Problem 8.61 A person of mass 71 kg stands at the center of a rotating merry-go-
ID: 1434911 • Letter: P
Question
Problem 8.61 A person of mass 71 kg stands at the center of a rotating merry-go-round platform of radius 3.4 m and moment of inertia 950 kgm2 . The platform rotates without friction with angular velocity 1.8 rad/s . The person walks radially to the edge of the platform.
Part A
Calculate the angular velocity when the person reaches the edge.
Express your answer using two significant figures.
Part B
Calculate the rotational kinetic energy of the system of platform plus person before and after the person's walk.
Express your answers using two significant figures. Enter your answers numerically separated by a comma.
Explanation / Answer
Angular momentum of the system remains conserved.
Initial angular momentum L(i) of the system is given by L=Iw
L= angular momentum,I=moment of inertia,w=angular velocity.
L(i)=950*1.8= 1710 kg.m^2.rad/s^2
(as inertia of man about axis is zero, L=0)
finally as there is no friction and as he goes to the end,
L(f) = (71*3.4*3.4+950)w=(1770.76)w
that was final angular momentum considering man also.
w=1.099 m/s^2 and that is approximately 1m/s^2
Initial rotational kinetic energies are given by E= Iw^2/2
{(950*3.4*3.4)/2}/ {(71*3.4*3.4+950)1*1}/2
= 6.20