Question

A skater extends her arms horizontally, holding a 5-kg mass in each hand. She is rotating about a vertical axis with an angular velocity of one revolution per second. If she drops her hands to her sides, what will the final angular velocity (in rev/s) be if her moment of inertia remains approximately constant at 5 kg⋅m2, and the distance of the masses from the axis changes from 1 m to 0.1 m?

Answer 1

Answer:

$1.98$ rev/s

Explanation:

$m$  = mass attached to each hand = 5 kg

$r_{i}$ = initial distance of masses in each hand = 1 m

$r_{f}$ = final distance of masses in each hand = 0.1 m

$I$ = moment of inertia of body = 5 kgm²

$I_{i}$ = initial total moment of inertia = $I + 2 mr_{i}^{2}$

$w_{i}$ = initial angular velocity = 1 rev/s

$I_{f}$ = final total moment of inertia = $I + 2 mr_{f}^{2}$

$w_{f}$ = final angular velocity = ?

Using conservation of angular momentum

$I_{i} w_{i} = I_{f} w_{f}$

$(I + 2 mr_{i}^{2}) w_{i} = (I + 2 mr_{f}^{2}) w_{f}$

$(5 + 2 (5) (1)^{2}) (1) = (5 + 2 (5) (0.1)^{2}) w_{f}$

$w_{f} = 2.94$ rev/s

Answer 2

The answer is attached