A unifrom rod of mass 300 g and length 50 cm rotates in ahorizontal plane about

Question

A unifrom rod of mass 300 g and length 50 cm rotates in ahorizontal plane about a fixed, vertical, frictionless pin throughits center. Two small, dense beads, each of mass m, are mounted onthe rod so that they can slide w/o friction along its length.Initially, the beads are held by catches at positions 10 cm on eachside of the center and the sys. is rotating at an angular speed of36 rad/s. The catches are released simultaneously and the beadsslide outward along the rod. Find: a) angular speed wf of the sys. at the instant the beadsslide off the ends of the rod as it depends on m b)what are the max. and min. possible values for wf and thevalues m to which they correspond? describe the shape of a graph ofwf versus m. A unifrom rod of mass 300 g and length 50 cm rotates in ahorizontal plane about a fixed, vertical, frictionless pin throughits center. Two small, dense beads, each of mass m, are mounted onthe rod so that they can slide w/o friction along its length.Initially, the beads are held by catches at positions 10 cm on eachside of the center and the sys. is rotating at an angular speed of36 rad/s. The catches are released simultaneously and the beadsslide outward along the rod. Find: a) angular speed wf of the sys. at the instant the beadsslide off the ends of the rod as it depends on m b)what are the max. and min. possible values for wf and thevalues m to which they correspond? describe the shape of a graph ofwf versus m.

Explanation / Answer

Given :    M = 300 g, L = 50 cm, r= 10 cm, = 36rad/s initial moment of inertia I = ML2/12 +2mr2 initial angular momentum = I final moment of inertia If = ML2/12 +2m(L/2)2 final angular momentum = Iff angular momentum conservation: (ML2/12 + 2mr2) =[ML2/12 + 2m(L/2)2]f f = (ML2/12 +2mr2)/[ML2/12 +2m(L/2)2] a) m = 0 f =(ML2/12)/[ML2/12] = 36 rad/s b) m --> M = 300 g, L = 50 cm, r= 10 cm, = 36rad/s f = (r2) /(L/2)2      = ( 10 m )2 ( 36rad/s ) (25)2= 5.766 rad/s f decreases smoothly from its maxvalue toward its minimum value as m approachesinfinity.
M = 300 g, L = 50 cm, r= 10 cm, = 36rad/s initial moment of inertia I = ML2/12 +2mr2 initial angular momentum = I final moment of inertia If = ML2/12 +2m(L/2)2 final angular momentum = Iff angular momentum conservation: (ML2/12 + 2mr2) =[ML2/12 + 2m(L/2)2]f f = (ML2/12 +2mr2)/[ML2/12 +2m(L/2)2] a) m = 0 f =(ML2/12)/[ML2/12] = 36 rad/s b) m --> M = 300 g, L = 50 cm, r= 10 cm, = 36rad/s f = (r2) /(L/2)2      = ( 10 m )2 ( 36rad/s ) (25)2= 5.766 rad/s f decreases smoothly from its maxvalue toward its minimum value as m approachesinfinity.
f = (ML2/12 +2mr2)/[ML2/12 +2m(L/2)2] a) m = 0 f =(ML2/12)/[ML2/12] = 36 rad/s b) m --> M = 300 g, L = 50 cm, r= 10 cm, = 36rad/s f = (r2) /(L/2)2      = ( 10 m )2 ( 36rad/s ) (25)2= 5.766 rad/s f decreases smoothly from its maxvalue toward its minimum value as m approachesinfinity.

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