# Help help help please .rotational motion

these dang problems keep gettin me//// 20 kg girl on a turntable 4m from center... the turntable has a moment of inertia = 200kgm^2. assume the girl is a point mass. assume the girl and the turn table are rotating at 1 rad/sec. if the girl walks inward to a radius of 3 m, what is the new angular velocity of the girl and the turntable.....

ok so here is my work and once again i dont know what i did wrong.......
I=200
at radius 4m Ipm= .5(20)(4)^2 = 160 so at radius 4 the turntable and the girl have inertia = 360

ok and now for radius 3 m .5(20)(3)^2= 90 so at radius 3m turntable and girl have inertia =290 sooooo then i used MEi = MEf

.5(360)(1)^2 = .5(290)(omega)^2 giving me an angular velocity of 1.12rad per sec.......alas the answer is 1.37 and thats where im stuck

.... i feel the part im missing is when the girl is walking in,,,, im thinking of using Wnet=deltaKE by doing this i get total work done is (1) (Vsq)/r which i get to equal 4J... however i equated this at radius 4m..... im sure i could use MEi = MEf,,,,, im just frustrated right now, so im trying everything.... i guess im just confused how you account for the girl walking in toward the center WHILE its turning.....thank you a ton for anyone who can help me///

OlderDan
Homework Helper
The first thing you are missing is how to calculate the moment of inertia of the girl. Fix that first. Then it is a conservation of angular momentum problem. I don't see why you would have an omega^2.

ummm, well isnt Ipm for the girl .5mR^2 which is .5 * 20kg * 4sq = 160

and omega is angular velocity the omega squared comes from KErotation which is .5 I (omega)squared

OlderDan
Homework Helper
physics noob said:
ummm, well isnt Ipm for the girl .5mR^2 which is .5 * 20kg * 4sq = 160

and omega is angular velocity the omega squared comes from KErotation which is .5 I (omega)squared

I assume "pm" means "point mass". The moment of inertia of a point mass at a distance R from the axis of rotation is not .5mR^2. That is the moment of inertia of a disk rotating around its symmetry axis. The kinetic energy is not conserved. It is the angular momentum that is conserved.

ok... you are right....its not .5mr(sq) Ipm = mr(sq).....so is that all i did wrong..... i will rework the problem using this....fingers crossed

ok dan, i got the right answer, but im still confused,,,,, how do i know i can use Li = Lf...? i mean i used it and got the right answer, but how can i be sure its ok to use it, can you explain what it is actually stating.....does it imply a constant angular acceleration? or just that sumation of torque = 0? what exactly does that mean? can i make the assumption that T=0 because it doesnt explictly state that there are any forces,,, ie torque acting on the turntable?? oh thanks for the clear up on Ipm....that was the hang up

OlderDan