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)?
4 answers
What do you think the answer is?
I keep getting 0.68 ?????
I get something close to that.
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.
= (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.
0 answers