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Benzoate
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Homework Statement
Calculate the gravitational force betwee two uniform(1-D) rods as shown below.(hint you have to integrate the force over the mass elements of both rods . we did the first integral in class. As before you only have to worry about the componet in the x-direction by symmetry. Demonstrate that , in the limit that the separation of the rods becomes large compared to their lengths, the force the correct limiting form.
Here is a drawing of the two rods on my course webpage. You might have to click on the homework tab of the main page and then proceed to click on assignment three if link directly
fails to link you to the homework page containing the problem
http://courses.ncsu.edu/py411/lec/001/
Homework Equations
principle of superposition for Forces
lambda=delta(m)/L=mass/length
The Attempt at a Solution
General information about the forces between two or more rods
F=lim(as delta(x)==> 0) sigma(-G*m0*lambda*delta(xi) (x-hat)/(x0-xi)^2
F0=\lmoustache{-L/2 , L/2}(-G*m0*lambda(x)*dx/((x0-x)^2)*(x-hat)
since object is uniform, all lambdas are constant and therefore all lambda's equal lambda(0).
F0=integral(from -L/2 to L/2)(-G*lambda(0)*m(0)*(x-hat)/(b-x)^2=-G*m0*lambda(0)(x-hat) *integral(from -L/2 to L/2) dx/(b-x)^2. From u-substitution, I can say that :
u=b-x
du=-dx
F0=-G*m0*lambda(0)*(x-hat)*integral(from b-L/2 to b+L/2)(-du/u^2)=-G*m0*lambda(0)*x-hat[-1/(b-L/2)+1/(b+L/2)]
1/(b+L/2)-((1)/(b-L/2))= -L/(b^2-(L/2)^2)
F0=-G*m0*lambda*L*x-hat/(b^2-(L/2)^2)
in limiting case, as b>>L, then F0 = G*m0*M/b^2. My solution
let M/L=lambda
with cos(alpha)=b/r and r=sqrt(b^2+y^2)
by symmetry, the x component is the only left since force is only x direction.
F=integral(from -L/2 to L/2)dy(-G*m(particle)*lambda*dy/(r^2)*cos(alpha))x-hat=-G*m(particle)*lambda*integral(from -L/2 to L/2)b*dy/r^3=-G*m(particle)*lambda*b*integral( from -L/2 to L/2)
dy/(b^2+y^2)^(1.5)==>
==>G*m(particle)*lambda*b^-1*integral(from -L/2 to L/2)*(dy/b)/(1+(y/b)^2)^(1.5)
using u-substitution , I find that y=b*tan(theta)
dy=b*sec^2(theta)*dtheta
F=-G*m(particle)*lambda*b^-1*integral(from arctan(L/2b) to arctan(-L/2b)*(sec^2(theta)*dtheta/(1+tan^2(theta)= G*m(particle)*lambda*b^-1 integral(arctan(-L/b) to arctan(L/b)) dtheta/sec(theta)=-G*m(particle)*lambda*cos(theta)*dtheta=-G*m(particle)*lambda*b^-1*sin(theta)*integral(arctan)
F=-G*m(particle)*lambda*b^-1 * [(2L/b)/(sqrt(1+(L/b)^2)
F=-G*m(particle)*lambda*L*(1/(b*sqrt(b^2+L^2))
Force between two rods is r=sqrt(b^2+(y'-y)^2)
I'm not sure where to go from here :(
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