Answer
given,
radius of merry - go - round = 4 m
mass of the disk = 120 kg
speed of the merry- go-round = 0 rad/s
speed = 2 m/s
mass of student = 60 kg
[tex]I_{disk} = \dfrac{1}{2}MR^2[/tex]
[tex]I_{disk} = \dfrac{1}{2}\times 120 \times 4^2[/tex]
[tex]I_{disk} = 960 kg.m^2[/tex]
initial angular momentum of the system
[tex]L_i = I\omega_i + mvR[/tex]
[tex]L_i =960 \times 0 + 60 \times 2 \times 4[/tex]
[tex]L_i =480\ kg.m^2/s[/tex]
final angular momentum of the system
[tex]L_f = (960 + 60\times 4^2)\omega_{f}[/tex]
from conservation of angular momentum
[tex]L_i = L_f[/tex]
[tex]480 = (1920)\omega_{f}[/tex]
now,
change in kinetic energy of the system
initially the merry-go-round was on rest KE_i = 0
[tex]\Delta KE = KE_f- KE_i[/tex]
[tex]\Delta KE = (\dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2)- 0[/tex]
[tex]\Delta KE = \dfrac{1}{2}\times 60 \times 2^2 + \dfrac{1}{2}\times 960 \times( {\dfrac{v}{r})^2[/tex]
[tex]\Delta KE =120 + 480 \times \dfrac{4}{16}[/tex]
[tex]\Delta KE = 240\ J[/tex]