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21 April, 08:28

A horizontal turntable (disc) of mass = 5kg, radius = 1.80m, is spinning about its center at the rate of 5 rad/s. A lump of mass 0.800 kg falls on it vertically and sticks to it at a distance of 1.20 m from the center.

A) Calculate the new angular speed of the disc.

B) What is the new rotational kinetic energy of the disc+lump

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Answers (1)
  1. 21 April, 10:18
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    A) 4.38 rad/s

    B) 88.75 J

    Explanation:

    A)

    M = mass of the disc = 5 kg

    R = radius of the disc = 1.80 m

    I = initial moment of inertia of the system = (0.5) MR² = (0.5) (5) (1.80) ² = 8.1 kgm²

    m = mass of the lump = 0.800 kg

    r = distance of the lump from axis of rotation = 1.20 m

    I' = initial moment of inertia of the system = I + m r² = 8.1 + (0.800) (1.2) ² = 9.252 kgm²

    w' = new angular speed

    w = initial angular speed = 5 rad/s

    Using conservation of angular momentum

    I w = I' w'

    (8.1) (5) = (9.252) w'

    w' = 4.38 rad/s

    B)

    new rotational kinetic energy of the disc+lump is given as

    KE' = (0.5) I' w'²

    KE' = (0.5) (9.252) (4.38) ²

    KE' = 88.75 J
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