physics-Kleppner and Kollenkow-solutions

physics-Kleppner and Kollenkow-solutions - Physics 141...

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Physics 141 Problem Set 2 Corrected Solutions Nataliya Yufa (2.6) Gravel mixer We want to find ω such that at least for a moment the particles are not stuck to the cylinder. This is most likely to happen at the top, where the normal force is in the direction of gravity. For the particle to leave the cylinder, N=0. Hence the y-component of the total force on the particle is F g + N = F g = mv 2 r . (1) Using the relation v = ωR we obtain mg = m v 2 r = 2 R, (2) ω c = r g R . (3) Thus for all values of ω less than ω c the particles will not be stuck to the walls all the time. (2.9) Particle in a cone The particle happens to be moving in a horizontal circle inside a cone of half-angle θ . We want to find the radius r of the circle in terms of v 0 and g . Consider the x- and y-components of the net force acting on the mass. Since there’s no movement in the y-direction, F y is zero, i.e. F y = N y - mg = 0 N y = mg. (4) From the geometry we find that tan θ = N y N x , hence N x = N y tan θ = mg tan θ . (5) Circular motion in the horizontal direction implies that F x = N x = mg tan θ = mv 2 0 r , (6) and solving for r we have r = v 2 o tan θ g . (7) 1
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Initially, the glass is accelerated by the force of friction due to the tablecloth for some time t max . In this time it will reach velocity v o v o = F f t max m = μgt max . (8) On the other hand, the glass will be slowed down by the friction due to the table (which, because the coefficients of friction happen to be the same for the table and the tablecloth, has the same magnitude as the frictional force of the tablecloth). Therefore, during both the acceleration and the deceleration the glass will travel exactly half of the available distance. We want the glass to slow down in a given distance d/ 2, thus v 0 = q μgd. (9) Combine the two expressions for v 0 to get v 0 = μgt max = q μgd (10) t max = s μgd μg = s d μg = s 0 . 5 ft 0 . 5 · 32 ft/s 2 = 1 4 2 s. (11) (2.25) Shortest possible period Intuitively, the shortest period of rotation will occur when the two solid spheres are as close as possible to each other. Let’s try to prove this. Because gravity is responsible for the circular motion, we have Gm 2 4 r 2 = mv 2 r (12) v = s Gm 4 r . (13) The period is given by T = 2 πr v = 4 π Gm r 3 / 2 . (14) Note that the smaller the radius, the shorter the period. However, we’re limited by the radius of the spheres, as they cannot come closer than their radius R . Hence the shortest possible period would be T = 4 π Gm R 3 / 2 . (2.28) Car going around a bend
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This note was uploaded on 02/03/2012 for the course PHY F1118 taught by Professor Landora during the Spring '11 term at Andhra University.

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physics-Kleppner and Kollenkow-solutions - Physics 141...

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