A 37-kg boy running at 4.5 m/s jumps tangentially onto a small circular merry-go
ID: 1463830 • Letter: A
Question
A 37-kg boy running at 4.5 m/s jumps tangentially onto a small circular merry-go-round of radius 2.0 m and rotational inertia 2.0×102 kgm2 pivoting on a frictionless bearing on its central shaft. Merry-go-round initially rotating at 1.0 rad/s in the same direction that the boy is running.
a)
Determine the rotational speed of the merry-go-round after the boy jumps on it.
Express your answer to two significant figures and include the appropriate units.
B)
Find the change in kinetic energy of the system consisting of the boy and the merry-go-round.
Express your answer to two significant figure and include the appropriate units.
C)
Find the change in the boy's kinetic energy.
Express your answer to two significant figure and include the appropriate units.
Part D
Find the change in the kinetic energy of the merry-go-round.
Express your answer to two significant figures and include the appropriate units
Explanation / Answer
initial angular momentum of mery go round = L1 = I*w1
initial angular momentumof boy = L2 = m*v1*r
after the boy jumps to merry go round
finaal angular momentum of mery go round = L1' = I*w2
final angular momentumof boy = L2' = m*r^2*w2
from momentum conservation
Li = Lf
(200*1) +(37*4.5*2) = (200*w2)+(37*2^2*w2)
w2 = 1.53 rad/s
_____
B)
KEi = 0.5*I*w1^2 + 0.5*m*v1^2 = (0.5*200*1^2)+(0.5*37*4.5^2) = 476.62 J
KEf = 0.5*I*w2^2 + 0.5*m*r^2*w2^2 = (0.5*200*1.53^2)+(0.5*37*2^2*1.53^2) = 407.32 J
change = KEi -KEf = 69.3 J
(C)
dK2 = 0.5*m*v1^2 - m*r^2*w2^2 = 201.4 J
(D)
dK1 = 0.5*I*w2^2 - 0.5*I*w1^2 = 134.09 J