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MOI Physics

 
 
JackK
 
Reply Tue 30 Sep, 2014 06:23 pm
Problem:
Consider an ice skater spinning in a circle. We will approximate the body as a cylinder with a radius of 0.25 m. The total mass of the skater is 50kg. When the skater has her arms extended we will approximate her body as a cylinder with two rods attached. Each arm is 0.75 m long and has a mass of 2.5 kg. When the skater brings her arms in, we will approximate her body as a cylinder.

[​IMG] (if the image doesn't show go here: http://s3.amazonaws.com/answer-board-image/200751634536331488390360812501606.jpg)

Questions:
a)What is the moment of inertia of one of the skater's arms?

b)What is the total moment of inertia of the skater with her arms outstretched?

c)What is the total moment of inertia of the skater after she brings in her arms?

d)When the arms are extended, the skater spins at a rate of 5 rad/s. What is the angular velocity of the skater when she brings in her arms?

e)How does the skaters energy increase from when her arms are extended to when they are compressed- where is the energy coming from?

Attempt:
a)It says to use the parallel axis theorem I=Icm+md^2, but I didn't know how to apply it, so I did
I=1/2(MR^2)+(2.5kg)(0,525)^2
=1.40625+0.689 ~ 2.095 kg/m^2

b)I=1.40625+2(0.689)~2.784 kg/m

c)I=1/2(MR^2)
=1/2(50)(0.25^2)
=1.563 kg/m^2

d)IoWo=IW
(2.784*5rad)/1.563=W
W=8.9 rad/s

e)I was able to determine that there is an increase in the skater's energy, but I need to figure out where the energy is coming from.

I am apparently supposed to use the parallel axis theorem I= Icom+ md^2
for part a. How would i do this; what equation would i set up.

I know that, for the arm, im supposed to use ml^2/12 but how would I set it up
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engineer
 
  1  
Reply Tue 30 Sep, 2014 06:59 pm
@JackK,
For a, you need to use the parallel axis theorem and the moment of inertia formula for a cylinder on its end. From this list of moments of inertia, the MOI is mL^2/3 for a cylinder rotating about its end. Use that in the I=Icm+md^2 formula.

In part e, the skater's energy DOES NOT INCREASE. Energy is conserved. Because the MOI changed, the velocity changes because energy is conserved, but no energy is added to the system. (Remember that energy = 1/2 Iw^2 so if I goes down, w has to go up.)
JackK
 
  1  
Reply Tue 30 Sep, 2014 07:36 pm
@engineer,
Thanks for the help, I will try to work it out and get back to you in the next 10 minutes. As for part e, energy is increased (tutor said so)- that part is right. The question is where the energy comes from.

If you think about it the angular velocity increases from when her arms are extended to when her arms are compressed.
JackK
 
  1  
Reply Tue 30 Sep, 2014 07:56 pm
@JackK,
Okay. Using the parallel axis theorem, the MOI of the arm is :

I = Icm + md^2
I = (ml^2)/3 + md^2
I = [2.5(0.75)^2]/3 + 2.5(1)^2
=0.468 + 2.5
= 2.968
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engineer
 
  1  
Reply Tue 30 Sep, 2014 08:07 pm
@JackK,
For part e, I'm going to disagree with your tutor, energy does not increase. Energy = 1/2 I w^2 where I is the moment of inertia and w is the angular velocity. When the skater pulls her arms in, her MOI goes down but energy in conserved, so if E is constant and I goes down, the angular velocity must go up.
JackK
 
  1  
Reply Tue 30 Sep, 2014 09:05 pm
@engineer,
mmm ok, is my work correct for part a?
JackK
 
  1  
Reply Tue 30 Sep, 2014 09:07 pm
@JackK,
also, could you explain how i would do part d? i don't know where i went wrong
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