A horizontal vinyl record of mass 0.115 kg and radius 0.0896 m rotates freely about a vertical axis through its center with an angular speed

Question

A horizontal vinyl record of mass 0.115 kg and radius 0.0896 m rotates freely about a vertical axis through its center with an angular speed of 5.58 rad/s and a rotational inertia of 4.84 x 10-4 kg·m2. Putty of mass 0.0484 kg drops vertically onto the record from above and sticks to the edge of the record.What is the angular speed of the record immediately afterwards?

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Thu Nguyệt 5 years 2021-07-14T03:51:05+00:00 1 Answers 15 views 0

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  1. minhkhoi
    0
    2021-07-14T03:52:57+00:00
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    Answer:

    Explanation:

    Angular momentum is the product of inertial and angular frequency

    L = I × ω

    Where

    L is angular momentum

    I is inertia

    ω is angular frequency

    So, given that

    Vinyl record has a massz

    M = 0.115kg

    Radius R = 0.0896m

    Angular velocity of vinyl record

    ω(initial) = 5.58 rad/s

    Rotational inertial of vinyl record.

    I(initial) = 4.84 × 10^-4 kgm²

    Putty drop on the record

    Mass of putty M’ = 0.0484kg

    Angular speed after putty drop ω’

    Using conversation of angular momentum

    Initial angular momentum is equal to final angular momentum

    I(initial) × ω(initial) = I(final) × ω(final)

    So, we need to find I(final)

    Inertia log putty can be determine using MR² by assuming a thin loop

    I(putty) = M’R² = 0.0484 × 0.0896

    I(putty) = 3.89 × 10^-4 kgm²

    I(final) = I(initail) + I(putty)

    I(final) = 4.84 × 10^-4 + 3.89 × 10^-4

    I(final) = 8.73 × 10^-4 kgm²

    Therefore,

    I(initial) × ω(initial) = I(final) × ω(final)

    ω(final) = I(initial) × ω(initial) / I(final)

    ω(final) = 4.84 × 10^-4 × 5.58 / 8.73 × 10^-4

    ω(final) = 3.1 rad/s

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