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

2 個解答

  • 最愛解答

    Let the mass of each of P and Q be m, then OP = PQ = r

    Then 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.

    資料來源: 原創答案
  • 天同
    Lv 7
    9 年前

    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