A uniform 200 g rod with length 69 cm rotates in a horizontal plane about a fixe

Question

A uniform 200 g rod with length 69 cm rotates in a horizontal plane about a fixed, vertical, frictionless pin through its center. Two small 21 g beads are mounted on the rod such that they are able to slide without friction along its length. Initially the beads are held by catches at positions 10 cm on each sides of the center, at which time the system rotates at an angular speed of 22 rad/s. Suddenly, the catches are released and the small beads slide outward along the rod. Find the angular speed of the system at the instant the beads reach the ends of the rod. Answer in units of rad/s.

Explanation / Answer

Mass of the rod, M = 200 g                                = 0.2 kg Length of the rod, L = 69 cm                                = 0.69 m Mass of each bead, m = 21 g                                    = 0.021 kg Initial distance of each bead from the center, r1 = 10 cm                                                                           = 0.1 m Final distance of each bead from the center, r2 = 34.5 cm                                                                        = 0.345 m Initial angular speed, 1 = 22 rad/s Final angular speed, 2 = ? -------------------------------------------------------------------- Moment of inertia of the rod about the center, I = ML^2 / 12                                                                         = 0.2 * 0.69^2 / 12                                                                         = 0.0079 kgm^2 Initial moment of inertia of both beads, I1 = 2mr1^2                                                                = 2*0.021*0.1^2                                                                = 0.00042 kgm^2 Final moment of inertia of both beads, I2 = 2mr2^2                                                                = 2*0.021*0.345^2                                                                = 0.00498 kgm^2 --------------------------------------------------------------------- According to law of conservation of angular momentum, (I+I1) 1 = (I+I2) 2 2 = (0.0079 +0.00042) * 22 / ( 0.0079 + 0.00498)       = 0.00832 * 22 / 0.01288       = 14.21 rad/s Final moment of inertia of both beads, I2 = 2mr2^2                                                                = 2*0.021*0.345^2                                                                = 0.00498 kgm^2 --------------------------------------------------------------------- According to law of conservation of angular momentum, (I+I1) 1 = (I+I2) 2 2 = (0.0079 +0.00042) * 22 / ( 0.0079 + 0.00498)       = 0.00832 * 22 / 0.01288       = 14.21 rad/s

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