2
$\begingroup$

A bead of mass $m$ is moving on the inner surface of a smooth circular cylinder of radius $R$ whose $Oz$-axis is directed vertically downwards.
A bead started to move from $x$-axis with velocity $\vec V$ parallel to $y$-axis.

I used cylindrical coordinates.
After calculation, using Conservation of energy law and Conservation of angular momentum law, I've got this equation $$\dot r^2+\dot z^2-2mgz=0$$

I have to show that $z$ changes as $$z(t)=\frac{gt^2}{2}$$

If I ignore $\dot r^2$, I get it. But I am not sure if I may do that.

Can anyone help me a bit?

$\endgroup$
2
$\begingroup$

The mass is moving on the surface of a cylinder, this is to say, in cylindrical coordinates you used, $r = \sqrt{x^2+y^2}$ does not change along with time, it remains a constant, namely the radius of the cylinder.

Remark: the reason why the mass will be always on the inner surface of this cylinder follows from physics, determined by the initial condition. Ignoring the velocity component in the $z$-direction, the inner surface can provide arbitrary large centripetal radially inward force to sustain the circular motion perpendicular to $z$-axis. Initially the mass is moving tangentially w.r.t. the surface, the magnitude of $x,y$-components of the velocity will remain the same. The trajectory will be a spiral, stretched more and more in the $z$-direction as time goes on.

$\endgroup$
2
  • $\begingroup$ @gov no problem, just in case the problem asks you to justify why the mass will remain on the inner surface, I updated a little bit more argument. $\endgroup$
    – Shuhao Cao
    Aug 20 '13 at 22:45
  • $\begingroup$ Great, I appreciate it! :) $\endgroup$
    – gov
    Aug 20 '13 at 22:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.