# Angular momentum of skaters

1. Dec 31, 2008

1. The problem statement, all variables and given/known data
Eight 60-kg skaters join hands and skate down an ice rink at 4.6 m/s. Side by side, they form a line 12m long. The skater at one end stops abruptly, and the line proceeds to rotate rigidly about that skater. Find the angular speed.

2. Relevant equations
$$L=I\omega$$
$$I_{rod} = \frac{1}{3}ML^2$$

3. The attempt at a solution
I said that initial angular momentum = $$rmv = 6m \times 480kg \times 4.6m/s = 13248 kg m^2 /s$$
final angular momentum = $$\frac{1}{3} \times 480 \times 12^2 \times \omega=23040\omega$$
so
$$13248=23040\omega$$
$$\omega=0.575 rad / sec$$
which according to the book is wrong. In the book it said 0.537 rad /sec. What have I done wrong here?

thanks

2. Jan 1, 2009

### djeitnstine

As far as I can tell...if he stops the group will rotate about his center not at the end of his body in which you are implying by r = 6. so find the length of each person etc...

3. Jan 1, 2009

### FedEx

You applied conservation of angular momentum. Thats good. But the equations are wrong.

Think : What is the initial axis of rotation and the final axis of rotation. And find the angular momentum about those axes.

4. Jan 1, 2009

### just__curious

Can you look at this like a weel? Where the stop person has velocity of zero and the far one has a velocity of 9.2m/s?

5. Jan 3, 2009

My guess was initial axis was the centre of mass of the skaters and the final axis was the end of the rod (I treated the line of skaters as a uniform rod)

6. Jan 3, 2009

### FedEx

Thats true. But the initial momentum is not mvr. There is a mistake in the final momentum also.

Hint 1: What is the moment of inertia of a rod through the center of mass?

Hint 2: The mass for the final momentum is not 480. Think. you have to substract the mass of one person. The person about which the rotation is taking place.

7. Jan 4, 2009

$$L=\frac{1}{12}ML^2$$
$$L=\frac{1}{3}ML^2$$