One end of a light spring force constant 100 N/m is attached to a vertical wall.

Question

One end of a light spring force constant 100 N/m is attached to a vertical wall. A light string is tied to other end of the horizontal spring. The string changes from horizontal to vertical as it passes over a solid pulley of diameter 4.00 cm. The pulley is free to turn on a fixed, smooth axle. The vertical section of the string supports a 200-g object. The string does not slip at its contact with the pulley. Find the frequency of oscillation of the object, assuming the mass of the pulley is (a) negligible, (b) 250 g, and (c) 750 g. I don't know how to start this problem, I need to know step by step. The most broke down solution would be very helpful.

Explanation / Answer

the velocity of the center of mass when the string is rollingover the pulley is vcm= [2gh/1 + (Icm/MR^2)]^(1/2) assuming that the pulley is in the form of a solid disk,themoment of inertia is Icm= (MR^2/4) or vcm= [2gh/1 + (Icm/MR^2)]^(1/2) or vcm= [2gh/1 + ((MR^2/4)/MR^2)]^(1/2) =(8gh/5)^(1/2) ----------(1) the distance through which the object moves down is F = k * h or (m + m1) * g = k * h or h = [(m + m1) * g/k] ------------(2) from (1) and (2) vcm= (8gh/5)^(1/2) = (8g[(m + m1) * g/k]/5)^(1/2) =g * (8 * (m + m1)/5k)^(1/2) or vcm= g * (8 * (m + m1)/5k)^(1/2) or (2r/T) = g * (8 * (m + m1)/5k)^(1/2) or 2r * f = g * (8 * (m + m1)/5k)^(1/2) or f = (1/2r) * g * (8 * (m + m1)/5k)^(1/2)-----------(3) r = (d/2) = (4.00/2) = 2.00 cm = 2.00 * 10^-2 m g = 9.8 m/s^2 m is the mass of the pulley = (a)negligible, (b)250 g = 250 * 10^-3 kg, (c)750 g = 750 * 10^-3 kg m1 is the mass of object = 200 g = 200 * 10^-3 kg k = 100 N/m substituting the various values of mass of the pulley inequation (3) we get various values of oscillation frequency.

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