# Rolling body kinematics

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

1. ### prometheusviva voce!Registered Senior Member

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As I said, it's simply a definition. I can choose zero to be anywhere I like.

Sometime. Believe it or not my life does not revolve around this forum.

That's their prerogative. As I said before, I got my inspiration from a book that I linked to previously. Why don't you file an errata with the publisher and see how that goes?

See what a stupid argument that is now?

3. ### TachBannedBanned

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I did not dispute that, I am just pointing out that you get the arc length to be a negative entity when $-\pi <\alpha<0$ with your approach. How do you plan to solve this?

Actually, this is not such a stupid idea. Did you file yours?

5. ### prometheusviva voce!Registered Senior Member

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2,045
Quite obviously you say that the distance from $-\pi$ to 0 is the same as minus the distance from 0 to $-\pi$, because I have defined zero to be in the middle of the trajectory. It's not complicated at all.

Yep. Go for it, and let us know how it goes.

7. ### TachBannedBanned

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Nope, the change (reversal) in traversing parameter space ($[-\pi,0]$ vs $[0,-\pi]$) does not warrant changing (reversing) the sign for s since s is an absolute quantity.

Arc length is a positive entity, by definition. Your solution to your error is to simply stick a minus sign in front of its expression doesn't cut it.

I asked you if you filed your error report, why the deflection? Did you file?

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

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What exactly is your problem? I showed in post 59 that
$ds = 2 R \cos \frac{\alpha}{2} d \alpha$ and hence $s = 2 R \int_{-\pi}^{\pi} \cos \frac{\alpha}{2} d \alpha = 8R$ in exact agreement with AN's post. There is no issue with negative arc length.

9. ### TachBannedBanned

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The problem is that you arrived to $s=4R sin(\frac{\alpha}{2})$ .

10. ### prometheusviva voce!Registered Senior Member

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why is that a problem?

11. ### TachBannedBanned

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We've been over this, for $-\pi <\alpha<0$ your $s$ is negative. Your attempt at reversing the parametrization doesn't fix the problem.

12. ### prometheusviva voce!Registered Senior Member

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Why is this a problem? For your parameterisation s is going to be something like $\cos \frac{\alpha}{2}$ which is negative for $\pi < \alpha <2 \pi$.

13. ### TachBannedBanned

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Nope, $s=4R sin( \theta/2)$ , positive for all $0<\theta< 2 \pi$

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

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$x = R(\alpha- \sin \alpha), \quad y = R(1+\cos \alpha)=2 R \sin^2 \frac{\alpha}{2}$

Equation of motion: $\frac{d^2 s}{dt^2} + \frac{g}{4R} s=0$

Solution subject to the boundary conditions: $s(0) = s_0, \quad \frac{ds}{dt}(0) = 0$ is $s = s_0 \cos(\omega t)$ where $\omega^2 = \frac{g}{4R}$

hence the velocity is $\frac{ds}{dt} = - s_0 \omega \sin (\omega t)$ and the kinetic energy is $KE = \frac{1}{2} m \omega^2 s_0^2 \sin^2 (\omega t)$

The potential energy is $PE = mgy$. Using $y = 2 R \sin^2 \frac{\alpha}{2}$ and $s = 4 R \sin\frac{\alpha}{2}$ we can show $y = \frac{s^2}{8R}$. Hence $PE = \frac{mgs^2}{8R} = \frac{mg s_0^2 \cos^2(\omega t)}{8R}$

Therefore the total energy having subbed for $\omega^2$ is $E = KE + PE = \frac{mgs_0^2}{8R}\left(\sin^2 (\omega t) + \cos^2 (\omega t) \right) = \frac{mgs_0^2}{8R}$ and $\frac{\partial E}{\partial t} = 0$

15. ### prometheusviva voce!Registered Senior Member

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2,045
By my reckoning for $x = R(\alpha - \sin \alpha), \quad y = R(1- \cos \alpha)$ we have $ds = 2 R \sin \frac{\alpha}{2} d\alpha$ and hence $s = -4 R \cos\frac{\alpha}{2}$ which is negative for $\alpha$ between 0 and $\pi$

16. ### TachBannedBanned

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Nope, the starting point is $ds^2=2R^2 (1-cos (\theta))$ .
What does this tell you?

17. ### TachBannedBanned

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Err, you surely meant $y = R(1+\cos \alpha)=2 R \cos^2 \frac{\alpha}{2}$, right?

Kind of oK, so far (modulo the trigonometry flub).

How can that be? At $\alpha=-\pi$ your $y=0$ and $PE=0$ which is clearly not the case.
At the bottom of the trajectory , i.e. at $\alpha=0$ your $PE=2mgR$ , which is clearly not the case.

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

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Yikes! That was a rotten typo. What I actually meant was $x = R (\alpha + \sin \alpha), \quad y = R(1-\cos \alpha)=2 R \sin^2 \frac{\alpha}{2}$ in line with my original solution. That also clears up the potential energy issue.

19. ### prometheusviva voce!Registered Senior Member

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You mean $ds^2=2R^2 (1-cos (\theta))d \theta^2$ which means $ds^2=4R^2 \sin^2 \frac{\theta}{2} d \theta^2 \Rightarrow ds=2R \sin \frac{\theta}{2} d \theta$ and s is as I wrote above.

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Yes.

Not exactly.

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Go on...

22. ### TachBannedBanned

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$ds^2=4R^2 sin^2(\theta/2) d \theta^2 => ds=2R d \theta |sin(\theta/2)|$

..and this should also fix the problems with your derivations.

23. ### prometheusviva voce!Registered Senior Member

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2,045
No...

$ds^2=4R^2 sin^2(\theta/2) d \theta^2 \Rightarrow ds= \pm 2R d \theta sin(\theta/2)$ which means there is always a region where s < 0.

I just did a big lol at this. Right, I think we've established that s can be positive or negative as a function of $\alpha$...