Posted by **Lacey** on Monday, October 25, 2010 at 3:45pm.

A 3.6 m diameter merry-go-round is rotating freely with an angular velocity of 0.78 rad/s. Its total moment of inertia is 1700 kg\m^2. Four people standing on the ground, each of mass 62 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)?

- PHYSICS ---HELP! -
**Allyson**, Monday, October 25, 2010 at 3:47pm
What do you think the answer is?

- PHYSICS ---HELP! -
**Mark**, Monday, October 25, 2010 at 3:56pm
I keep getting 0.68 ?????

- PHYSICS ---HELP! -
**Lacey**, Monday, October 25, 2010 at 3:57pm
I get something close to that.

- PHYSICS ---HELP! -
**MathMate**, Monday, October 25, 2010 at 4:13pm
Total kinetic energy

= (1/2)Iω²

ω1=0.78 rad/s

I1=1700 kg-m²

I2=1700+4*(62*1.8²)=2503.5

(1/2)I1ω1²=(1/2)I2ω2²

Solve for

ω2

=√(I1/I2)ω1

=0.64 rad/s.

Check my calculations.

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