Gravitational potential energy bead of mass

AI Thread Summary
The discussion revolves around calculating the gravitational potential energy of a bead sliding on a smooth rod between two spheres. The gravitational force acting on the bead is derived using the equation F = 2GMm/(r^2) cos(theta), with theta defined in terms of the bead's position. The potential energy is determined by integrating the force from the bead's initial position at x = 3a to the origin. The participant expresses uncertainty about achieving a potential energy of zero when r equals a, indicating a need for clarification on the calculations. The thread emphasizes the importance of correctly applying gravitational force equations to find potential energy and speed as the bead passes through the origin.
ttja
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Homework Statement


A bead of mass m slides without friction on a smooth rod along the x-axis. The rod is equidistant between two spheres of mass M. The spheres are located at x=0 , y= \pm a.

a. Find the potential energy of the bead.

b. The bead is released at x = 3a with an initial velocity vo toward the origin. Find the speed as it passes through the origin

c. Find the frequency of small oscillations of the bead about the origin.



Homework Equations



F= GMm/(r^2)
U = \intF(r)dr

The Attempt at a Solution



First i found the net gravitational force for the mass m at a point, d, which equals: 2*GMm/(r^2) cos theta.

I thought cos theta to be d/r. Therefore, i have the final equation for force: 2*GMm/(r^2)*(d/r).

d = sqrt(r^2 - a^2)

F = 2*GMm/(r^2)*sqrt(r^2 - a^2)/r

taking the derivative from r to a where the variable is the radius from m to M, i got:

2GMm int( sqrt((r^2 - a^2)/(r^3)), r, r, 0)

= 2GMm*[ arctan( a / sqrt( r^2 - a^2 ))/(2a) - sqrt( r^2 - a^2 )/(2r^2) ]

Now, i don't know if this is even remotely correct, since i cannot get a potential energy of 0 when r = a.

Please help (with part a)
 
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ttja said:
First i found the net gravitational force for the mass m at a point, d, which equals: 2*GMm/(r^2) cos theta.

What is the net force in terms of x? (hint: \cos \theta = \frac{x}{r} = \frac{x}{(x^2+a^2)^{1/2}})

Then calculate:

\int_{x=3a}^{x=0}Fdx

to find the work done by the gravitational force.

AM
 
Thank you Andrew
 
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