Gravitational potential energy bead of mass

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SUMMARY

The discussion centers on calculating the gravitational potential energy of a bead of mass m sliding on a smooth rod between two spheres of mass M. The potential energy is derived from the gravitational force equation F = GMm/(r^2), leading to the integral U = ∫F(r)dr. The participant also explores the bead's speed as it passes through the origin and the frequency of small oscillations about the origin, emphasizing the need for accurate integration techniques and understanding of gravitational forces in a two-body system.

PREREQUISITES
  • Understanding of gravitational force equations, specifically F = GMm/(r^2)
  • Knowledge of potential energy calculations using integrals
  • Familiarity with the concept of small oscillations in physics
  • Basic trigonometry, particularly the use of cosine in relation to angles and distances
NEXT STEPS
  • Study the derivation of gravitational potential energy in multi-body systems
  • Learn about the principles of small oscillations and their mathematical modeling
  • Explore advanced integration techniques for calculating work done by forces
  • Investigate the effects of initial velocity on motion in gravitational fields
USEFUL FOR

Physics students, educators, and anyone interested in classical mechanics, particularly those studying gravitational interactions and oscillatory motion.

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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