# Finding a change in angular velocity, with out I equation

1. Oct 29, 2009

### Shepherd7

1. The problem statement, all variables and given/known data
A playground carousel is rotating about its center. The radius of the carousel is 1.5m, its initial angular speed is 3.14 rad/s and its moment of inertia is 125 kg m^2. A 40kg person climbs onto the carousel near its outer edge. What is the angular speed of the carousel after the person climbs aboard?

2. Relevant equations

Krot = 1/2 I W^2
L=IW
I = intergral r^2 dm
I = Icm + Md^2
Here's the kicker we can't use the equations for the moment of inertia about a rotating disk. Can't use I = 1/2MR^2

3. The attempt at a solution
I'm just lost I've tried solving this problem multiple ways with different equations.
This is my last attempt

Li=Lf
IiWi = IfWf
IiWi/Wf = If

1/2 Ii Wi^2 = 1/2 If Wf^2

1/2 Ii Wi^2 = 1/2 IiWi/Wf * Wf^2

Ii Wi^2 / Ii Wi = Wf
(125 * 3.14^2) / (125*3.14) = Wf = 3.14

Am I any where close!!!

2. Oct 29, 2009

### Troels

It seems that you are trying to apply conservation of angular momentum as well as conservation of rotational energy. you could do both, but not at the same time, as they are redundant; the former is simpler:

$$L_\mathrm{before}=L_\mathrm{after}=L$$

so

$$I_\mathrm{before}\omega_\mathrm{before}=I_\mathrm{after}\omega_\mathrm{after}$$

as you stated, yourself. To complete the argument however, you need to know the moment of inertial after the person gets on. Fortunately, that is a simple matter if you assume the person to be a point mass:

$$I_\mathrm{after}=I_\mathrm{before}+m_\mathrm{person}r^2$$

putting is together:

$$I_\mathrm{before}\omega_\mathrm{before}=\left(I_\mathrm{before}+m_\mathrm{person}r^2\right)\omega_\mathrm{after}$$

or

$$\omega_\mathrm{after}=\frac{I_\mathrm{before}}{I_\mathrm{before}+m_\mathrm{person}r^2}\omega_\mathrm{before}$$

As this qoutient is less than one, the angular velocity will decrease to conserve the angular momentum.