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Solutions for class #7 from Yosumism website
Problem 44:
Mechanics }Lagrangians The kinetic energy, in general, is given by. The potential energy is just
. The Lagrangian is given by. Now, given the constraint , one can differentiate it and plug it into the Lagrangian above to reexpress the Lagrangian in terms of just y, for example. Differentiating, one has. Square that to get , where one replaces the through the given relation. Plug that back into the Lagrangian above to get exactly choice (A).
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Problem 45:
Mechanics }Conservation of Energy Conservation of energy gives , where is the velocity of the ball before it strikes the ground. Thus,. Afterward, the ball bounces back up with. Apply conservation of energy again to get . Plugging in , one has , which is choice (D).
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Problem 66:
Mechanics }Work Work is defined by. The force here is just due to gravity, thus , where is the density of the chain. The chain is wound upwards, so work is , as in choice (C). (The approximation is made.)
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Problem 74:
Mechanics }Small Oscillations The small oscillations of the hoop has the same frequency as that of a simple pendulum. Thus,
. However, in this case, is the distance from the center of mass to the oscillation point--- which is just the radius of the loop. Since , the period does not depend on mass. Since. The ratio of periods is. Thus, the period of Y is just . (Note, the technique of leaving out constants requires that 's are
used instead of 's. Practice a few times with this technique, as this will save time on the actual exam.)
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Problem 78:
Mechanics }Multiple Particles The angular momentum equation gives the angular frequency. , which relates the angular momentum to the moment of inertia, the angular velocity, the radius of gyration and the linear velocity. The system spins about its center of mass, which is conserved. Since the pole is massless and the skaters are off the same mass,. The moment of inertia of the system is just . Thus, the angular momentum equation gives , since the cross-products point in the same direction. Now that one has the angular velocity, one eliminates all but choices (B) and (C). Now, take the time-derivative of x for choices (B) and (C), then evaluate it at. For B, one has for. For C, one has for. Since the top skater is initially at , only choice (C) has the right initial condition. Choose choice (C). (FYI: The center of mass velocity is given by. One can also arrive at (C) by noting conservation of center of mass velocity, since there is no net force.)
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Alternatively, one has, for a circular orbit, the equality between centripetal force and the attractive force,. Thus, the kinetic energy is just. Since , where the extra negative sign for the potential energy is due to an attractive potential. Thus, the total energy , which is choice (C). (Alternate solution is due to user crichigno.)
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Problem 93:
Mechanics }Boundary Condition Getting low on time, one should begin scoring points based more of testing strategy than sound rigorous physics. At the initial release point, the acceleration is due to gravity and the tension is 0 (no centripetal acceleration). The only choice that gives is choice (E).
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Problem 100:
Mechanics }Sum of Moments
Take the sum of the moments (or torque) about the triangular pivot fulcrum and set it to 0. , where is the distance from the fulcrum to the 20kg mass, is the distance from the fulcrum to the 40kg mass and is the distance from the fulcrum to the center of mass of the rod. From conservation of length, one also has and. Plug everything into the moment equation. Shake and bake at 300 K. Solve for q to get choice (C). Alternatively, one can solve this problem in one fell swoop. Taking as the distance from the fulcrum to the center of mass of the rod, one sums the moment about the fulcrum to get
. Solve for to get choice (C). (This is due to the user astro_allison.)
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