Solve Angular Velocity of Circular Platform with 73.3kg Student

[chaotixmonjuish]

A horizontal circular platform (M = 92.1 kg, r = 3.49 m) rotates about a frictionless vertical axle. A 73.3 kg student walks slowly from the rim of the platform toward the center. The angular velocity of the system is 3.5 rad/s when the student is at the rim. Find the angular velocity of the system when the student is 1.61 m from the center.

So I have these calculations:

I=mr^2
I=92.1*3.49^2+73.3*3.49^2=2014.59
I=92.1*3.49^2+73.3*1.61^2= 1311.79

Rotational Kinetic Energy:
1/2*2014.59*3.5^2=1/2*1311.79*x^2

I haven't been able to get the right answer of 6.78 rad/s. I'm reviewing for a test by reworking my homework, but I'm not sure how I got that answer.

 

[hage567]

Use conservation of angular momentum.

$$L_i = L_f$$

so $$I_i \omega_i = I_f \omega_f$$

 

[chaotixmonjuish]

I tried that and still got a wrong answer:5.375

 

[hage567]

OK I just noticed you are using the same moment of inertia expression for both "objects". You need to use this

for the wheel $$I = \frac{1}{2}MR^2$$

for the student, take as a point mass so $$I = mr^2$$