Question

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? Express to 3 sig figs.

Answer 1

w₂ = 22.6 rad/s

Step-by-step explanation:

This exercise the system is formed by platform, man and bricks; For this system, when the bricks are released, the forces are internal, so the kinetic moment is conserved.

Let's write the moment two moments

initial instant. Before releasing bricks

L₀ = I₁ w₁

final moment. After releasing the bricks

`L_(f)` = I₂W₂

L₀ = L_{f}

I₁ w₁ = I₂ w₂

w₂ = I₁ / I₂ w₁

let's reduce the data to the SI system

w₁ = 1.2 rev / s (2π rad / 1rev) = 7.54 rad / s

let's calculate

w₂ = 6.0/2.0 7.54

w₂ = 22.6 rad/s