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Solutions for class #11 from Yosumism website
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Problem 5:
Mechanics }Centripetal Force acts in the direction as shown and the centripetal acceleration acts in the direction of. Centripetal acceleration is a net force, however, and thus, is in the positive direction and is in the negative direction. Thus the force of the road is.
YOUR NOTES:
Problem 6:
Mechanics }Inclined Plane Set up the usual coordinate system with horizontal axis parallel to incline surface. The equations are, (since the mass slides down at constant speed), Friction is given by , where the normal force is determined from the equation. For constant velocity one also has, To find the work done by friction, one calculates , where. Thus , as in the almost-too-trivial, but right, choice (B).
Problem 7:
Mechanics }Elastic Collisions One determines the velocity of impact of the ball from conservation of energy, Conservation of momentum gives, Conservation of kinetic energy gives, Plug in the momentum and kinetic energy conservation equations to solve for and in terms of to get Write yet another conservation of energy equation for the final energy, where the condition that the mass slides on a frictionless plane is used. Thus, , where the previous result and is used.
Problem 21:
Mechanics }Moment of Inertia To solve this problem, one should remember the parallel axis equation to calculate the moment of inertia about one end of the hoop: , where is the distance from the pivot point to the center of mass, which in this problem, is just equal to. (In the last equality, note that the moment of inertia of a hoop of radius R and mass m about its center of mass is just .) The problem gives the period of a physical pendulum as. Thus, plugging in the above result for the moment of inertia, one has, , which is closest to choice (C). (Since was rounded to 3, the period should be slightly longer than 1.2s.)
Problem 22:
Mechanics }Geometry The harder part of this problem involves determining the radius of Mars. It's an approximate geometry problem. The problem gives a vertical drop of 2m for every 2600m tangent to the surface. The tangent to the surface is approximately one leg of a triangle whose hypotenuse is the radius of Mars, since the radius is much larger than the tangent distance. The other leg of the right triangle is just , where is the radius of Mars. In equation form, what was just said becomes. The square terms cancel out, and dropping out the 4, one has . (The above deduction was due to Ayanangsha Sen.) The easier part comes in the final half of the problem: applying the centripetal force to the force of gravity. , which is closest to , as in choice (C).
YOUR NOTES:
Problem 23:
Mechanics }Stability of Orbits The gravitational force suspect to a bit of perturbation is given as . One can narrow down most choices by recalling some basic facts from central force theory: (A) No mention is made of frictional effects, and thus energy should be conserved. (B) Angular momentum is always conserved since the net torque is 0 (to wit: the force and moment arm are parallel). (C) This is just Kepler's Third Law applied to this force. (Recall the following bromide: The square of the period is equal to the cube of the radius---for the inverse square law force. For a perturbed force, the bromide becomes: The square of the period is equal to the power of the radius.) (D) Recall Bertrand's Theorem from Goldstein. Stable non-circular orbits can only occur for the simple harmonic potential and the inverse-square law force. This is of neither form, and thus this choice is FALSE.
Problem 31:
Mechanics }Frictional Force (A) A falling object experiencing friction falls faster and faster until it reaches a terminal speed. Its kinetic energy increases proportional to the square of the velocity and approaches a asymptotic value. (B) The kinetic energy increases to a maximum, but it does not decrease to 0. See (A). (C) The maximal speed is the terminal speed. (D) One has the equation. Without having to solve for , one can tell by inspection that will depend on both and. (E) See (D). This is the remaining choice, and it's right.
YOUR NOTES:
Problem 32:
Mechanics }Moment of Inertia The inertia through the point A is. From geometry, one deduces that the distance between each mass and the centerpoint A is. T he moment of inertia about A is thus The inertia about point B can be obtained from the parallel axis theorem ( , where d is the displaced distance from the center of mass). Because , one has. Since the angular velocity is the same for both kinetic energies, recalling the relation for kinetic energy , one has , as in choice (B).
Problem 44:
Mechanics }Chain Rule Recall that. , as in choice (A).
YOUR NOTES:
Problem 65:
Mechanics }Conservation Laws From conservation of momentum, one has. The man does work on both himself and the boat. Thus, the work-kinetic energy theorem has
