What is the angular velocity of the merry-go-round now?

[AlvinC]

Homework Statement

A 4.3 m diameter merry-go-round is rotating freely with an angular velocity of 0.78 rad/s. Its total moment of inertia is 1760 kg m^2. Four people standing on the ground, each of mass 67 kg, suddenly step onto the edge of the merry-go-round.

1)What is the angular velocity of the merry-go-round now?

2)What if the people were on it initially and then jumped off in a radial direction (relative to the merry-go-round)?

Homework Equations

I_i*w_i=I_f*w_f

w_f=(I_i/I_f)*w_i

The Attempt at a Solution

For question 1 I plugged the numbers in and got 0.46 rad/s.
w_f=(1760/(1760+4*67*2.15^2)*.78 =0.46 rad/s

For question 2 since the people were on it initially i made I_f in my previous question to I_i and I_i to I_f. So it looked something like this:
w_f=(2998.83/1760)*.78 = 1.3 rad/s which was wrong. Not sure what I did wrong.

 

[cupid.callin]

For question 2 since the people were on it initially i made I_f in my previous question to I_i and I_i to I_f. So it looked something like this:
w_f=(2998.83/1760)*.78 = 1.3 rad/s which was wrong. Not sure what I did wrong.

i also get same answer

you sure that w_i in second case is .78 and not .56 like we found in second case ?

 

[AlvinC]

Oh yess! I forgot to make .46 as my new initial angular velocity. Much Thanks!