Question

A disk-shaped merry-go-round of radius 2.53 m and mass 185 kg rotates freely with an angular speed of 0.601 rev/s . A 55.4 kg person running tangential to the rim of the merry-go-round at 3.11 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?        

Answer 1

From Law of Conservation of angular momentum
The merry go round has I=m*r^2/2
=185*2.53^2/2 =592.08 kgm^2

the person has I=m*r^2
=55.4*3.11^2
= 535.8 kgm^2

Given speed is 0.601 rev/sec is
0.601*2*3.14 rad/sec
= 3.774 rad/sec

for the person v/r=w
w=3.774/ 3.11
w = 1.213 rad/sec

So, we write,

592.08*3.774+535.8*1.213=(592.08+535.8)*w

solve for w,

Final Angular Speed is
w=2.557 rad/sec