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A man stands on a platform that is rotating (without friction) with an angular speed of 1.2 rev/s; his arms are outstretched and he holds a brick in each hand.The rotational inertia of the system consisting of the man, bricks, and platform about the central vertical axis of the platform is 6.0 k g times m squared. If by moving the bricks the man decreases the rotational inertia of the system to 2.0 k g times m squared, what is the resulting angular speed of the platform in rad/s

Respuesta :

Answer:

resulting angular speed = 3.6 rev/s

Explanation:

We are given;

Initial angular speed; ω_i = 1.2 rev/s

Initial moment of inertia;I_i = 6 kg/m²

Final moment of inertia;I_f = 2 kg/m²

From conservation of angular momentum;

Initial angular momentum = Final angular momentum

Thus;

I_i × ω_i = I_f × ω_f

Making ω_f the subject, we have;

ω_f = (I_i × ω_i)/I_f

Plugging in the relevant values;

ω_f = (6 × 1.2)/2

ω_f = 3.6 rev/s

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