Diameter of a disk? Please help, THANKS!?

Question

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 p, in the shape of a uniform disk with a thickness of L.

A.)What would the diameter of such a disk need to be if it is to store an amount of kinetic energy of E when spinning at an angular velocity of w about an axis perpendicular to the disk at its center?


B.) What would be the centripetal acceleration of a point on its rim when spinning at this rate?

Answer 1

Consider a ring of infinitesimal width dr
The area of the ring would be pi * ((r+dr)^2 - r^2) = pi * ( 2rdr + dr^2)
dr^2 is zero in the limit (when we apply it). neglect it.
area of the ring = pi * 2rdr
Multiply this by the thickness and density to get its mass.
mass = p*L*pi*2rdr.
The kinetic energy for this ring would be
(1/2) * m * v^2
v = wr, where w is the angular velocity.
=> k.e. of this ring = (1/2) * p*L*pi*2rdr * (wr)^2
= pi * p * L * w^2 * r^3 * dr
Integrate this from 0 to R where R is the radius of the disk
This yields,
E = pi * p * L * w^2 * R^4 / 4which is the k.e. of the disk.
=> R = (4E / (pi * p * L * w^2))^(1/4)

U can calculate the acceleration as v^2 / R