Why do the principles of physics get more complex when the frame of reference undergoes acceleration?

heelallev5
2022-08-10
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asked 2022-09-27

Two frames of references $S$ and ${S}^{\prime}$ have a common origin $O$ and ${S}^{\prime}$ rotates with constant angular velocity $\omega $ with respect to $S$.

A square hoop $ABCD$ is made of fine smooth wire and has side length $2a$. The hoop is horizontal and rotating with constant angular speed $\omega $ about a vertical axis through $A$. A small bead which can slide on the wire is initially at rest at the midpoint of the side $BC$. Choose axes relative to the hoop and let $y$ be the distance of the bead from the vertex $B$ on the side $BC$. Write down the position vector of the bead in your rotating frame. Show that $\ddot{y}-{\omega}^{2}y=0$ using the expression for the acceleration. Hence find the time which the bead takes to reach a vertex $C$.

I showed that $\frac{{d}^{2}\overrightarrow{r}}{d{t}^{2}}=(\frac{{d}^{2}\overrightarrow{r}}{d{t}^{2}}{)}^{\prime}+2\overrightarrow{\omega}\times (\frac{d\overrightarrow{r}}{dt}{)}^{\prime}+\overrightarrow{\omega}\times (\overrightarrow{\omega}\times \overrightarrow{r})$ where ′ indicates that it's done in rotating frame. $\overrightarrow{r}$ is position vector of a point $P$ measured from the origin.

So,

$\overrightarrow{r}=r\mathrm{cos}\theta \overrightarrow{i}+y\overrightarrow{j}$

${\overrightarrow{r}}^{\prime}=(\dot{r}cos\theta -r\dot{\theta}\mathrm{sin}\theta )\overrightarrow{i}+\dot{y}\overrightarrow{j}$

${\overrightarrow{r}}^{\u2033}=(\ddot{r}cos\theta -\dot{r}\dot{\theta}sin\theta -\dot{r}\dot{\theta}\mathrm{sin}\theta -r\ddot{\theta}\mathrm{sin}\theta -r{\dot{\theta}}^{2}cos\theta )\overrightarrow{i}+\ddot{y}\overrightarrow{j}$

$\omega \times {\overrightarrow{r}}^{\prime}=-\omega \dot{y}\overrightarrow{i}+(\omega \dot{r}\mathrm{cos}\theta -\omega r\dot{\theta}\mathrm{sin}\theta )\overrightarrow{j}$

$\overrightarrow{\omega}\times (\overrightarrow{\omega}\times \overrightarrow{r})=-{\omega}^{2}r\mathrm{cos}\theta \overrightarrow{i}-{\omega}^{2}y\overrightarrow{j}$

A square hoop $ABCD$ is made of fine smooth wire and has side length $2a$. The hoop is horizontal and rotating with constant angular speed $\omega $ about a vertical axis through $A$. A small bead which can slide on the wire is initially at rest at the midpoint of the side $BC$. Choose axes relative to the hoop and let $y$ be the distance of the bead from the vertex $B$ on the side $BC$. Write down the position vector of the bead in your rotating frame. Show that $\ddot{y}-{\omega}^{2}y=0$ using the expression for the acceleration. Hence find the time which the bead takes to reach a vertex $C$.

I showed that $\frac{{d}^{2}\overrightarrow{r}}{d{t}^{2}}=(\frac{{d}^{2}\overrightarrow{r}}{d{t}^{2}}{)}^{\prime}+2\overrightarrow{\omega}\times (\frac{d\overrightarrow{r}}{dt}{)}^{\prime}+\overrightarrow{\omega}\times (\overrightarrow{\omega}\times \overrightarrow{r})$ where ′ indicates that it's done in rotating frame. $\overrightarrow{r}$ is position vector of a point $P$ measured from the origin.

So,

$\overrightarrow{r}=r\mathrm{cos}\theta \overrightarrow{i}+y\overrightarrow{j}$

${\overrightarrow{r}}^{\prime}=(\dot{r}cos\theta -r\dot{\theta}\mathrm{sin}\theta )\overrightarrow{i}+\dot{y}\overrightarrow{j}$

${\overrightarrow{r}}^{\u2033}=(\ddot{r}cos\theta -\dot{r}\dot{\theta}sin\theta -\dot{r}\dot{\theta}\mathrm{sin}\theta -r\ddot{\theta}\mathrm{sin}\theta -r{\dot{\theta}}^{2}cos\theta )\overrightarrow{i}+\ddot{y}\overrightarrow{j}$

$\omega \times {\overrightarrow{r}}^{\prime}=-\omega \dot{y}\overrightarrow{i}+(\omega \dot{r}\mathrm{cos}\theta -\omega r\dot{\theta}\mathrm{sin}\theta )\overrightarrow{j}$

$\overrightarrow{\omega}\times (\overrightarrow{\omega}\times \overrightarrow{r})=-{\omega}^{2}r\mathrm{cos}\theta \overrightarrow{i}-{\omega}^{2}y\overrightarrow{j}$

asked 2022-07-14

A puck is placed on a friction less disk that is rotating with angular velocity $\overrightarrow{\omega}$. What is the equation of motion with respect to the rotating frame of the puck?

The contradiction that I can not resolve is that it seems really obvious that the puck will rotate with angular velocity $-\overrightarrow{\omega}$ in the frame of the disk. But if look at the puck from the frame of reference of the disk, we can see that there is a pseudo centrifugal force acting on the puck. This is the only force that acts on the puck in this frame.

Hence, the puck should accelerate radially outward in this frame...

The contradiction that I can not resolve is that it seems really obvious that the puck will rotate with angular velocity $-\overrightarrow{\omega}$ in the frame of the disk. But if look at the puck from the frame of reference of the disk, we can see that there is a pseudo centrifugal force acting on the puck. This is the only force that acts on the puck in this frame.

Hence, the puck should accelerate radially outward in this frame...

asked 2022-05-08

Can the astronaut tell at what speed is he moving? and how do you define speed in that case when you have no outside references. Can you refer to the speed relative to the empty space?

asked 2022-05-10

What is the coordinate in this system then and how do they all connect to each other? I've read the mentioning of an observer and the observer's state of motion and I don't understand how that relates to a frame of reference.

asked 2022-08-19

Does the mass of a moving object in empty space with a constant velocity change within its own frame of reference from its rest mass?

asked 2022-05-20

The Fourier rate equation of heat conduction states:

$\overrightarrow{q}=-k\mathrm{\nabla}T$

But I'm wondering if this is valid in every frame of reference, because the heat flux $\overrightarrow{q}$ does obviously change when the area under consideration is moving. The equation is often applied to moving fluids, and because there is no objective way to determine a non-moving frame of reference in a moving substance, I guess that the equation is valid for every inertial frame of reference, but I just wanted to be sure.

$\overrightarrow{q}=-k\mathrm{\nabla}T$

But I'm wondering if this is valid in every frame of reference, because the heat flux $\overrightarrow{q}$ does obviously change when the area under consideration is moving. The equation is often applied to moving fluids, and because there is no objective way to determine a non-moving frame of reference in a moving substance, I guess that the equation is valid for every inertial frame of reference, but I just wanted to be sure.

asked 2022-08-12

Cosmic microwave background temperature is about $2.7\text{}K$. But what temperature we will measure in space using a simple Kelvin thermometer in the shadow? Can it be lower than $2.7\text{}K$?