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Unformatted text preview: Physics 325 homework 11 - Solutions Homework due on Thursday 2 December 2010 1.) (15 points) Calculate the gravitational potential due to a thin rod of length l and mass M at a distance R from the center of the rod and in a direction perpendicular to the rod. The expression for calculating the gravitational force is the following; F = G m 1 m 2 r 2 This formula is defined for two point-like objects only. In order to calculate a force due to a finite-size object, we must break the object into infinitesimal volumes, calculate the effect of each, then add these together to get the result.. For this reason, let us consider a small line element dx , located at distance x from the center of the rod. It’s mass is given by dm = λdx , where λ is the line mass density: λ = M/L . The potential due to this line element is then dφ =- G dm √ x 2 + R 2 =- G λdx √ x 2 + R 2 where we are considering a unit mass. Adding contributions from all of the pieces, we get φ = Z rod dφ =- Gλ Z L/ 2- L/ 2 1 √ x 2 + R 2 dx Substitute x = R sinh( y ), φ =- GM L Z sinh- 1 ( l 2 R ) sinh- 1 (- l 2 R ) dy =- GM L sinh- 1 ( l 2 R )- sinh- 1 (- l 2 R ) =- 2 GM L sinh- 1 ( l 2 R ) or equivalently, φ =- 2 GM L ln L 2 R + s 1 + L 2 R 2 2.) (20 points) Calculate the gravitational field vector due to a homogeneous cylinder ar exterior points on the axis of the cylinder. Perform the calculation... In both cases, all we need to do is (i) break down the cylinder to infinitesimal volume elements dV , (ii) calculate the force/potential due to each piece, and (iii) sum them up to obtain the net force/potential. The most important thing is the choice of coordinate system; you should pick one that respects the symmetry of the system. Here we have a (homogeneous) cylinder, so cylindrical coordinates are good to work with. We takea (homogeneous) cylinder, so cylindrical coordinates are good to work with....
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