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A mass $m$ is placed inside a frictionless tube of length $R$ which rotates with constant angular velocity $\omega$ around an axis to it perpendicular passing through one of its extremes. The mass begins at rest and accelerates outwards because of the apparent centrifugal force. What is the velocity of the mass once it leaves the tube? I have tried approaching this problem by solving the second-order ODE $$\ddot{x} = \omega^2x$$ which yields $$x(t) = ce^{\omega t}-ce^{-\omega t}$$ and therefore $$v(t) = \dot{x} = c\omega e^{\omega t}+c\omega e^{-\omega t}$$ however I would have to solve $x(t) = R$ and then plug this value of $t$ into $v(t)$ which is a complicated task to do by hand. Is there a better way which doesn't involve finding the value of $t$?

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  • $\begingroup$ you can use basic calculus , as well as work energy theorem , $\endgroup$ – maverick Jun 23 at 10:01
  • $\begingroup$ It would appear that you are assuming a vertical axis of rotation. $\endgroup$ – R.W. Bird Jun 23 at 18:26
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Following the suggestion of @maverick I solved the problem using the work-energy theorem using $$\frac{1}{2}mv^2 = \int_{0}^{R} m\omega^2 r \; \rm{d}r$$ $$\implies \frac{1}{2}mv^2 = \frac{1}{2}m\omega^2 R^2 \\ \implies v = \omega R $$

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  • $\begingroup$ You have found the radial component of the exit velocity (parallel to the tube). There is also a tangential component (ωR) caused by the normal force within the tube. $\endgroup$ – R.W. Bird Jun 23 at 18:32
  • $\begingroup$ But is there a way to find the modulus of the normal force? $\endgroup$ – Lorenzo Catani Jun 23 at 22:38
  • $\begingroup$ You don't need a modulus. You know that the tangential velocity of the mass at the end of the tube is the same as that of the end of the tube. $\endgroup$ – R.W. Bird Jun 24 at 13:53
  • $\begingroup$ Upon further thought, it appears that I have introduces some confusion. If you work with a pseudo force in a rotating coordinate system, there is no velocity at a right angle to the rotating radius. If you stay in the inertial frame, only the normal force (which can be shown to be = mrω^2) does work. In this case increasing the tangential velocity. An unbounded mass moving tangentially increases its distance from the instantaneous axis of rotation, so the mass moves outward along the tube. The mass leaves the tube tangentially and continues to increase its distance from the axis. $\endgroup$ – R.W. Bird Jun 24 at 18:08
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If you could ignore direction, this would just be a constant acceleration for a known time.

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  • $\begingroup$ no since $a = \omega^2r$ so it depends on the mass' position on the tube. $\endgroup$ – Lorenzo Catani Jun 23 at 12:02
  • $\begingroup$ Divide position into distance and direction. solve only for distance. Does that work? $\endgroup$ – J Thomas Jun 24 at 6:41

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