Two children (m = 40 kg each) stand opposite each other on the edge of a merry-go-round. The merry-go-round, which has a mass M of 200 kg and a radius of 2.0 m, is spinning at a constant rate of 12 rad/s. Treat the two children and the merry-go-round as a system. Both children walk half the distance toward the center of the merry-go-round. Calculate the final angular speed of the system.

Explanation:

Initially the two boys were sitting on the periphery , total moment of inertia

= 1/2 M  r² + 2mr²     ; M is mass of the merry go round , m is mass of each boy and r is the radius

1/2 x 200 x 2² + 2 x 40 x 2²

= 400 + 320

I₁ =  720 kg m²

Finally  the two boys were sitting at the middle  , total moment of inertia

= 1/2 M  r² + 2m( r/2)²     ; M is mass of the merry go round , m is mass of each boy and r is the radius

1/2 x 200 x 2² + 2 x 40 x 1²

= 400 + 80

I₂  = 480

Now the system will obey law of coservation of angular momentum because no torque is acting on the system.

I₁ω₁ =  I₂ω₂ ,         I₁  and ω₁  are moment of inertia and angular velocity of first case and  I₂  and ω₂ are of second case.

720 X 12 = 480 ω₂

ω₂ = 18 rad / s

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y