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21 May, 18:13

A tall, cylindrical chimney falls over when its base is ruptured. Treat the chimney as a thin rod of length 58.8 m. At the instant it makes an angle of 30.2° with the vertical as it falls, what are (a) the radial acceleration of the top, and (b) the tangential acceleration of the top. (Hint: Use energy considerations, not a torque.) (c) At what angle θ is the tangential acceleration equal to g? Assume free-fall acceleration to be equal to 9.81 m/s2.

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  1. 21 May, 22:07
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    Using conservation of mechanical energy for rotational motion of chimney

    gain of rotational KE = loss of potential energy

    1/2 Iω² = mgl/2 (1 - cosθ); I is moment of inertia, ω is angular velocity, m is mass of chilney, l is its length.

    I = 1/3 m l²

    1/6 m l²ω² = mgl/2 (1 - cosθ)

    1/6 m v² = mgl/2 (1 - cosθ)

    v² = 3gl (1 - cosθ)

    radial acceleration at the top = v² / r

    =3gl (1 - cosθ) / l

    = 3g (1 - cosθ)

    = 3 x 9.8 (1 - cos 30.2)

    = 4 m / s²

    b) Let angular acceleration be α.

    Torque acting at that time

    mgsinθ x l / 2 = I α

    mgsinθ x l / 2 = 1/3 m l² α

    α = 3g sinθ / 2l

    tangential acceleration

    α x l = 3gsinθ / 2

    = 3 x 9.8 sin 30.2 / 2

    = 7.4 m / s²

    c) tangential acceleration equal to g

    3gsinθ / 2 = g

    sinθ = 2 / 3

    42 degree.
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