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:

$E_L = E_r$

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

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

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

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

Answer 2

The rotational speed of the couple about the center of mass is: 4.4 rad/s

Meaning of Couple

A couple can be defined as an effect or force created by two parallel forces that are equal in masses but opposite in sense or opposite in direction and do not share the same line of action.

The couple can also be used to signify two individuals.

In conclusion, the rotational speed of the couple about the center of mass is: 4.4 rad/s

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