# Homework Help: Problem with a Differential in a Non Inertial System.

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1. Sep 28, 2014

### paalfis

A ball of mass 'm' is inside of a tube that rotates in a horizontal plane around the vertical axis (Drawing a circunference). Attached to the ball (inside of the tube) there is a massless, inextensible rope that goes to the midpoint of the circle described by the rotating tube. The other end of the rope is attached to a hanging block of mass 'M'. Describe the motion of the ball in terms of its position as a function of time.
2. Relevant equations

My non inercial reference system was x parallel to the tube, z perpendicular to the plane (in the opposite direction of the hanging block) and y in the plane but perpendicular to x.
After solving the dynamics of the problem, the differential I got (I think it is correct) was:

m * w2 * x - M*g = (m+M) * x''

where w is the angular speed of the tube and x is the position of the ball along the tube starting from the center of the circumference described by it. x'' is the acceleration of the ball along the tube, which is the same of the acceleration of the hanging block. Of course, the first term is the centrifugal force and the second one is the weight of the block.

3. The attempt at a solution

Now my problem is the next one. When solving this differential (proposing for the homogeneous solution: A * eL*t , and for the particular solution x''=0) I only got a solution where x increases exponential with time, but what about the other case, in which the hanging block is heavier and x decreases with time?

Thanks!

2. Sep 28, 2014

### paalfis

Actually, after putting in the initial conditions (x'(t=0)=0 and x(t=0)=R/2 , where R is the length of the tube) the solution (wich was something like A1*eL*t+A2*E-L*t+particular solution ) has A1=A2= 1/2 * [ (R/2 - [ (M*g)/(m*w2) ] ] which makes sense, it is negative for heavier blocks. Is this correct?

3. Sep 28, 2014

### Orodruin

Staff Emeritus
The equilibrium (x = Mg/(mw^2)) is unstable and regardless of how you deviate from it. Thus, you should expect to get exponentially increasing solutions on both sides. In one case x will grow exponentially and in the other it will decrease exponentially away from the equilibrium point. The only thing is that in the decreasing solution, there is an obvious change of dynamics when the ball reaches x = 0.

4. Sep 28, 2014

### paalfis

Right, right.. Can you please clarify a little bit more about the exponentially decreasing part? do you mean something like -eL*t or something like -e-L*t ?

5. Sep 28, 2014

### paalfis

I would guess you mean -eL*t , right?

6. Sep 28, 2014

### Orodruin

Staff Emeritus
This was what I intended, e.g., when the ball starts out at a radius shorter than the equilibrium point at zero velocity. As you have already discovered, the solutions are given in terms of exponential functions also with exp(-Lt). A more convenient parametrization of these are cosh(Lt) and sinh(Lt).

7. Sep 28, 2014

Thanks!