1

This question struck me a few minutes back, I was at a table with a pear. It was more narrow than round.I proceeded to rotate this pear in one swift movement. It rotated for a few seconds, and suddenly changed its orientation so that its axis of rotation was "vertical" instead of horizontal, and it stayed vertical till it stopped rotating. The same repeated every single time I rotated the object. but oddly this did not work when the initial force I applied was a small one. Does this happen with all objects? Why do objects tend to change their axis of rotation?

.Rudimentary Illustration.

Hritik Narayan
  • 7,318
  • 3
  • 33
  • 46
  • 3
    For the Dzhanibekov effect, the tennis racquet theorem, and the intermediate axis theorem, see e.g. http://physics.stackexchange.com/q/17504/2451, http://physics.stackexchange.com/q/34364/2451, http://physics.stackexchange.com/q/67957/2451, and links therein. – Qmechanic Nov 30 '14 at 13:35
  • You can buy a cute wooden top which spins, precesses, and then flips over to spin on the opposite end of the same axis. The answer lies in ugly equations of angular momentum in 3 dimensions. Thankfully I finished my schooling long ago and don't have to try to solve these situations any more :-) – Carl Witthoft Nov 30 '14 at 13:36
  • Is there any non mathematical explanation for this? (I'd work the mathematical ones but I prefer the earlier :D ) – Hritik Narayan Nov 30 '14 at 13:38

1 Answers1

-1

If we look at the symmetry of the pear of mass $m$, we can see on the x axis that momentum is conserved because the derivative of its motion in space $x$ is the same from any point we measure it's motion from:

$$p_x=m \frac{dx}{dt}$$

On the y axis however, the system is affected by an external field gravity, $g$ which pulls the pear downwards. This removes symmetry from the vertical axis and means that vertical momentum is not conserved.

The moment of inertia of m is given by:

$$I=mr^2$$

Where $r$ is the radius. Here, the pear itself also does not have an evenly distributed radius. Its center of mass in all 3 dimensions is given by:

$$x_{cm}\frac{1}{M}\int{x dm}$$

$$y_{cm}\frac{1}{M}\int{y dm}$$

But we already know that $p_y=m \frac{dy}{dt}$ is not conserved. Where have an axis of rotation when we start spinning the pear, because the pear has a momentum that is not conserved in the vertical direction this pulls the pear downward.

Magpie
  • 537