Conservation of angular momentum in a spinning figure skater

• chaotiiic
In summary: You omitted the rotational inertia of the body when the arms are outstreached from the calculation. The 2.16 should be 0.40+ 2.16 = 2.56. In summary, the conversation discusses the calculation of a figure skater's final angular speed during a spin, taking into account the moment of inertia of his outstretched arms and hands and the remainder of his body. The correct formula for the moment of inertia for a slender rod pivoting about its center is 1/12ML^2, and for a thin cylindrical shell is 1/2M(R^2). The final angular speed is found using conservation of angular momentum, which takes into account the initial and final
chaotiiic

Homework Statement

The outstretched hands and arms of a figure skater preparing for a spin can be considered a slender rod pivoting about an axis through its center. When the skater's hand and arms are brought in and wrapped around his body to execute the spin, the hands and arms can be considered a thin-walled hollow cilinder. His hands and arms have a combine and arms have a combined mass of 8kg. When outstreched, they span 1.8m; when wrapped, they form a cylinder of radius 25cm. The moment of inertia about the axis of rotation of the remainder of his body is constant and equal to .40kg*m^2. If the skater's original angular speed is .40rev/s, what is his final angular speed?

Homework Equations

moment of inertia for slender rod, axis through center I= 1/12ML^2
moment of inertia for hollow cylinder 1/2M(R_1^2 + R_2^2)
conservation of angular momentum I_i*ω_i = I_f*ω_f

The Attempt at a Solution

I_i = 1/2*8*1.8^2 = 12.96
ω_i = 2.51
I_f = 1/2*8*(.9^2 + .125^2) = 3.3 +.4 = 3.7

putting those numbers in the conservation of energy forumla and i get ω_f = 8.7 rad/s. the back of my book says the answer is 7.16 rad/s.

First of all you do not use conservation of energy for this problem. You use conservation of angular momentum.

I_i = 1/2*8*1.8^2 = 12.96
Above is not correct formula. You miscopied it from your previous.

Secondly you need the formula for a thin cylindrical shell, not a hollow cylinder. You do not know the inner radius so how can you use it?

Last edited:
LawrenceC said:
First of all you do not use conservation of energy for this problem. You use conservation of angular momentum.

I_i = 1/2*8*1.8^2 = 12.96
Above is not correct formula. You miscopied it from your previous.

Secondly you need the formula for a thin cylindrical shell, not a hollow cylinder. You do not know the inner radius so how can you use it?
im still off.
I_i = 2.16
I_f = M(R^2) = .5 + .4 = .9 i used thin walled hollow cylinder
2.16*2.5= 5.42
5.42 = .9ω_f
did i do something wrong? am i suppose to use thin walled hollow sphere? it gets me 7.39

Last edited:
What about inertia of torso? You forgot it...

Last edited:
LawrenceC said:
What about inertia of torso? You forgot it...
is there an equation for that or is it already given in the problem. my book says nothing about inertia of torso

Torso means body!

Habeas corpus!

oh well it doesn't matter. i just took my test tonight and probably failed.

You omitted the rotational inertia of the body when the arms are outstreached from the calculation. The 2.16 should be 0.40+ 2.16 = 2.56.

1. What is conservation of angular momentum?

Conservation of angular momentum is a physical law that states that the total angular momentum of a system remains constant as long as there is no external torque acting on the system.

2. How does a spinning figure skater demonstrate conservation of angular momentum?

A spinning figure skater demonstrates conservation of angular momentum by changing their body position while spinning. When they extend their arms or bring them closer to their body, they are changing the distribution of their angular momentum but the total amount of angular momentum remains constant.

3. What is the relationship between angular velocity and moment of inertia in conservation of angular momentum?

The angular velocity of a system is inversely proportional to the moment of inertia in conservation of angular momentum. This means that as the moment of inertia increases, the angular velocity decreases and vice versa.

4. Can a figure skater change their angular momentum without any external forces acting on them?

No, a figure skater cannot change their angular momentum without any external forces acting on them. Conservation of angular momentum states that the total angular momentum of a system remains constant, so any change in the figure skater's angular momentum must be caused by an external torque.

5. How does friction affect conservation of angular momentum in a spinning figure skater?

Friction can cause a decrease in the angular velocity of a spinning figure skater. This is because friction acts as an external torque, causing the figure skater to slow down and decrease their angular momentum.

• Introductory Physics Homework Help
Replies
3
Views
429
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
10
Views
1K
• Introductory Physics Homework Help
Replies
335
Views
10K
• Introductory Physics Homework Help
Replies
9
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
538
• Introductory Physics Homework Help
Replies
9
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
382
• Introductory Physics Homework Help
Replies
13
Views
1K
• Introductory Physics Homework Help
Replies
12
Views
1K