Timeline for Comparing Newtons 2nd law and Tsiolkovskys
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
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| Nov 2, 2022 at 19:12 | comment | added | David Hammen | @user253751 Here's a silly example: Consider a long rod sliding horizontally on a frictionless track. Once the rod crosses a certain line, we'll arbitrarily say the part of the rod that has crossed the line as outside the system boundary. The rod is moving, so $v$ is non-zero, and the mass is decreasing thanks to the silly system boundary. Yet the acceleration of any point on the rod that remains within that system boundary is zero despite $\dot m v$ being non-zero. | |
| Nov 2, 2022 at 9:39 | comment | added | David Hammen | @preferred_anon That expression works fine for an object with a fixed mass where $dp/dt = ma$. You do not want to use that $F_\text{net} = dp/dt$ for a variable mass object as this yields wildly different results based on the choice of the inertial frame. Acceleration should be the same in all inertial frames in Newtonian mechanics. Use $F_\text{net} = ma$. Compensating for variable mass (and the center of mass moving within the rocket body) is non-trivial, but $F_\text{net} = ma$ is a good start, much better than $F_\text{net} = dp/dt$. | |
| Nov 2, 2022 at 9:24 | comment | added | preferred_anon | If the mass is not assumed to be constant, isn't this also an incorrect statement of Newton II? I am used to seeing $\sum F = \frac{dp}{dt}$. | |
| Nov 2, 2022 at 1:20 | history | answered | David Hammen | CC BY-SA 4.0 |