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

`\omega=3.135rad/s`

Step-by-step 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=(v)/(r)\\\\\omega=(2.50)/(0.0800)`

`\omega=3.135rad/s`