# Rolling body kinematics

Discussion in 'Physics & Math' started by arfa brane, Nov 26, 2011.

1. ### prometheusviva voce!Registered Senior Member

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Seriously, this is very funny indeed.

3. ### TachBannedBanned

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I never participated in this nonsense, s is always increasing monotonically from 0. This is by the very definition of arc length.

5. ### TachBannedBanned

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Flubs happen to the very best.

Last edited: Dec 9, 2011

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Why?

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Wrong post.

9. ### prometheusviva voce!Registered Senior Member

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You have again forgotten that $\alpha$ is a parameter.

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ok.

11. ### TachBannedBanned

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No, I haven't , while $\alpha$ increases monotonically, so does $s$ except that $s$ must be positive at all times. By definition.

12. ### prometheusviva voce!Registered Senior Member

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Ok, I understand how this works now. What you are calling the arc length s is not actually the length of an arc. It is how s varies as a function of the cycloid generating angle $\alpha$. The actual arc length is given by $\int_{s_0}^{s_1} ds = 2R \int_{\alpha_0}^{\alpha_1} d\alpha \sin \frac{\alpha}{2}$ with your parametrisation of the cycloid. It's basically the difference between definite and indefinite integrals, so while I agree that $2R \int_{\alpha_0}^{\alpha_1} d\alpha \sin \frac{\alpha}{2}$ will be a monotonically increasing function of $\alpha_1$ for fixed $\alpha_0$ within the bound $\left[0, 2\pi \right]$, I do not agree that $2R \int d\alpha \sin \frac{\alpha}{2}$ will be a monotonically increasing function of $\alpha$.

Consider an example where we take $\alpha_0 = 0$ - evaluating the integral gives an arc length of $8 R \sin^2 \left(\frac{\alpha_1}{4}\right)$ which is a monotonically increasing function of $\alpha_1$ in the appropriate range. Presumably for my method the same technique will apply with the different range.

13. ### TachBannedBanned

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Correction: $\int_{s_0}^{s_1} ds = 2R \int_{\alpha_0}^{\alpha_1} d\alpha | \sin \frac{\alpha}{2} |$.

I already pointed that out to you.

14. ### prometheusviva voce!Registered Senior Member

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You've just stuck the absolute value signs in by hand with no justification. If you consider the bob in the appropriate range (as you should for a pendulum / track) then the absolute value is unnecessary.

15. ### TachBannedBanned

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No, it is the same exact reason why you had to stick the minus sign in the LHS of the equation of motion $-mg \sin\frac{\alpha}{2} = m \frac{d^2 s}{dt^2}$.

For $-\pi< \alpha <0$ the LHS must be positive (gravitation accelerates the pendulum bob).
For $0< \alpha <\pi$ the LHS is negative (gravitation slows down the pendulum bob)

This is how I knew instantly that your expression for s is not correct.

16. ### prometheusviva voce!Registered Senior Member

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The bob is undergoing SHM, therefore the sign of the force must be negative as it is a restoring force. Face it Tach, I've provided an analysis you didn't think of and you're butthurt by that.

You still haven't justified the presence of the absolute value function (which is not actually necessary anyway).

17. ### TachBannedBanned

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I don't think that you understood what I told you, the minus sign needs to be put in by hand in the interval $[-\pi,0]$ in order to make the force positive, as it should be, since it works "with" the pendulum.
On the interval $[0,\pi]$ the minus sign signifies that the gravitation works "against" the pendulum.

You need to do the same exact thing to $ds$ in order to get s to be a positive quantity, independent of the interval defining $\alpha$

Getting personal and insulting only shows that you don't understand what you've been told <shrug>.
Besides, proof that I understood your analysis is that I pointed out your mistakes within minutes you posted it.

Last edited: Dec 9, 2011
18. ### prometheusviva voce!Registered Senior Member

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It clearly doesn't because the sine function is odd, so $\sin (-x) = - sin(x)$. For $\alpha < 0$ the LHS of that equation is positive without the need for an absolute value.

19. ### prometheusviva voce!Registered Senior Member

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As I explained before, $F(\alpha) = \int f(\alpha) d\alpha$ is not required to be monotonically increasing. $\mathcal{F}(\alpha_1,\alpha_0) = \int_{\alpha_0}^{\alpha_1} f(\alpha) d\alpha, \quad \alpha_0<\alpha_1$ is.

20. ### TachBannedBanned

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You aren't paying attention to what I've been telling you. You need the minus sign for different reasons on each interval.

21. ### prometheusviva voce!Registered Senior Member

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I don't see why that would be.

22. ### prometheusviva voce!Registered Senior Member

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I missed this earlier.

What signs you get in various places in the parametric equations depend on what you want the range of your parameter to be and whether you want your cycloid to be right way up or upside down. I think all permutations of signs in front of the trig functions and outside the brackets will give you various cycloids, as opposed to something that's not a cycloid.

23. ### TachBannedBanned

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Let me help you. You wrote the e.o.m for the ascending part of the pendulum. The LHS has a minus sign because gravitation works in the opposite direction of the swing.
Try writing the e.o.m for the descending part of the pendulum and you will understand why you need to stick in a minus sign by hand in the LHS.