Question

Twin skaters approach each other with identical speeds. Then, the skaters lock hands and spin. Calculate their final angular velocity, given each had an initial speed of 2.50 m/s relative to the ice. Each has a mass of 70.0 kg, and each has a center of mass located 0.800 m from their locked hands. You may approximate

Answer 1

Answer:

$\omega=3.135rad/s$

Explanation:

From the question we are told that:

initial Speed $V_1=2.50$

Mass $m=70.0kg$

Center of mass $d=0.0.800m$

Generally the equation for angular velocity is is mathematically given by

$\omega=\frac{v}{r}\\\\\omega=\frac{2.50}{0.0800}$

$\omega=3.135rad/s$