Angular Velocity Problem - Merry Go Round

[vm310]

A 4.8m diameter merry-go-round is rotating freely with an angular velocity of 0.8rad/s. Its total moment of inertia is 1950(kg)(m²). Skid, Mitch, Larry, and Greezy all jump on at the same time. They each have a mass of 65kg.What is the angular velocity now?



Relevant equations
[tex]\omega=\frac{v}{r}[/tex]

[tex]I=\frac{1}{2}[/tex]mv²

The Attempt at a Solution

I'm totally lost. Someone please give me a hint

Thanks

 

[cepheid]

[tex]I=\frac{1}{2}[/tex]mv²

This is not actually correct. 1/2 mv^2 is the formula for kinetic energy. The moment of inertia of a rotating body is something totally different, and it depends on both the mass of the body and its shape. But you don't have to worry about how to calculate it, because the problem *gives* you its numerical value right from the start.

I'm totally lost. Someone please give me a hint

There is a relationship between moment of inertia, angular velocity, and angular momentum. If you look at it closely, it should become clear what to do.

EDIT: Hint 1 - One of these three quantities changes after the kids jump on, as compared to before, which results in a change in another one of the quantities.

Hint 2 - As is often the case in physics, a general conservation law is what allows us to understand how the system will respond after the change has occurred.

 

[vm310]

Thank you for the quick response. I know that,

[tex]L=I\omega[/tex]

and that,

[tex]L=mvr[/tex]

I know I'm supposed to sum the masses of everyone who jumps on, but am I suppose to sum the radius also?

 

[cepheid]

Thank you for the quick response. I know that,

[tex]L=I\omega[/tex]

Right, and the quantity that changes (before vs. after) is the moment of inertia of the merry go round, because now it has the additional individual moments of inertia of the people standing on it. If you can figure out by how much I changes, you can figure out how much omega changes (because angular momentum is conserved).

 

[vm310]

Thanks cepheid I got it!