Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
Help with physics problem please!?
A horizontal circular platform (m = 80 kg, r = 4m) rotates about a frictionless vertical axle. A student (m = 63kg) walks slowly from the rim of the platform toward the center. The angular velocity, omega, of the system is 4.5 rad/s when the student is at the rim. Find omega when the student is 3m from the center.
I've tried several ways of going about solving this but i end up at a dead end. Please leave an in depth detailed explanation. Im talking step by step. Just answers wont help me Ive changed the values around a little bit from the original problem so i don't get accused of cheating. Thanks!
1 Answer
- electron1Lv 78 years agoFavorite Answer
When an object is a specific distance from the center, I = m * r^2
The total moment of inertia is the sum of the individual moments of inertia.
This is a conservation of angular momentum problem.
For the platform, I = ½ * 80 * 4^2 = 640
For the student at the rim, I = 63 * 4^2 = 1008
Total I = 640 + 1008 = 1648
Initial angular momentum = 1648 * 4.5 = 7416
For the student at the 3 m, I = 63 * 3^2 =567
Total I = 640 + 567 = 1207
Final angular momentum = 1207 * ωf
1207 * ωf = 7416
ωf = 7416 ÷ 1207 ≈ 6.144 rad/s
I hope this helps you understand how to solve this type of problem. If you have any specific questions, you can contact me at morrison60957@yahoo.com .
