A diver comes off a board with arms straight up and legs straight down, giving her a moment of inertia about her rotation axis of 18kg⋅m2. S

Question

A diver comes off a board with arms straight up and legs straight down, giving her a moment of inertia about her rotation axis of 18kg⋅m2. She then tucks into a small ball, decreasing this moment of inertia to 3.6kg⋅m2. While tucked, she makes two complete revolutions in 1.2s.

Required:
If she hadn’t tucked at all, how many revolutions would she have made in the 1.5 s from board to water?

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Thiên Di 5 months 2021-08-11T11:07:58+00:00 1 Answers 106 views 0

Answers ( )

    0
    2021-08-11T11:09:15+00:00

    Answer:

    θ₁ = 0.5 revolution

    Explanation:

    We will use the conservation of angular momentum as follows:

    L_1=L_2\\I_1\omega_1=I_2\omega_2

    where,

    I₁ = initial moment of inertia = 18 kg.m²

    I₂ = Final moment of inertia = 3.6 kg.m²

    ω₁ = initial angular velocity = ?

    ω₂ = Final Angular velocity = \frac{\theta_2}{t_2} = \frac{2\ rev}{1.2\ s} = 1.67 rev/s

    Therefore,

    (18\ kg.m^2)\omega_1 = (3.6\ kg.m^2)(1.67\ rev/s)\\\\\omega_1 = \frac{(3.6\ kg.m^2)(1.67\ rev/s)}{(18\ kg.m^2)}\\\\\omega_1 = \frac{\theta_1}{t_1} =  0.333\ rev/s\\\\\theta_1 = (0.333\ rev/s)t_1

    where,

    θ₁ = revolutions if she had not tucked at all = ?

    t₁ = time = 1.5 s

    Therefore,

    \theta_1 = (0.333\ rev/s)(1.5\ s)\\

    θ₁ = 0.5 revolution

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