Timeline for Does car have more kinetic energy when turning?
Current License: CC BY-SA 3.0
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| Apr 5, 2014 at 16:08 | comment | added | DarioP | @user2174870 keep in mind that this effect is quite tiny in a car: the rotation energy around the centre of curvature is a factor $(R/r)^2$ bigger than the one around its centre of mass where $R$ is the radius of the curvature and $r$ is the average size of the car. This effect can be seen much better when an ice skater jumps and start turning very fast on himself: his linear speed goes down. | |
| Apr 5, 2014 at 10:40 | comment | added | user2174870 | Love you, my hero, you saved my day!!! That what I meant when I asked the question | |
| Apr 5, 2014 at 10:28 | vote | accept | user2174870 | ||
| Apr 4, 2014 at 16:43 | history | edited | Tobias | CC BY-SA 3.0 |
added 165 characters in body
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| Apr 4, 2014 at 15:47 | comment | added | DarioP | What you did is correct for sure, I was just saying that if $R$ is kept through the computation the final result becomes more interesting. | |
| Apr 4, 2014 at 15:41 | comment | added | Tobias | @DarioP In the example we have $R=1$ and if we omit pointwise evaluation of $H$ we have $H^2=H$. In fact, I wanted to use the definition $$H(s) = \begin{cases}0&@ s<0\\ 1 &@ s\geq 0\end{cases}$$ at first but adopted then en.wikipedia.org/wiki/Heaviside_step_function. In the distributional sense they are the same (modification on zero measure). What counts is the limit for the dirac impulse and that may vary for a product with an essentially discontinous function. One could interprete $\dot s$ there as multivalued. The right value must be selected by some other criterium (e.g. energy). | |
| Apr 4, 2014 at 14:28 | comment | added | DarioP | I deleted it because I realised what you were doing. It is correct but I think you misunderstood the question. However in the Lagrangian I would put $(RH)^2$ instead of $H$. I found it interesting to propagate the contribution of the bending radius. | |
| Apr 4, 2014 at 14:09 | comment | added | Tobias | @DanioP Why did your comment disappear? Now, my previous comment is out of context. | |
| Apr 4, 2014 at 14:02 | comment | added | Tobias | @DanioP I interprete $J$ as moment of inertia of the car for rotations about the center of mass of the car. In this case my way is correct. If you model the car as point mass you get $J=0$. Consider a heavy Lorry... It will have a moment of inertia w.r.t. the center of mass. | |
| Apr 4, 2014 at 13:08 | history | answered | Tobias | CC BY-SA 3.0 |