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19 June, 20:47

A uniform circular disk of moment of inertia 8.0 kg m2 is rotating at 4.0 rad/s. A small lump of mass 1.0 kg is dropped on the disk and sticks to it at a distance of 1.0 m from the axis of rotation. What is the new rotational speed of the combined system?

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  1. 19 June, 21:03
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    The new rotational speed of the combined system is 32 rad/s

    Explanation:

    Given-

    Moment of Inertia, I = 8 kg m²

    speed of rotation, ω = 4 rad/s

    Mass of the lump = 1 kg

    distance = 1 m

    New rotational speed = ?

    We know,

    When the lump is dropped on the rotating disc then the angular momentum remains conserved.

    Angular Momentum of the disc, Lₐ = Iω

    Angular momentum of the lump, L ₓ = mvR

    Since the angular momentum remains conserved,

    Lₐ = Lₓ

    8 X 4 = 1 X v X 1

    New rotational speed, v = 32 rad/s

    Therefore, the new rotational speed of the combined system is 32 rad/s.
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