- #1
chaotixmonjuish
- 287
- 0
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.
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.