Moment of inertia ice skater problem

In summary: I see, so I think that instead of using mL^2/2 as the MOI from the center of the rod, I just used the cylinder MOI which was completely wrong. I will try again using the correct MOI for the arms. Thank you for your help!In summary, the problem involves finding the moment of inertia of an ice skater spinning with her arms extended and compressed. The skater's body is approximated as a cylinder with a radius of 0.25m and a total mass of 50kg. When her arms are extended, they are approximated as two rods with a length of 0.75m and a mass of 2.5kg each. The parallel axis theorem is used
  • #1
jasonmoon
13
0
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.

http://s3.amazonaws.com/answer-boar...o figure out where the energy is coming from.
 
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  • #2
It's hard to follow your steps if you plug in numbers straight away. A good habit to get into is keeping everything in symbols until the final step. It has many advantages.
Let the cylinder radius be R, the length of each arm L, etc. and post your equation in those terms. (In particular, the 0,525 looks wrong.)
 
  • #3
haruspex said:
It's hard to follow your steps if you plug in numbers straight away. A good habit to get into is keeping everything in symbols until the final step. It has many advantages.
Let the cylinder radius be R, the length of each arm L, etc. and post your equation in those terms. (In particular, the 0,525 looks wrong.)
I see. Do you know how to apply the parallel axis theorem to solve this problem? This was my mistake and I didn't know how to do so.
 
  • #4
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.

http://s3.amazonaws.com/answer-boar.... The parallel axis th. formula is I=Icm+md^2
 
  • #5
a)It says to use the parallel axis theorem I=Icm+md^2, but I didn't know how to apply it ...
... that's a problem then.
You need to learn how to apply it - you will have some course notes about it, some examples, and there are numerous examples online.

You know the formula for the moment of inertia of a cylinder taken about different axes that go through the com right?
http://en.wikipedia.org/wiki/List_of_moments_of_inertia

You also know which com axis is needed for the problem, and you can work out how far (d) that axis is from the rotation axis.
http://en.wikipedia.org/wiki/Parallel_axis_theorem

I=1/2(MR^2)+(2.5kg)(0,525)^2
... MR^2/2 is the moment of inertia of a solid cylinder rotated about an axis passing through the center of it's endcaps.
Which way do the arms rotate for (a)?
 
  • #6
jasonmoon said:
I see. Do you know how to apply the parallel axis theorem to solve this problem? This was my mistake and I didn't know how to do so.
Ok. If the moment of inertia about the mass centre of an object of mass M is I, then its moment about a point at distance X from its mass centre is ##I+MX^2##
 
  • #7
Simon Bridge said:
... that's a problem then.
You need to learn how to apply it - you will have some course notes about it, some examples, and there are numerous examples online.

You know the formula for the moment of inertia of a cylinder taken about different axes that go through the com right?
http://en.wikipedia.org/wiki/List_of_moments_of_inertia

You also know which com axis is needed for the problem, and you can work out how far (d) that axis is from the rotation axis.
http://en.wikipedia.org/wiki/Parallel_axis_theorem... MR^2/2 is the moment of inertia of a solid cylinder rotated about an axis passing through the center of it's endcaps.
Which way do the arms rotate for (a)?
I see, so I think that instead of using mL^2/2 as the MOI from the center of the rod, I just used the cylinder MOI which was completely incorrect.
 
  • #8
That's right - the arms would count as cylinders rotating end-over-end rather than around their central axis.
The only way arms would rotate round the center is if you were unscrewing them (or turning somersaults) ;)
 
  • #9
So, for part a I did

I = I cm + md^2
= 1/2 mr^2 + md^2
=1/2*45*0.25^2+2,5*0.625^2
=1.71+0.97
=2.70

part b was
I = 1/2mr^2 + 2md^2
=1.71 + 5*0.625^2
=3.67

Is this correct?
 
  • #10
Always include your reasoning with your working.

For (a) you appear to have still use ##I_{com}=\frac{1}{2}mr^2## for a cylinder rotating end-over-end - you have already been told that this is incorrect. If you do not take advise, we cannot help you.

For (b) - isn't the person's body included in their total moment of inertia? What about two arms?
 
  • #11
I thought that for b I calculated the moo of the body and the arms
I used I=mr^2 for her body and 2md^2 for her armd
 
  • #12
[QUOIowa lTE="jasonmoon, post: 4866739, member: 520129"]I thought that for b I calculated the moo of the body and the arms
I used I=mr^2 for her body and 2md^2 for her armd[/QUOTE]
Should I have done I=ml^2/2 for Icom?
 
  • #13
jasonmoon said:
I thought that for b I calculated the moo of the body and the arms
I used I=mr^2 for her body and 2md^2 for her armd
... but that is not consistent with the equation for the moment of inertia of just one arm, according to your (incorrect) calculation in part (a) above. Therefore it cannot be correct.

[If m is the mass of the body, then how can md^2 be an arm: are the arms the same mass as the body? You have used mr^2/2 for the com moi for an arm so how does that suddenly become 2md^2 for two arms? However, the arms have a different orientation to the body, so the com moi must have a different equation. mr^2 is the moi of a point mass going in a circle - not a cylinder.]


Lets get (a) correct first. Try drawing a picture - it's somewhat like the Android robot with arms outstretched.

The arms are small cylinders with mass m, length l, and radius r.
The body is a large cylinder with mass M, length L, and radius R.

Which arm-cylinder (com) axis is parallel to the axis of rotation?
What is the equation for the moment of inertia for a cylinder rotating about that axis? (Look it up.)
In terms of arm and body dimensions, what is the equation for the distance between the arm-cylinder com axis and the center of rotation?
Therefore: what is the equation for the moment of inertia of just one of the arms? ##I_{arm}=##?
 
Last edited:
  • #14
So Iarm=1/2ml^2
 
  • #15
Iarm=mL2/12

therefore,

I= (2.5 kg)*(0.75 m)2/12
I= (2.5)*(0.5625)/12
I=0.117
 
  • #16
From my answer above, the MOI of a skater's arm via the || axis theorems is
0.117 + 0.97 = 1.08
(0.97 is from my solution of md^2 above)
 
  • #17
jasonmoon said:
Iarm=mL2/12
... that's better.
Only I wanted you to use L was the length of the body.
The reason for that was so we can understand each other, and you have had this habit of using the same variable name for two things.
So for the arm, ##I_{com,arm} = \frac{1}{12}ml^2##

Don't do the numbers right away.
When you keep the variables, you make your reasoning more clear.
To give you full marks - the marker needs to be able to, easily, see why you did what you did.

jasonmoon said:
From my answer above, the MOI of a skater's arm via the || axis theorems is
0.117 + 0.97 = 1.08
(0.97 is from my solution of md^2 above)
... OK - so: $$I_{arm}=I_{com,arm} + md^2 = \frac{1}{12}ml^2 + ?$$
Please express d in terms of the variable names in post #13?

Basically getting this question right takes concentration and communication.
 
  • #18
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

ml^2/3 is the actual MOI for the arm because the axis is at the end of the cylinder - ml^2/12 is the formula for COM in the center of the cylinder, and it is not applicable here
 
  • #19
I = Icm + md^2
I = (ml^2)/3 + md^2
... do you mean to write: ##I_{cm}=ml^2/3##? Because that is what the above says.
... what is d in terms of the variable names in post #13?
I = [2.5(0.75)^2]/3 + 2.5(1)^2
... you seem to have put d=1m ... where did you get that figure from?

ml^2/3 is the actual MOI for the arm because the axis is at the end of the cylinder
... but earlier in the same post you stated that the actual MOI for the arm was:
I = (ml^2)/3 + md^2
... which is it?

You have to be more careful about your communication - if you pull this in an exam, the marker will get confused about what you mean.
Confusing the marker costs you marks.

Could it be that you want to use the "axis about the end of the rod" formula instead of the com formula? In which case, you cannot call it Icm: that's misleading. You can use either formula, it just changes "d".

Note:
If you do not follow suggestions, I cannot help you.
If you do not answer questions, I cannot help you.
 
  • #20
Yes, I do mean to write Icm = ml^2/3 - I meant to say that this should be used instead of ml^2/12
What I wrote for the MOI is what I think is the right formula, taking into account for the second half of the formula

d= distance between com and previous com
If the axis is in the center of the arm, then d is 0.625, and the formula for I is ml^2/12.
If the axis is at the end of one arm, then d is 1.
If the axis is right next to her body; in between of her arm and her body, then d is 0.25.

What I am trying to ask is which axis should I use - what is the correct one to go with? I am leaning on the axis being in the center of the arm as it would be the center of mass for the rod itself.
 
  • #21
What I am trying to ask is which axis should I use - what is the correct one to go with? I am leaning on the axis being in the center of the arm as it would be the center of mass for the rod itself.
What difference does it make?
Do it both ways and see.
 

Related to Moment of inertia ice skater problem

1. What is the "Moment of Inertia Ice Skater Problem"?

The "Moment of Inertia Ice Skater Problem" is a physics problem that involves calculating the moment of inertia of an ice skater as they change their rotational speed by bringing their arms and legs closer or further from their body. It is used to demonstrate the concept of conservation of angular momentum.

2. How do you calculate the moment of inertia of an ice skater?

The moment of inertia of an ice skater can be calculated by using the formula I = mr^2, where m is the mass of the skater and r is the distance from the axis of rotation to the skater's center of mass. This formula can be applied to each segment of the skater's body (arms, legs, torso) and then added together to get the total moment of inertia.

3. What happens to the moment of inertia when an ice skater brings their arms and legs closer to their body?

When an ice skater brings their arms and legs closer to their body, their moment of inertia decreases. This means that they can rotate faster without changing their angular momentum, as the product of moment of inertia and angular velocity must remain constant according to the law of conservation of angular momentum.

4. How does the moment of inertia affect an ice skater's rotational speed?

The moment of inertia has an inverse relationship with an ice skater's rotational speed. As the moment of inertia decreases, the skater's rotational speed increases. This relationship is demonstrated in the "Moment of Inertia Ice Skater Problem" as the skater's arms and legs are brought closer to their body, resulting in a faster rotation.

5. What other real-life examples demonstrate the concept of conservation of angular momentum?

Other real-life examples that demonstrate the concept of conservation of angular momentum include a figure skater performing a spin, a diver tucking their body to increase their rotational speed, and a planet orbiting around a star. In all of these cases, the moment of inertia changes, but the angular momentum remains constant.

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