# Help help help please .rotational motion

#### physics noob

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///

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#### 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.

#### physics noob

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.

#### physics noob

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

#### physics noob

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

Homework Helper
physics noob said:
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
The key to the problem is to recognize that there are no external torques acting. Under those conditions, the angular momentum of the system will be conserved. This is analogous to conservation of linear momentum in a system where only internal forces are acting. These internal forces and torques can and do dissipate energy, but they cannot change momentum or angular momentum.

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