2011Q7
A light rigid rod is hinged smoothly at O at one end.Two small beads of the same mass are fixed to the mid-point P and Q of the rod respectively as shown. Initially the rod is held horizontally and it is then releaesd form rest .What is the acceleration of the bead at Q when the rod reaches the position veritcally below O?
Ans 12/5 g
AL PHY(URGENT)
2012-04-16 11:40 pm
回答 (2)
2012-04-17 12:03 am
✔ 最佳答案
Let the mass of each of P and Q be m, then OP = PQ = rThen when the beads fall to the position vertically below O:
Total gpe loss = mgr + mg(2r) = 3mgr
Suppose that ω is their common angular velocity at this position, then:
(1/2)m(rω)2 + (1/2)m(2rω)2 = 3mgr
(5/2)mr2ω2 = 3mgr
rω2 = 6g/5
So the acceleration of Q at this position is given by (2r)ω2 = 12g/5 which is the centripetal acceleration.
參考: 原創答案
2012-04-17 12:11 am
Let m be the mass of each bead,
Moment of inertia of the two beand = mL^2 + m(2L)^2 = 5 mL^2
where L is the length OP
Loss of potential energy = mgL + mg(2L) = 3mgL
Gain in kinetic energy = (1/2).(5mL^2)w^2
where w is the angular speed of the beads when they are vertically below O
Hence, 3mgL = (5/2).mL^2w^2
3g = (5/2).Lw^2
i.e. Lw^2 = 6g/5
Centripetal acceleration of bead at Q
= (2L).w^2 = 2 x 6g/5 = 12g/5
Moment of inertia of the two beand = mL^2 + m(2L)^2 = 5 mL^2
where L is the length OP
Loss of potential energy = mgL + mg(2L) = 3mgL
Gain in kinetic energy = (1/2).(5mL^2)w^2
where w is the angular speed of the beads when they are vertically below O
Hence, 3mgL = (5/2).mL^2w^2
3g = (5/2).Lw^2
i.e. Lw^2 = 6g/5
Centripetal acceleration of bead at Q
= (2L).w^2 = 2 x 6g/5 = 12g/5
收錄日期: 2021-04-29 17:46:15
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