Timeline for Doubt in fictitious forces chapter in Morin
Current License: CC BY-SA 4.0
5 events
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| Sep 25, 2023 at 13:26 | comment | added | Neeladri Reddy | Thanks for this example @QPhysl. I see your point - I was saying $\frac{d\mathbf{r}}{dt}$ can be the velocity in the new frame, but you easily disproved it by showing how when $\frac{d\mathbf{r^{'}}}{dt}$ = $\frac{d\mathbf{r}}{dt}$ - clearly the velocity in the non-inertial frame isn't zero. I came up with an example of the ball rotating in an orbit, so velocity in the non-inertial frame would be zero(orbit angular velocity equal to the rotational frame's), but clearly then $\frac{d\mathbf{r}}{dt}$ is not zero. So rotation needs to be taken into account. | |
| Sep 24, 2023 at 14:25 | comment | added | QPhysl | Emphasizing what @NeeladriReddy commented, if we set $g_a=0$ we have this. An inertial observer sees the ball doesn't move, since it feels no force, but in the rotating frame, we (at the green point) see it rotating (because our frame is rotating). If the rotating observer believes they are at rest (the same way we don't feel the rotation of the earth) then since the object is seen rotating, the rotating observer must (falsely) conclude a force acts on the ball (hence the introduction of the fictitious forces). | |
| Sep 23, 2023 at 18:40 | comment | added | Neeladri Reddy | $\frac{d\mathbf{r}(t)}{dt} = (\frac{dr_x}{dt}\hat{x} + \frac{dr_y}{dt}\hat{y} + \frac{dr_z}{dt}\hat{z}) + (r_x\frac{d\hat{x}}{dt} + r_y\frac{d\hat{y}}{dt} + r_z\frac{d\hat{z}}{dt} )$ The first term here is called the velocity wrt to the asteroid frame, while I say the total can also be the velocity, and we can just project that onto the axes of the asteroid frame. | |
| Sep 23, 2023 at 18:32 | comment | added | Neeladri Reddy | Thank you for taking the time to write the answer. But I did understand that the unit vectors on the asteroid change, and we need to use the chain rule. My question is why $\frac{d\mathbf{r}(t)}{dt}$ can't be found, and then projected on to the asteroid axes? The book instead finds $\frac{d\mathbf{r}(t)}{dt}$, but separates it into two parts - with the first part being called the velocity wrt the asteroid. Further in the next comment, as character limit being near. | |
| Sep 23, 2023 at 18:02 | history | answered | QPhysl | CC BY-SA 4.0 |