physics and merry go round?

Question

A 4.8 m diameter merry-go-round is rotating freely with an angular velocity of 0.50 rad/s. Its total moment of inertia is 1900 kgM2. Four people standing on the ground, each of 50 kg mass, suddenly step onto the edge of the merry-go-round.

(a) What is the angular velocity of the merry-go-round now?
w= rad/s

(b) Assume that the people were on it initially and then jumped off in a radial direction (relative to the merry-go-round). What would be the angular velocity of the merry-go-round?
w= rad/s


this stuff goes way over my heard....can anyone help?

Answer 1

This is solved using conservation of angular momentum for both cases.

Conservation of angular momentum states that
I1*w1=I2*w2
To solve the problems you just consider the events that are known and how the variables on each side are affected

a. Before the event of the four people stepping on the merry-go-round, it is happy with angular momentum of
1900*.5

What changes when the people step on is that I now includes four point masses of 50Kg at 4.8/2 m from the center of mass of the original
so
1900*.5=
(1900+4*50*4.8/2)*w2
w2=1900*.5/(1900+4*50*4.8/2)
w2=0.4 rad/sec

Note: This assumes that the people stepped on without pushing off or dragging on the ground. Tricky to do.

b) Same sort of in reverse
The moment of inertia gets changed by the people jumping off radially, meaning there impulse did not increase the angular kinetic energy. Rather, we look at the change in I to determine the new w
(1900+4*50*4.8/2)*.5=1900*w2
w2=0.6

j