Question

A man stands at the center of a platform that rotates without friction with an angular speed of 3.95 rev/s. His arms are outstretched, and he holds a heavy weight in each hand. The moment of inertia of the man, the extended weights, and the platform is 12.5 kg*m^2. When the man pulls the weights inward toward his body, the moment of inertia decreases to 2.5 kg*m^2. What is the resulting angular speed of the platform?

Answer 1

Answer:

$W_f$= 124.05 rad/s

Explanation:

Using the conservation of the angular momentum:

$L_i = L_f$

so:

$I_iW_i = I_fW_f$

where $I_i$ is the initial moment of inertia, $W_i$ the initial angular velocity, $I_f$ the final moment of inerta and $W_f$ the final angular velocity.

Note: Wi = 3.95 rev/s = 24.81 rad/s

Then, replacing values, we get:

$(12.5)(24.81rad/s) = (2.5)W_f$

Finally, solving for $W_f$:

$W_f$= 124.05 rad/s