A Couple more "Rotation of Rigid Bodies" questions

Tags:
1. Nov 8, 2015

sophixm

1. The problem statement, all variables and given/known data
1.
It has been argued that power plants should make use of off-peak hours (such as late at night) to generate mechanical energy and store it until it is needed during peak load times, such as the middle of the day. One suggestion has been to store the energy in large flywheels spinning on nearly frictionless ball-bearings. Consider a flywheel made of iron, with a density of 7800 kg/m3 , in the shape of a uniform disk with a thickness of 11.5cm . (also KE=10.6MJ, angular velocity=89rpm)
What would be the centripetal acceleration of a point on its rim when spinning at this rate?
2. A uniform, solid disk with mass m and radius R is pivoted about a horizontal axis through its center. A small object of the same mass m is glued to the rim of the disk.
If the disk is released from rest with the small object at the end of a horizontal radius, find the angular speed when the small object is directly below the axis.
Express your answer in terms of the variables m, R, and appropriate constants

2. Relevant equations
1. Centripetal acceleration= v^2/r
v=rω
2. Conservation of Energy
U=mgR
K=1/2Iω^2

3. The attempt at a solution
1. so in part one I found the diameter to be 7.26m through sheer luck. I'm going into office hours this week to understand this part of the problem, but I wont have time to before the assignment is due. The second part I don't have seems easy enough, I converted 89.0rpm to rps by dividing by 60, so 1.48rps, times the radius 3.63, square the answer, and divided by the radius. I know my answer is wrong though because the answer to the problem in the book is 327m/s^2 with slightly different values, and i get a very small value.
2. Ive done some searching online, and I'm pretty sure I'm just supposed to set the initial potential energy equal to the final kinetic energy, so mgR=(1/2)Iω^2. I think I'm just simplifying wrong, but with my incorrect answers I've gotten the feedback that the correct answer does not depend on: Rm, m, or mR. Using different solutions if come up with different sources, ive tried √(19.6m/Rm), √(2g/3R), and √(9.8m/.5mR), and obviously none of those are right.
Thanks for any help:)

2. Nov 8, 2015

SammyS

Staff Emeritus
Angular velocity is usually expressed in units of radians per second, rather than revolutions per second .