A hollowed bead is attached to a long straight rod that is rotating with an angu

Question

A hollowed bead is attached to a long straight rod that is rotating with an angular frequency of = 3.04rad/s in the horizontal plane (gravity is perpendicular to the plane). Excitedly, we find that the equation of motion for this system can simply be written as, 0 = mr2 - (d/dt)(m dr/dt), where m is the mass and r is the distance from the axis of rotation. The general solution of this equation is found to be r(t) = Aet + Be-t, where A and B are constants. If we assume that the bead started from some initial distance, r0, from the point of rotation at the initial time t = 0, then the radial distance from the axis of rotation can be described by the equation r(t) = (r0/2)(et + e-t). (a) What is the location, r(t), for a bead that started at r0 = 5.23cm over the course of 1.15s? By definition, the radial velocity is vr = dr/dt. (b) What is the radial velocity of the bead in units of cm/s at the same time, t = 1.15s, when started from the same r0 = 5.23cm value?

Explanation / Answer

Given that,

w = 3.04 rad/s

r(t) = (ro / 2)*(e^wt + e^-wt)

At ro = 5.23 cm , t = 1.15 s

r(t) = (5.23 / 2) * [e^(3.04*1.15) + e^(-3.04*1.15)]

r(t) = 86.33 cm

(b)

radial velocity of the bead,

v = dr / dt

v = d( (ro / 2)*(e^wt + e^-wt)) / dt

v =  (ro / 2)*(e^wt - e^-wt)*w

v = (5.23 / 2) * [e^(3.04*1.15) - e^(-3.04*1.15)]*3.04

v = 261.96 cm/s

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