Question

A skater makes one revolution per second, f = w/(2p) = 1, with her arms and leg stretched (left). By lowering her leg and bringing her arms close to her chest (right) she reduces her moment of inertia from I to I′ = 0.4I. How many revolutions per second, f′ = w¢/(2p), is she making now? Show all your work.

Answer 1

Answer:

f'=w'/2p=2.49rev/s

Explanation:

The angular momentum L must be equal for both cases.

Hence you have that

L before = L after

$L=I\omega$

$\omega=2\pi f$

I = moment of inertia

w =angular momentum

Hence

$L=L'\\I\omega =I'\omega'\\\omega'=\frac{I\omega}{I'}=\frac{I*2\pi *1}{0.4I}=15.70\frac{rad}{s}$

f'=w'/2p=2.49rev/s

hope this helps!!

Answer 2

Answer:

Answer is given in the attachment.

Explanation:

Download pdf