Question

a disk shaped merry-go-round of radius 2.6m and mass 155kg rotates freely with an angular speed of .641 rev/s.a 60 kg person running torwards the edge at 3.41 m/s jumps on the edge. the person was moving in the same direction of the rotation, what is the angular velocity of the ride now? (angular momentum=momentum*radius)

Answer 1

Use conservation of angular mimentum law

Answer 2

Momentum

Answer 3

Angular momentum L=mvr=m*omega*r^2

Answer 4

L1i+L2i=L1f+L2f

Answer 5

Use Conservation of Angular Momentum
Initial Angular momentum of the System=Final Angular Momentum of the System
Initial Angular Momentum of Merry-Go-Round=
Inertia*Omega
Inertia of Disc(Merry-Go-Round) = (1/2)*155*(2.6)^2=523.9 kg-m^2
Omega of DiscMerry-Go-Round) = 2*(22/7)*0.641= 4.03 rad/sec
Initial Angular Momentum of Merry-Go-Round= 523.9*4.03 = 2110.87
Initial Angular Momentum of Man = mass*velocity*Perpendicular distance from axis of Rotation
Velocity of man = 3.41 + omega*r (Since 3.41 is relative to merry-go-round)
=3.41 +4.03*2.6 = 13.89 m/sec
Initial Angular Momentum of Man = 60*13.89*2.6 = 2166.84

Initial Angular Momentum of the System = 2110.87+2166.84 = 4277.71

Final Angular Momentum of the System = Final Angular Momentum of the disc
= 523.9*omega
Initial Angular Momentum = Final Angular Momentum
4277.71 = 523.9*omega

Omega = 8.165rad/sec
In rev/sec = (8.165*7)/(2*22) = 1.299 rev/sec

Answer 6

Thank you both, I got it now! This really helped me study for my test today:)