Question

skater spins over a point at a speed of 3.0 rotations per second then the momentum of inertia is 0.60 kg.M2, what is its angular momentum?

Answer 1

`L=11.3\ kg-m^2/s`

Step-by-step explanation:

Given that,

Angular speed of a skater,
`\omega=3\ rot/s=18.84\ rad/s`

The moment of inertia of the skater, I = 0.6 kg-m²

We need to find the angular momentum of the skater. The formula for the angular momentum of the skater is given by :

`L=I\omega`

Substitute all the values,

`L=0.6* 18.84\\\\L=11.3\ kg-m^2/s`

So, its angular momentum is equal to
`11.3\ kg-m^2/s`.