# How to identify cosntants of motion/ which constants of motion are independent of mass?

I was asked to identify a constant of motion which does not depend on the mass of the object whilst in contact with the surface.

I found the equation of motion of the material point but I don't know how to identify/find a constant of motion independent of mass? How should I approach the problem?

equations of motion projected on axes:

the answer for "find a constant of motion which does not depend of mass while the point is in contact with the surface of the cone" is the following but I don't understand what it represents:

• what does r^2 times dphi/dt represent? – SoHCahToha Jan 19 '17 at 11:37
• I don't know what "cste" & "selon" are. – JMac Jan 19 '17 at 11:40
• cste means constant and "selon" is french for "with respect to " – SoHCahToha Jan 19 '17 at 11:54

## 1 Answer

The final equation of motion corresponds to the conservation of angular momentum. So $\dot{phi}$ is a first integral. Do you know how to solve the last equation for the first integral? Divide by $m$ and $sin\theta$ and then you can integrate directly to obtain $h=r^2 \dot{\phi}$ where $h$ is a constant of motion. Basically what you are looking for is $\frac{d}{dt}$(some expression)$=0$. Then when you integrate with respect to time you will have a constant on the RHS.

This was tough to write on an iPhone. Good luck!

• Thanks! but how do you integrate it once you divide by msintheta? – SoHCahToha Jan 19 '17 at 12:07
• Can you get to $\ddot{\phi}/\dot{\phi} + 2\dot{r}/r = 0$? – Rumplestillskin Jan 19 '17 at 12:26
• Once you integrate this you'll get $\ln(\dot{\phi}) + 2\ln(r) = \ln(h)$ where $h$ is a constant. Which is equal to $r^2 \dot{\phi}. = h$. – Rumplestillskin Jan 19 '17 at 12:29
• thank you very much! I now know how to look for a constant of motion( look for expressions such that d/dt = 0 and integrate it to get the expression = const ! :) – SoHCahToha Jan 19 '17 at 12:33
• How do you know that the final equation corresponds to the conservation of angular momentum ? the answer is rangularvelocity (m^2 /s) should it be rangularvelocity*m to be the angular momentum? Ahhh or is it because we divided by m at the start ? – SoHCahToha Jan 19 '17 at 16:41