A disk-shaped merry-go-round of radius 2.93 m and mass 135 kg rotates freely with an angular speed of 0.621 rev/s . A 62.4 kg person running tangential to the rim of the merry-go-round at 3.21 m/s jumps onto its rim and holds on. Before jumping on the merry-go-round, the person was moving in the same direction as the merry-go-round's rim. What is the final angular speed of the merry-go-round? rad/s

2.55 rad/sec

Explanation:

Use conservation of angular momentum

The merry go round has

I = ½mr²

I = ½ * 135 * 2.93²

I = 579 kgm²

the person has

I = mr²

I = 62.4 * 2.93²

I = 536 kgm²

Converting 0.621 rev/sec to rad/sec we have 0.621 * 2π

0.621 * 2 * 3.14 rad/sec

3.9 rad/sec

for the person v/r = w

w = 3.21 / 2.93

w = 1.10 rad/sec

So

579 * 3.9 + 536 * 1.1 = (579+536)*w

2258 + 589.6 = 1115w

2847.6 = 1115w

solve for w

w = 2847.6 / 1115

w = 2.55 rad/sec

Thus, the final angular speed of the merry go round is 2.55 rad/sec

where is the circuit diagram ?

option number c is the right answer.