Mauro Mavrinac, a 47 year old male from Ottawa asks on October 13, 2002,Given: a horizontal rigid tube with one end on an axis at a constant rotation. Insert a solid ball into the tube at the axis end. The ball will accelerate and leave the tube. How do you calculate the velocity of the ball inside the tube (and only relative to the tube) for any point inside the tube if the rotation is constant?
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I presume we are to assume that the ball slides down the tube without friction. The answer is that the ball does not accelerate at all if it starts exactly at the centre of rotation; but if it is pushed a small distance r_0 into the tube before being released, its subsequent motion relative to the tube will be (of course) strictly in the radial (r) direction and the velocity will be given by r_0 omega exp(omega t), where omega is the constant angular velocity, t is time and exp is the exponential function. There are several ways to see why this must be so; the one most people will like is that the "centrifugal force" increases linearly with r and so the acceleration is proportional to how far it's travelled, which always gives an exponential dependence. Of course there is absolutely no such thing as a "centrifugal force", but that doesn't prevent everyone from using it all the time. This is one of the very, very, very few cases (always involving rotating reference frames that we want to pretend aren't rotating) where I can excuse the use of this bogus notion for mnemonic purposes.
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