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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. The child grabs and clings to a bar that is 1.45 m from the center of the merry‑go‑round, causing the angular velocity of the merry‑go‑round to abruptly drop from 55.0 rpm to 17.0 rpm. What is the moment of inertia of the merry‑go‑round with respect to its central axis?

Respuesta :

Answer:

[tex]I_1 = 32.5 kg m^2[/tex]

Explanation:

Here as we know that total angular momentum is always conserved for child + round system

so here by angular momentum conservation we have

[tex]I_1 \omega_1 = (I_1 + I_2)\omega_2[/tex]

here we have

[tex]I_2 = mR^2[/tex] (inertia of boy)

[tex]I_2 = (34.5)(1.45^2)[/tex]

[tex]I_2 = 72.5 kg m^2[/tex]

now we have

[tex]I_1(2\pi 55) = (I_1 + 72.5)(2 \pi 17)[/tex]

[tex]I_1 = (I_1 + 72.5)(0.31)[/tex]

[tex]I_1(1 - 0.31) = 22.4[/tex]

[tex]I_1 = 32.5 kg m^2[/tex]

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