Question

A. Question: Jessica stretches her arms out .6 m from the center of her body while holding a 2 kg mass in each hand. She then spins around on an ice rink at 1.1 m/s. What is the combined angular momentum of the masses?

Work: For this question, I got L =2.64 kg*m^2/s. Is this right?

B. Question: If she pulls her arms in to 0.15 m, what is her linear speed if the angular momentum remains constant?

Work: This is the equation I used, L/(m)(r^2)
Which would make L= 2.64 from previous question. r= .15 m. m= 2 kg. v2= unknown
Substitution:
v2= 2.64/(.15)(2)(2) Is this step right?
What is the linear speed? (I tried to show all my work, so, you know where I got stuck.)

Answer 1

1. 2.64 kg m/s

The angular momentum of each mass is given by:

$L=mvr$

where

m = 2 kg is the mass

v = 1.1 m/s is the speed

r = 0.6 m is the distance from the center of rotation

Substituting the numbers, we get

$L=(2 kg)(1.1 m/s)(0.6 m)=1.32 kg m/s$

Since we have two masses in two different hands, the combined angular momentum will be twice the angular momentum of each mass:

$L=2 \cdot 1.32 kg m/s=2.64 kg m/s$

2. 4.4 m/s

The angular momentum must remains constant: this means that the only quantity that will change in the formula will be v, the speed.

The formula for the total angular momentum is:

$L=2mvr$

In this case, we have:

L = 2.64 kg m/s

m = 2 kg

r = 0.15 m (because she pulled her arms to this distance)

Re-arranging the equation, we can find the new speed v:

$v=\frac{L}{2mr}=\frac{2.64 kg m/s}{2(2 kg)(0.15 m)}=4.4 m/s$