Note: Drawings are nol to scale) You are given two objects, cach with the same m

Question

Note: Drawings are nol to scale) You are given two objects, cach with the same mass. One object is a solid, uniform disk of outer radius R The other object is a thin, uniform rod of unknown length. 1. Initially these objects are suspended as shown (each from a frictionless pin located on its uppermost edge). When allowed to oscillate freely as pendulums (through small angular displacements), they have equal frequencies. disk rod Then the rod is removed from its pin and attached instead to the end of a massless ideal spring suspended from the ceiling, with a y-axis defined as positive downward (and y = 0 at the ceiling), as shown. You may ignore all friction, air drag and spring fatigue. The rod-spring system is set into vertical oscillation so that all of the following are truc: . The position of the rod's upper cnd is given by y() = Acos(on + ) + H, where is in seconds. . The spring is compressed for some time interval during each cycle of oscillation. At a certain position in each cycle, the rod experiences a maximum net force magnitude Ft y and-letting Ug for the system be defined as zcro at that same position-the total mechanical energy of the system is a constant known value, E . At 1 = 0.123 s, the lower end of the rod reaches its greatest distance, D, from the ceiling Lrtelintume H = 0.536 m E , 1.60 J D 0.987 m g = 980 me The known data: R=0-124 m a. Find the relaxed length of the spring b. Find the arm length of a simple pendulum whose period is three times the period of the rod-and-spring system. c. Find the y-velocity (both magnitude and direction) of the rod at time 465 s.

Explanation / Answer

from the given situation

position of rods upper end is given by

y(t) = Acos(wt + phi) + H

maximum net force is observed by the rod at the highest point, Ug = PE of system = 0 at this point

total mechanical energy of the system = Em

at t= 0.123 s, lower end of the rod is at its greatest distance for the first time from the cieling

hence

at t= 0.123 s, y'(t) = 0

cos(wt + phi) = -1

now,

R = 0.124 m

H = 0.536 m

D = 0.987 m

F max = 34.2 N

E m = 1.6 J

g = 9.8 m/s/s

a. now,

at t = 0

y(0) = Acos(phi) + H

Em = 1.6 J = 0.5*k*A^2

k*A^2 = 3.2

Fmax = k*A = 34.2 - mg N

now, for the disc of mass m and radius R

time period of osscilation T = 2*pi*sqrt(0.5mR^2/mgR)

for the rod

T = 2*pi*sqrt(2mL^2/12*m*g*L)

equating both

0.5R/g = 2L/12g

0.5R = L/6

L = 3R

hence

L = 0.372 m

L = D - y(0.123) = 0.987 - A*cos(w*0.123 + phi) - 0.536

0.079 = Acos(0.123w + phi)

also, w = sqroot(k/m)

k = 3.2/A^2

w = sqroot(3.2/m)/A

m = (34.2 - 3.2/A)/g = 3.48623 - 0.3261/A

w = sqroot(3.2/A(3.48623A - 0.3261))

also, at t = 0.123 s, y'(0) = 0

-A*0.123*sin(0.123w + phi) = 0

0.123w + phi = pi

phi = (pi - 0.123w)

0.079 = Acos(0.123w + pi - 0.123w)

A = -0.079 m

hence

w = 8.2061 rad/s

m = 7.6140781 kg

k = 512.7383 N/m

phi = 2.1322423 rad = 122.1684877 deg

a. hence relaxed length of spring = H + mg/k = 0.681 m

b. time period = 2*pi*sqroot(m/k) = 0.7654693965466 s

T' = 3T = 2.2964 s

hence

length of pendulum = l

2*pi*sqroot(l/g) = 2.2964

l = 1.310410742496 m

c. at t= 4.65 s

y'(t) = -A*w*sin(wt + phi) = 0.314449317 m/s

pointing upwards

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