Question

Two skaters skate toward each other, each moving at 3.3 m/s. Their lines of motion are separated by a perpendicular distance of 1.5 m. Just as they pass each other (still 1.5 m apart), they link hands and spin about their common center of mass. What is the rotational speed of the couple about the center of mass? Treat each skater as a point particle, each with an inertia of 51 kg.

Answer 1

Answer:

4.4 rad/s

Explanation:

When the 2 skaters is spinning with a distance of 1.5m, their rotation radius is half of that distance, which is 1.5/2 = 0.75m.

Then their moments of inertia, given that their mass being 51 kg, is (treating them as point mass particle):

$I = Mr^2 = 2Mr^2 = 51*0.75^2 = 38.25 kg.m^2$

When they change from linear motion to rotational motion, their energy must be conserved:

$E_L = E_r$

$2*(0.5Mv^2) = 2*(0.5I\omega^2)$

$\omega^2 = \frac{Mv^2}{I}$

$\omega = v\sqrt{\frac{M}{I}}$

$\omega = 3.3\sqrt{\frac{51}{38.25}} = 4.4 rad/s$