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 ? = 3.34rad/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 = mr?2 - (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) = Ae?t + 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)(e?t + e-?t). (a) What is the location, r(t), for a bead that started at r0 = 5.63cm over the course of 1.05s? 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 = 1.05s, when started from the same r0 = 5.63cm value? answer for part (a) in (cm) answer for part (B) IN (cm/s)

Explanation / Answer

(a) r=(5.63/2)(e3.34*1.05 + e-3.34*1.05) = 94 cm (b) dr/dt=(r0/2)(e?t - e-?t)*? = 313.26 cm/s

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