A brave child decides to grab onto an already spinning merry‑go‑round. The child is initially at rest and has a mass of 34.5 kg. 34.5 kg. Th

Question

A brave child decides to grab onto an already spinning merry‑go‑round. The child is initially at rest and has a mass of 34.5 kg. 34.5 kg. The child grabs and clings to a bar that is 1.20 m 1.20 m from the center of the merry‑go‑round, causing the angular velocity of the merry‑go‑round to abruptly drop from 37.0 rpm 37.0 rpm to 19.0 rpm . 19.0 rpm. What is the moment of inertia of the merry‑go‑round with respect to its central axis?

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Hải Đăng 4 years 2021-07-21T07:57:19+00:00 2 Answers 68 views 0

Answers ( )

  1. thaiduong
    0
    2021-07-21T07:58:33+00:00
    Reply

    Answer:

    52.44 kgm^2

    Explanation:

    First convert from revolution per minute to rad per second knowing that a revolution is 2π rad and every minute has 60 seconds

    \omega_1 = 37 rpm = 37*2\pi/60 = 3.875 rad/s

    \omega_2 = 19 rpm = 19*2\pi/60 = 1.99 rad/s

    According to the law of momentum conservation, the total angular before and after the child jump on must be the same:

    \omega_1 I_w = \omega_2(I_w + I_c) (1)

    where I_w is the moments of inertia of the merry-go-round, which we are looking for, and I_c is the moments of inertia of the child, which can be calculated if we treat him as a point mass:

    I_c = mr^2 = 34.5*1.2^2 = 49.68 kgm^2

    Therefore from eq. (1) we have:

    3.875I_w = 1.99(I_w + 49.68)

    I_w(3.875 - 1.99) = 1.99*49.68 = 98.85

    I_w = \frac{98.85}{3.875 - 1.99} = 52.44 kgm^2

  2. Eirian
    0
    2021-07-21T07:59:08+00:00
    Reply

    Answer:

    the moment of inertia of the merry go round is 38.04 kg.m²

    Explanation:

    We are given;

    Initial angular velocity; ω_1 = 37 rpm

    Final angular velocity; ω_2 = 19 rpm

    mass of child; m = 15.5 kg

    distance from the centre; r = 1.55 m

    Now, let the moment of inertia of the merry go round be I.

    Using the principle of conservation of angular momentum, we have;

    I_1 = I_2

    Thus,

    Iω_1 = I’ω_2

    where I’ is the moment of inertia of the merry go round and child which is given as I’ = mr²

    Thus,

    I x 37 = ( I + mr²)19

    37I = ( I + (15.5 x 1.55²))19

    37I = 19I + 684.7125

    37I – 19 I = 684.7125

    18I = 684.7125

    I = 684.7125/18

    I = 38.04 kg.m²

    Thus, the moment of inertia of the merry go round is 38.04 kg.m²

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