Finding Acceleration in a Rotating Tube

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In a frictionless rotating tube, a cylindrical particle's acceleration can be determined using the equation a = r double dot - r * U^2, where r is the distance from the particle to the tube's center. To find r double dot, a function representing r is needed, which is not provided in the scenario. The tangential velocity is related to angular velocity by v = U * r, allowing for the calculation of centripetal acceleration as a = v^2/r. The formula a = r double dot - r * U^2 can be simplified to a = r * U^2, but r double dot represents the second derivative of r, indicating the particle's acceleration along the tube. Understanding these relationships is crucial for solving the problem effectively.
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Question goes something like:
A cylindrical particle is placed in a certain metre long, frictionless tube, where it can slide back and forth. If the particle is not positioned in the centre of the tube, then as the tube is turning at a constant angular velocity of U, determine the acceleration of the particle along the tube.

I understand from my textbook that the equation to use in such case should be
a_r = r "doubledot" (second derivative w.r.t time) - r x U^2
where r = the distance between the particle to the centre of the tube.

But I do not know how to work out r "doubledot".

Any help is much appreciated.
 
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you need a function that represents the distance r so you can differentiate it twice to get the acceleration (which is r double dot)
 
But I wasn't provided with any sort of function in relation to r. It's just a particle sliding inside a tube. ><
 
ah ok sorry i didnt see the angular velocity bit.

ok

Tangental velocity, v, is equal to angular velocity, w, times radius, r. Ie v/r = w. So if centripetal acceleration, a, is (v^2)/r you should be able to work out a in terms of r and/or w
 
Oh orite.

So what do I do with the formula

a = r double dot - r*w^2?

can I just ignore it and use a = r*w^2 instead?
 
its the same thing :) r double dot is simply the 2nd time derivative of r, which being a spatial value, is acceleration of that value.
 
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