Question

A uniform rod of mass 250 g and length 100 cm is free to rotate in a horizontal plane around a fixed vertical axis through its center, perpendicular to its length. Two small beads, each of mass 26 g, are mounted in grooves along the rod. Initially, the two beads are held by catches on opposite sides of the rod's center, 16 cm from the axis of rotation. With the beads in this position, the rod is rotating with an angular velocity of 13.0 rad/s. When the catches are released, the beads slide outward along the rod.(Please show all steps)(a) What is the rod's angular velocity (in rad/s) when the beads reach the ends of the rod? (Indicate the direction with the sign of your answer.)() rad/s(b) What is the rod's angular velocity (in rad/s) if the beads fly off the rod? (Indicate the direction with the sign of your answer.)()rad/s

Answer 1

Answer:

a.  ω' = 19.84 rad /s

b.  ω'' = 32.22 rad /s

Explanation:

Given:

m = 250 g = 0.25 kg , l = 100 cm = 1 m, mₙ = 26 g = 0.026 kg , r₁ = 16 cm = 0.16 m, ω = 13.0 rad /s , r₂ = 0.50 m

a.

Irad * ω + I beat * ω = I rad *ω' + I beat * ω'

0.25 * 1² / 12 * 13 + (2*0.026* 0.5² m) * 13 = 0.25 * 1² / 12 * ω' + (2*0.026*0.16²) ω'

ω' = 19.84 rad /s

b.

Irad * ω' + I beat * ω' = Ibeat * ω''

0.25 kg * 1² / 12 * 19.84 + (2*0.026kg*0.5² m) * 19.84 = 0.25 * 1² / 12 * ω''

ω'' = 32.22 rad /s