# Escape velocity of mass $m$ free to move inside frictionless tube [closed]

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 $$t$$?

• you can use basic calculus , as well as work energy theorem , – maverick Jun 23 at 10:01
• It would appear that you are assuming a vertical axis of rotation. – R.W. Bird Jun 23 at 18:26

Following the suggestion of @maverick I solved the problem using the work-energy theorem using $$\frac{1}{2}mv^2 = \int_{0}^{R} m\omega^2 r \; \rm{d}r$$ $$\implies \frac{1}{2}mv^2 = \frac{1}{2}m\omega^2 R^2 \\ \implies v = \omega R$$
• no since $a = \omega^2r$ so it depends on the mass' position on the tube. – Lorenzo Catani Jun 23 at 12:02