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
Messages
9
Reaction score
0

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)
 
Physics news on Phys.org
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
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top