First Equation also have:

\[\ddot{r}-\frac{h^2}{r^3}=\frac{F_r(r)}{m}\]

Using substitution, with the fact that \(\ddot{r}-r \dot{\theta}^2 & =F_r(r) / m\), prove that the above holds.

Small Perbuations

• This section requires further knowledge of SHM and Taylor exapansion. If you are a uni student, try to understand it. If you are not, just skip it.

I enjoy the physics of small perbuations, especially in classical mechanics. First, assume a small perbuation around a circle with radius \(a\) i.e. \(r(t)=a+\rho(t)\) with \(\rho(0) \ll a\). substitute into our orbit equation we have:

\[\ddot{\rho}-\frac{h^2}{(a+\rho)^3}=\frac{F_r(a+\rho)}{m}\]

Here we’ll use binomial expansion for negative coefficients. For more details view the wikipedia page for Negative Binomial Distribution

However there’s a nice trick by differentiating the taylor series.

\[\ddot{\rho}-\frac{h^2}{a^3}\left(1-3 \frac{\rho}{a}\right)=\frac{F_r(a)}{m}+\frac{d F_r(a)}{d r} \frac{\rho}{m}\]

Further rearranging will get you:

\[\ddot{\rho}+\left[\frac{3 h^2}{a^4}-\frac{1}{m} \frac{d F_r(a)}{d r}\right] \rho=0\]

Right-Hand-Side surprisingly reduces to 0, can you spot why?

Above describes a simple harmonic oscillator with period \(\omega\), in the form of \(\ddot{\rho}+{\omega}^2 \rho=0\). Now we’ll ensure \(\omega^2>0\), s.t. \(\omega\) is real, i.e.

\[\frac{3 h^2}{a^4}-\frac{1}{m} \frac{d F_r(a)}{d r}>0\]

Hence if the above stands we have:

\[\omega^2 \equiv \frac{3 h^2}{a^4}-\frac{1}{m} \frac{d F_r(a)}{d r}\]

This describes a simple harmonic motion with \(\rho(t)=A \cos (\omega t+\phi)\).

It will look like this:

We have used simple math to analyse the orbit equation! How cool is that!

Eccentricity and Energy

From Wikipedia, we can obtain: \(r(\theta)=\frac{L^2}{m^2 \mu} \frac{1}{1+e \cos \theta}\)

The symbols are different from what we have defined. So we will fix this. \(L\) in wikipedia is defined as \(mr^2\dot{\theta}\), which is equal to \(mh\). Hence \(L^2=m^2h^2\). Then we have:

\[r(\theta)=\frac{h^2}{\mu} \frac{1}{1+e \cos \theta}\]

As \(\mu\) is the constant for which \(\frac{\mu}{r^2}\) equals the acceleration of the smaller body, \(\mu=\frac{k}{m}\). Hence:

\[\begin{equation} r(\theta)=\frac{mh^2}{k} \frac{1}{1+e \cos \theta} \end{equation}\]

Knowing \(E=T+V\), where \(T\) is kinetic energy and \(V\) is potential energy, we can deduce:

\[\begin{equation} E=\frac{m}{2}\left(\dot{r}^2+\frac{h^2}{r^2}\right)-\frac{k}{r} \end{equation}\]

After that it’s just some annoying algebraic manipulations. We will get:

\[E=\frac{mh^2}{2l^2}\left[\epsilon^2-1\right]\]

From this website, we can see:

\[\begin{equation} E = \frac{G^2m^3M^2}{2L^2}\left[\epsilon^2-1\right] \end{equation}\]

We will compare it with our equation: \(l=\frac{mh^2}{k}=\frac{L^2}{mk}\), so \(l^2=\frac{L^4}{m^2k^2}\)

\[E=\frac{m^3h^2k^2}{2L^4}\left[\epsilon^2-1\right]\]

As \(k=GMm\), \(k^2=G^2M^2m^2\). Hence

\[\begin{aligned} E&=\frac{m^3h^2 G^2M^2m^2}{2L^4}\left[\epsilon^2-1\right] \\ &= \frac{m^3h^2 G^2M^2m^2}{2(L^2)(m^2h^2)}\left[\epsilon^2-1\right] \\ &= \frac{m^3 G^2M^2}{2L^2}\left[\epsilon^2-1\right] \\ &= \frac{G^2m^3M^2}{2L^2}\left[\epsilon^2-1\right] &\blacksquare \end{aligned}\]

It is important to note that the eccentricity of a circle is \(0\) and that of a parabola is \(1\). When \(E=0\), it simply means the magnitude of kinetic energy and magnitude of potential energy are the same.

Free Fall to the surface of planets

Part II

Part II deals with Statics and Dynamics, i.e. objects that moves

Gravitational Force

Difficult Question

Two point masses \(m_1\) and \(m_2\) are spaced apart in a vacuum in space, with no other forces present other than the gravitational force between them. At what time will they meet in the center?

Solution

I am heavily inspired by 1 and 2:

With object 1 on the left and object 2 on the right, we can define these below:

\[\begin{aligned} F_{21}&=G \frac{m_1 m_2}{\left(x_2-x_1\right)^2}=m_1 \ddot{x}_1 \\ -F_{21}&=-G \frac{m_1 m_2}{\left(x_2-x_1\right)^2}=m_2 \ddot{x}_2 \end{aligned}\]

Therefore the accelerations are given by:

\(\begin{aligned} G \frac{m_2}{\left(x_2-x_1\right)^2}&=\ddot{x}_1 -G \frac{m_1}{\left(x_2-x_1\right)^2}&=\ddot{x}_2 \end{aligned}\) Subtract these two equations will get us:

\[\ddot{x}_2-\ddot{x}_1=\frac{d^2}{d t^2}\left(x_2-x_1\right)=-G \frac{m_1+m_2}{\left(x_2-x_1\right)^2}\]

Define \(r=x_2-x_1\):

\[\ddot{r}=-G \frac{m_1+m_2}{r^2}\]

Multiply both sides by \(\dot{r}\):

\[\dot{r}\ddot{r}=-G \frac{m_1+m_2}{r^2} \dot{r}\] \[\begin{aligned} \frac{d}{dt}\left(\frac{1}{2}\dot{r}^2\right)&=-G \frac{m_1+m_2}{r^2} \left(\frac{dr}{dt}\right) \\ \int_{0}^{T}\frac{d}{dt}\left(\frac{1}{2}\dot{r}^2\right) dt&= \int_{0}^{T}-G \frac{m_1+m_2}{r^2} \left(\frac{dr}{dt}\right) dt \end{aligned}\]

This is not a calculus article so I won’t discuss the rigor behind next time. However, if you want to be rigorous you can do it yourself!

\[\begin{aligned} \frac{1}{2}\dot{r}^2&= \int_{R}^{r}-G \frac{m_1+m_2}{r^2} dr \\ \left(\frac{dr}{dt}\right)^2&= 2G(m_1+m_2)\left[\frac{1}{r}-\frac{1}{R}\right] \\ \frac{dr}{dt}&=-\sqrt{ 2G(m_1+m_2)\left[\frac{1}{r}-\frac{1}{R}\right]} \end{aligned}\]

Where we take the negative square root as \(v\) is positive

\[\begin{aligned} \frac{dr}{-\sqrt{ 2G(m_1+m_2)\left[\frac{1}{r}-\frac{1}{R}\right]}} &=dt \\ \sqrt{\frac{Rr}{R-r}} dr &=-\sqrt{2G(m_1+m_2)} dt \\ \sqrt{\frac{r}{R-r}} dr &= -\sqrt\frac{R} dt \\ \int_{R}^{0}\sqrt{\frac{r}{R-r}} dr &= -\sqrt\frac{R} T \end{aligned}\]

From Wolframalpha the left integral is equal to \(\frac{-R\pi}{2}\). Hence:

\[\begin{aligned} \frac{-R\pi}{2}&= -\sqrt\frac{R} T \\ T&=\frac{R\pi}{2} \sqrt\frac{2G(m_1+m_2)} \\ &= \frac{\pi}{2} \sqrt{\frac{R^3}{2G(m_1+m_2)}} \end{aligned}\]

(Angular)Momentum Conservation

We deduced that $$r^2 \dot{\theta} \equiv h=\frac{

\underline{L}

}{m}\(, or equivalently\)

\underline{L}

=mr^2 \dot{\theta}\(Combining the fact that\)v=r\dot{\theta}\(we have\)

\underline{L}

=mrv$$

Angular Momentum is something you will encounter while skating (or spinning your chair). If you spread your arms and contract your arms, you will realize that you will spin in a different speed, because angular momentum has to be conserved (assumed no friction)

Compared to \(p=mv\) in HKDSE this rings a bell. However, it is always incorrect to say “angular momentum is converted to linear momentum” because this violates dimensional analysis.

Free Fall of objects

Inspired by a conversation in Shenzhen, I am trying to compare the terminal velocity of a human and a plane under turbulent flow:

Consider \(a=\dot{v(t)}\), we have \(F=m\dot{v}\). Now consider downwards as positive. Assume air resistance is \(kv^2\), where \(k \in \mathbb{R}\). Then we have:

\[\begin{aligned} m\dot{v}&=mg-kv^2 \\ m \frac{dv}{dt}&=mg-kv^2 \\ mdv&=(mg-kv^2)dt \\ \frac{mdv}{mg-kv^2}&=dt \\ \int_{0}^{v} \frac{mdv}{mg-kv^2} &= \int_{0}^{t}dt \\ t&=\int_{0}^{v} \frac{mdv}{mg-kv^2} \end{aligned}\]

Where we assumed \(v(0)=0\)

\[\begin{equation} \int \frac{d v}{m g-k v^2} =\frac{1}{m g} \int\frac{d v}{1-\frac{k}{m g} v^2} \\ \end{equation}\]

Evaluate the indefinite integral first, If we assume \(u=\sqrt\frac{k}{mg} v\), hence \(\sqrt\frac{mg}{k} du=dv\), we have:

\[\begin{aligned} &\frac{1}{\sqrt{m g k}} \int \frac{d u}{1-u^2} \\ & =\frac{1}{\sqrt{m g k}} \tanh ^{-1}(u) \\ &=\frac{1}{\sqrt{m g k}} \tanh ^{-1}\left(v \sqrt{\frac{k}{m g}}\right) \end{aligned}\]

The reader can easily check that the definite integral is the same, i.e.

\[t=\frac{1}{\sqrt{m g k}} \tanh ^{-1}\left(v \sqrt{\frac{k}{m g}}\right)\]

After this we can do some algebraic manipulations to get : \begin{equation} v=\sqrt {\frac{mg}{k}}\tanh[(\sqrt{mgk})(t)] \end{equation}

As \(\tanh(x)\) approaches \(1\), \(v_\text{Terminal}=\sqrt{\frac{mg}{k}}\)

Next, we have to find the value of \(k\): Using drag equation , assuming turbulent flow:

\(F_{\mathrm{d}}=\frac{1}{2} \rho c_{\mathrm{d}} A v^2\) Hence:

\[\begin{aligned} v_\text{Terminal}&=\sqrt{\frac{mg}{k}} \\ &=\sqrt{\frac{2mg}{\rho c_d A}} \end{aligned}\]

As \(m_\text{human}\)«<\(m_\text{plane}\), the difference in \(c_d\) and \(A\) can’t compensate, and hence terminal velocity of planes are much higher during free fall.

Dropping to center of earth

• This section requires the knowledge of advanced vector operations. High school students can skip! How long does it take to drop to the center of earth and what will happen after that? To do this we need to know simple harmonic motion and Gauss’s law for gravitation!

The content for Gauss’s law for gravitation (as below) is heavily inspired (and some parts copied) from here

But first, we need to revise some vector calculus. If you are a mathematician, I highly recommend you to check out this this page written by Toby Lam, on Levi Civita. This is the notation physicists use often.

Theorem

\(\nabla \cdot\left(\frac{\mathbf{r}}{|\mathbf{r}|^3}\right)=4 \pi \delta(\mathbf{r})\)

Proof

As we are dealing with gravitational fields, we shall use cylindrical coordinate system.

Now, let \(\mathbf{A}=\frac{\mathbf{r}}{\mathbf{r}^3}\)

\[\nabla \cdot \mathbf{A}=\frac{1}{r^2} \frac{\partial\left(r^2 A_r\right)}{\partial r}+\frac{1}{r \sin \theta} \frac{\partial}{\partial \theta}\left(A_\theta \sin \theta\right)+\frac{1}{r \sin \theta} \frac{\partial A_{\varphi}}{\partial \varphi}\]

Above is a standard result.

As the field is \(\theta\) and \(\phi\) independent, we have:

\[\nabla \cdot \mathbf{A}=\frac{1}{r^2} \frac{\partial\left(r^2 A_r\right)}{\partial r}\]

Where \(A_r=\frac{1}{r^2}\)

Now, we will develop some theory in vector calculus:

\[\begin{aligned} & \nabla \cdot(\phi \mathbf{u}) \\ & =\partial_i\left([\phi \mathbf{u}]_i\right) \\ & = \partial_i\left(\phi \mathbf{u}_i\right) \\ & =\left(\partial_i \phi\right) \mathbf{u}_i+\phi\left(\partial_i \mathbf{u}_i\right) \\ & =[(\nabla \phi) \cdot \mathbf{u}]_i+[\phi(\nabla \cdot \mathbf{u})]_i \\ & =(\nabla \phi) \cdot \mathbf{u}+\phi(\nabla \cdot \mathbf{u}) \end{aligned}\]

Hence:

\[\nabla \cdot \frac{\mathbf{r}}{|\mathbf{r}|^3}=\left(\frac{1}{r^2}\right)(0)=0\]

Note that this is only valid for \(\mathbf{r} \neq \mathbf{0}\) as we are dividing by 0.

On the other hand, we have:

\[\begin{gathered} \iiint_{\text {volume } \tau} \nabla \cdot\left(\frac{\mathbf{e}_r}{r^2}\right) \mathrm{d} \tau=\iint_{\text {surface enclosing } \tau} \frac{\mathbf{e}_r}{r^2} \cdot \mathbf{e}_r \mathrm{~d} \sigma \\ =\int_{\phi=0}^{2 \pi} \int_{\theta=0}^\pi \frac{1}{r^2} r^2 \sin \theta \mathrm{d} \theta \mathrm{d} \phi=4 \pi \end{gathered}\]

The equation above is copied here

This means at \(\mathbf{r}=\mathbf{0}\), the divergence is equal to \(4 \pi\). Alternatively, we can write:

\[\nabla \cdot\left(\frac{\mathbf{r}}{|\mathbf{r}|^3}\right)=4 \pi \delta(\mathbf{r})\]

Now,

\[\begin{aligned} \nabla \cdot\mathbf{g}(\mathbf{r})&=-G \int \rho(\mathbf{s}) \left [\nabla \cdot\frac{(\mathbf{r}-\mathbf{s})}{|\mathbf{r}-\mathbf{s}|^3} \right] d^3 \mathbf{s} \\ &=-4 \pi G \int \rho(\mathbf{s}) \delta(\mathbf{r}-\mathbf{s}) d^3 \mathbf{s} \end{aligned}\]

Dirac Delta Function: \(\delta(x)\) is defined to be infinitity at \(x=0\), and 0 elsewhere.

Sifting Property:

\[\int_{-\infty}^{\infty} f(t) \delta(t-T) d t=f(T)\]

Using the “sifting property” of the Dirac delta function , we arrive at

\[\nabla \cdot \mathbf{g}(\mathbf{r})=-4 \pi G \rho(\mathbf{r})\]

Divergence Theorem:

\[\iiint_V(\nabla \cdot \mathbf{F}) d V= \oiint_S (\mathbf{F} \cdot \hat{\mathbf{n}}) dS\]

Using Divergence Theorem, we can easily prove that:

\[\oiint_{\partial V} \mathbf{g} \cdot d \mathbf{A}=-4 \pi G M\]

The proof is left as an exercise for the reader (One line argument).

Force acting on the ball:

\[F=G\frac{M'm}{r^2}\]

where:

\[M'=M\frac{r^3}{R^3}\] \[\begin{aligned} m\ddot{r}&=G\frac{Mm}{R^3}r \\ \ddot{r}&=G\frac{M}{R^3}r \end{aligned}\]

Any physics student should recognise this as a simple harmonic motion with frequency:

\[\omega=\sqrt{\frac{GM}{R^3}}\]

This corresponds to a period of:

\[T=2 \pi \sqrt{\frac{R^3}{GM}}\]

Lagrangian For Wedges

I clearly remember doing this 5 years ago and getting quite stuck.

Imagine a wedge of Mass \(M\) and a small particle of Mass \(m\). There is an angle of \(\phi\) between the mass and the wedge. The surface between the block and the surface between the wedge is smooth. Now we put the small particle on the wedge.

Intuitively thinking, the small particle moves down the wedge and the center of mass of the system doesn’t change, so the wedge slides backwards.

Define x along the axis at which the wedge moves and y along the axis at which the particle moves.

\[\begin{aligned} T_{\text {block }} & =\frac{1}{2} M \dot{x}^2 \\ T_{\text {particle }} & =\frac{1}{2} m\left(v_1^2+v_2^2\right)=\frac{1}{2} m\left[(\dot{x}+\dot{y} \cos \phi)^2+(\dot{y} \sin \phi)^2\right] . \end{aligned}\]

For \(T_\text{particle}\), we have transformed our basis vectors.

The potential is:

\[V=-m g y \sin \phi+ c\]

Where \(c \in \mathbb{R}\). \(c\) comes from the GPE of the wedge, which has no time dependence. By convention we choose \(c=0\), hence \(V=-mgy \sin \phi\)

Since \(\mathcal{L}=T-V\)

\[\mathcal{L}=\frac{1}{2} M \dot{x}^2+\frac{1}{2} m\left[\dot{x}^2+\dot{y}^2+2 \dot{x} \dot{y} \cos \phi\right]+m g y \sin \phi\]

Using the Lagrangian equation for \(x\):

\[\begin{aligned} \frac{d}{d t}\left(\frac{\partial \mathcal{L}}{\partial \dot{x}}\right) & =\frac{\partial \mathcal{L}}{\partial x} \\ \Rightarrow \quad \frac{d}{d t}(M \dot{x}+m(\dot{x}+\dot{y} \cos \phi)) & =(M+m) \ddot{x}+m (\cos \phi )\ddot{y}=0 \\ \Rightarrow \ddot{y} &=\frac{-(m+M) \ddot{x}}{m (\cos \phi )} \end{aligned}\]

On the other hand, we have:

\[\frac{d}{d t}\left(\frac{\partial \mathcal{L}}{\partial \dot{y}}\right)=m(\ddot{x} \cos \phi+\ddot{y})\]

From above, we can substitute \(\ddot{y}\):

\[m\left[\cos \phi-\frac{m+M}{m \cos \phi}\right] \ddot{x}=\frac{\partial \mathcal{L}}{\partial y}=m g \sin \phi\]

We have therefore found that the acceleration of the block is

\[\ddot{x}=\frac{g \sin \phi}{\cos \phi-\frac{(m+M)}{(m \cos \phi)}}\]

Hence:

\[\ddot{y}=-\left[\frac{(m+M)}{m (\cos \phi )}\right] \left[\frac{g \sin \phi}{\cos \phi-\frac{(m+M)}{(m \cos \phi)}} \right]\]

This is easier than Newtonian methods because it’s harder to make mistakes. Plus, if we consider friction terms then Newtonian methods can be quite complicated.

The End

This is the end of Article 1! \(\blacksquare\)