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Professor Chris Hammel, Ohio State University (OH), Physics, Classical Mechanics, Problem Set 1 Solution, Mechanics, Gravitational Law, inverse square law, radius of the planet, force, heavy object, Gauss Law, electric field, inverse- square law,frictional force, Normal Modes,symmetric mode, Torque, Mass of Earth,Fourier Series, Centripetal Force, Energy, Angular Kinematics,Lagrangians.
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Mechanics }Gravitational Law Recall the famous inverse square law determined almost half a millennium ago, where. The ratio of two inverse- square forces ( , where is the radius of the planet or huge heavy object) would be Thus, , which is choice (C).
Mechanics }Gauss Law The inverse- square law doesn't hold inside the Earth, just like how Coulomb's law doesn't hold inside a solid sphere of uniform charge density. In electrostatics, one can use Gauss Law to determine the electric field inside a uniformly charged sphere. The gravitational version of Gauss Law works similarly in this mechanics question since , where is the mass density of. In short, the gravitational field plays the analogous role here as that of Thus,. So, for , , where one assumes is constant. To express the usual inverse- square law in terms of , one can apply the gravitational Gauss Law again for , Since Therefore, which is choice (C).
Mechanics }Normal Modes For normal mode oscillations, there is always a symmetric mode where the masses move together as if just one mass. There are three degrees of freedom in this system, and ETS is nice enough to supply the test- taker with two of them. Since the symmetric mode frequency is not listed, choose choice it!---as in (A).
Mechanics }Torque The problem wants a negative component for. Recall that whenever and are parallel (or antiparallel). Thus, choices (A), (B), (E) are immediately eliminated. One can work out the cross-product to find that (D) yields a positive , thus (C) must be it.
Mechanics }Mass of Earth If one does not remember the mass of the earth to be on the order of , one might remember the mass of the sun to be. Since the earth weighs much less than that, the answer would have to be either (A) or (B). The problem gives the radius of the earth, and one can assume that the density of the earth is a few thousand and deduce an approximate mass from. The answer comes out to about , which implies that the earth is probably a bit more dense than one's original assumption. In either case, the earth can't be, on average, uniformly dense. Thus (A) is the best (and correct) answer. An alternate solution is provided by the user SlickAce21. Equating the mass of some object with the gravitational force, one has .
Advanced Topics }Fourier Series There's no need to go through the formalism of integrating out the coefficients. One can tell by inspection that the function is odd. Thus, one would use the Fourier sine series. This leaves choices (B) and (A). Choice (A) is trivially zero since for all integer n,. Choice (B) remains.
Mechanics }Angular Kinematics Kinematics with angular quantities is exactly like linear kinematics with (length to angle) (linear acceleration to angular acceleration) (linear velocity to angular velocity) (mass to moment of inertia) (force to torque). Thus, one transforms. Plugging in the given quantities, one gets. The torque is given by , whose magnitude is given by choice (D).
Mechanics }Lagrangians Recall the Lagrangian equations of motion. If then , since its time- derivative is 0. One can relate energy to momentum from elementary considerations by , where L is the kinetic energy. Thus, the generalized momentum defined for a generalized coordinate is just. From the above deductions, the generalized momentum is constant, as in choice (B). (Incidentally, the ignorable or cyclic coordinate would be and not since it does not appear in the Lagrangian.)
