Physics

October 29, 2023

Fundamentals of physics #

https://www.youtube.com/playlist?list=PLFE3074A4CB751B2B

Summary of lecture 1 - 10.

Newton’s 3 laws #

1. You need a force to change a velocity. Whereas aristoteles would have said, you need a force to move an object.
2. \( F = ma \) This doesn’t survive QM. You need an inertial observer. (ie, one who has no accerelation)
3. \( F_{12} = - F_{21}\)

How we can remove time from the equation. #

\begin{align*} & \frac{dv}{dt} = a \\ & v \frac{dv}{dt} = a v \\ & \frac{d}{d t}\left(\frac{v^2}{2}\right)=a \frac{d x}{d t} \\ & d\left(\frac{v^2}{2}\right)=a d x \\ & \int d \left(\frac{v^2}{2}\right)=\int a d x \\ & \frac{v^2}{2}-\frac{v_0^2}{2}=a\left(x-x_0\right) \\ \end{align*}

Conservation of energy #

Energy is work done over distance.

\begin{align*} & \frac{v^2}{2}-\frac{v_0^2}{2}=a d \\ & \frac{v^2}{2}-\frac{v_0^2}{2}=\frac{F}{m} d \\ & m\frac{v^2}{2}- m\frac{v_0^2}{2}=F d = W \text{and we name } m\frac{v^2}{2} \text{ kinetic energy}\\ & \sum{\Delta K} = \sum{F \Delta x} \\ & K_{2} - K_{1} = \int_{x_{1}}^{x_{2}} F(x) dx = G_{2} - G_{1} \text{where } \frac{d G}{d x} = F \\ & m g h_{1} - m g h_{2} \end{align*}

Circular motion #

We are considering a motion where rotation center is fixed. Then, analogue of distance in linear motion is the angle in the circular motion. After all, radian is a unit of distance (in unit circle).

\begin{alignat*}{2} v = \frac{d x}{d t} &\quad& w = \frac{d \theta}{d t} \\ a = \frac{d v}{d t} &\quad& \alpha = \frac{d w}{d t} \\ \end{alignat*}

\begin{align*} & \vec{R}=R(\vec{i} \cos \omega t+\vec{\jmath} \sin \omega t) \\ & \frac{d \vec{R}}{d t}=R(-\omega \vec{i} \sin \omega t+ \omega \vec{\jmath} \cos \omega t) \\ & \frac{d^{2} \vec{R}}{d t^{2}}=R(-\omega^{2} \vec{i} \cos \omega t - \omega^{2} \vec{\jmath} \sin \omega t) \\ & = -\omega^{2} \vec{R} \end{align*}

Torque #

We have an analogue of \( F = m a\) in a circular motion.

Kinetic energy of an object with circular motion is \[ K = \frac{1}{2} m v^{2 }= \frac{1}{2} m r^{2} w^{2} = \frac{1}{2} I w^{2} \] And we define \( I = m r^{2} \)

We can expand our anlaogy table

\begin{alignat*}{2} m &\quad& I \\ p = m v &\quad& w L = I w \\ F = m a &\quad& ? = I \alpha \\ \end{alignat*}

Note, we have only used \( F = m a \) and \( W \) upto now, and made an anlogy. Where newton’s law would have described the circular motion with constantly changing velocity. We now have an expression where the same circular motion is expressed with the constant angular velocity.

I.e. we have two systems where a term ‘changing’ converted to ‘constant’ and vice versa.

Now we look for “?” in \( ? = I \alpha\). Kinetic energy is conserved for the rotating object.

\begin{align*} & \Delta W = F r \Delta \theta = \Delta K \\ & \Delta K = \frac{1}{2} I (W^{2} - W_{0}^{2}) = \frac{1}{2} I 2 \alpha \Delta \theta \\ & F r \Delta \theta = \frac{1}{2} I 2 \alpha \Delta \theta \\ & F r = I \alpha \end{align*}

Conservation of angular momentum #

Without external torque, angular momentum is preserved.

\begin{alignat*}{2} p = m v &\quad& L = I w \\ \frac{d p}{d t} = 0 &\quad& \frac{d L}{d t} = 0 \\ m v_{1} = m v_{2} && I w_{1} = I w_{2} \\ \end{alignat*}

Center of Mass #

\[ m_{1} \ddot{x_{1}} + m_{2} \ddot{x_{2}} = F_{1e} + F_{2e} + F_{12} + F_{21} = F_{e} \] \[ M \frac{m_{1} \ddot{x_{1}} + m_{2} \ddot{x_{2}} }{M} = F_{1e} + F_{2e} = F_{e} \]

With \( X = \frac{m_{1} x_{1} + m_{2} x_{2}}{M} \), we can think of external force \( F_{e} \) is acting on the center of mass \( X\)

\[ M \ddot{X} = F_{e} \]

Question #

Newton’s 2nd law is a conjecture. and testing it needs to use the conjecture as well. If we use spring as discussed in the lecture to get the \(F = -kx\), deriving it (by experimentation) would also have used the law.

So using the two equation as in the lecture beats the purpose of showing \( F = ma \), in a sense that we don’t know what the true \( F \) is for a string.

\begin{align*} F &= ma F &= -kx \end{align*}

What it achieved is that, the \( F = ma \) is consistent in describing the motion if we accept the equation as the law.

Question #

To make an object to orbit, we use \( ma = m \frac{v^2}{R} \), so how do we make sure the calculation is exact in a non idealized (with friction) environment? Do the object has a way of changing its verlocity when needed?

Question #

in inelastic collision, energy might be lost and conservation of energy might not hold. But momentum is always preserved (when there’s no external net force). Why? well that’s what Newton’s 1st law is saying. The velocity doesn’t change. So 1st law survives. What happens to internal forces and their effect? they all cancel out.

Question #

conservation of linear momentum says \( mv \) is constant conservation of angular momentum says \( Iw \) is constant But \( Iw = m r^2 * \frac{v}{r} = m r v\), so it says a different quantity is constant.

Now if i were, given an instantaneous information of m and v, I can tell its linear momentum, but I can’t tell the angular momentum unless I need a frame of reference (whree the rotation axis is).

So it feels strange to have two laws where one says \( mv \) is constant, and another \( m r v \) is constant, and this two statement can’t be true at the same time, if the first statement is also conditional statement.

It turns out, you need also need a frame of reference for the linear momentum. You need an inertial frame of reference for linear momentum. You need an rotaional(?) frame of reference for angular momentum.

Also, as pointed out later, \( m r v \sin\theta \) is conserved.

And so it happens there are deeper explanation which I don’t yet understand: Noether’s theorem. The conservation of linear momentum corresponds to the translational symmetry of space, while the conservation of angular momentum corresponds to its rotational symmetry

Question #

newton’s first law says linear momentum is preserved.

Whereas conservation of angular momentum says the angular momentum is conserved unless there’s no external torque. But then, the rotating object is changing its velocity and according to newton’s 2nd law, it needs a force, so what’s going on?

Angular motion is another way of saying there’s a force acting perpendicular to the motion. So, angular motion and the related law is describing an object where a there is a perpendicular force acting on it. And this centripetal force does change the velocity of the object.

So we are considering this special situation where perpendicular force is acting on an object, and linear velocity is changing (either only direction or both direction and magnitude). But since this force is perpendicular, the perpendicular velocity isn’t changing and consequently perpendicular momentum which is angular momentum isnt’s changing.

It’s easy to see when center of rotation is fixed, can this thought be still regarded true when center of rotation is changing by attributing the change to linear motion so as to keep the r in \(m r v \) to remain constant?

on to next question

Question #

A thought experiment where the mass of the Sun changes but the gravitational force exerted on the earth remains exactly the same due to a corresponding change in the distance between the sun and earth.

If sun changes its mass and distance but keep the gravitational force equal, would earth move to a different path than it would have moved?

Put it differently, does the same force (with same magnitude and same direction) on earth would change the course of Earth’s movement when the force is exerted from different distance d1 and d2?

I’d like to guess it doesn’t change the path, and that was the core idea of understanding angular momentum and torque, but I can’t yet prove it by equation. (My thought experiment applies same force so earth has to move the same, that’s the argument)

Question #

If the answer to above question is yes, can we talk about torque without thinking of the axis of rotation? \( r \) If there’s no axis of rotation, we can’t talk about rotation, so torque is not defined?

1. if there’s no rotation, you can take any point as axis of rotation.

   \[ \sum F_{i} (x_{i} +a ) = \sum {F_{i} x_{i}} + a \sum F_{i} \] i.e. you can translate by any a, when \( \sum F_{i} = 0 \) and \( \sum F_{i} x_{i} = 0 \)

2. angular momentum can be defined when there’s no rotation.

   \begin{align*} & \vec{L} = \vec{r} \times \vec{p} \\ & \frac{d \vec{L}}{d t} = \frac{d \vec{r}}{d t} \times \vec{p} + \vec{r} \times \frac{d \vec{p}}{d t} = \vec{r} \times \frac{d \vec{p}}{d t} \\ & L = | \vec{r} * \vec{p} | = r p \sin \theta \end{align*}

statement #

(not proven) Angular momentum conservation is achieved, only when internal force is exerted in the line joining the two objects. All the natural forces have this property.