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A uniform 220 g rod with length 45 cm rotates?
A uniform 220 g rod with length 45 cm rotates n a horizontal plane about a fixed, vertical, frictionless pin through its center. Two small 18 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 13 cm on each sides of the center, at which time the system rotates at an angular speed of 12 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.
1 Answer
- Anonymous9 years agoFavorite Answer
MI rod = 1/12 . m . L^2 = 1/12 . 0.22 . 0.45^2 = 3.7125 . 10^–3 kgm^2
MI beads = 2 . 0.018 . 0.13^2 = 6.084 . 10^–4 kgm^2
MI beads at finish = 2 . 0.018 . 0.225^2 = 1.8225 . 10^–3 kgm^2
Angular momentum is conserved
(37.125 + 6.084) . 10^–4 . 12 = (37.125 + 18.225 ) . 10^–4 . w
w = 9.368 = 9.4 rad/s
Source(s): Old teacher