Question

You stand on a frictionless platform that is rotating with an angular speed of 5.1 rev/s. Your arms are outstretched, and you hold a heavy weight in each hand. The moment of inertia of you, the extended weights, and the platform is 9.9 kg · m2 . When you pull the weights in toward your body, the moment of inertia decreases to 2.6 kg · m 2 . What is the resulting angular speed of the platform? Answer in units of rev/s. 017 (part 2 of 2) 10.0 point

Answer 1

Answer:

$\omega'=19.419\ rev.s^{-1}$

Explanation:

Given:

angular speed of rotation of friction-less platform, $\omega=5.1\ rev.s^{-1}$

moment of inertia with extended weight, $I=9.9\ kg.m^2$

moment of inertia with contracted weight, $I'=2.6\ kg.m^2$

Now we use the law of conservation of angular momentum:

$I.\omega=I'.\omega'$

$9.9\times 5.1=2.6\times \omega'$

$\omega'=19.419\ rev.s^{-1}$

The angular speed becomes faster as the mass is contracted radially near to the axis of rotation.