A merry-go-round with a radius R = 1.6 m and moment of inertia I = 150 kg-m2 is
ID: 1698177 • Letter: A
Question
A merry-go-round with a radius R = 1.6 m and moment of inertia I = 150 kg-m2 is spinning along with a person of mass m = 74 kg standing at the rim with an angular speed of ? = 2 rad/s. Once the person gets half way around, he decides to simply let go of the merry-go-round to exit the ride.
What is the linear velocity of the person right as he leaves the merry-go-round?
What is the angular speed of the merry-go-round after the person lets go?
Explanation / Answer
oment of inertia of merry -go- round = I1 =150kgm^2
moment of inertia when the person when he moves nearer to edge = I2= I1+ I -man
I-man = mR^2 = 74(1.6)^2 kgm^2
=189.44 kgm^2
thus, I2= 150+ 189.44 kgm^2
=339.44 kgm^2,
angular speed of the merry-go-round w1= 2 rad/s
angular speed of the merry-go-round after the person lets go w2=?
acc. to conservation of linear momentum ,
I1W1=I2W2
W2= I1W1/I2
= 150(2)/339.44
=300/339.44
=0.8838 rad/s
the linear velocity of the person as he leaves the merry- go- round is
v=RW2
= 1.6(0.8838) m/s
=1.414 m/s or sqrt(2)
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