Question

It has been argued that power plants should make use of off-peak hours to generate mechanical energy and store it until it is needed during peak load times. One suggestion has been to store the energy in large flywheels spinning on nearly frictionless ball bearings. Consider a flywheel made of iron (density 7800 kg/m3 ) in the shape of a 10.0 cm thick uniform disk.

(a) What would the diameter of such a disk need to be if it is to store 10.0 megajoules of kinetic energy when spinning at 90.0 rpm 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

Step-by-step explanation:

90 rpm = 90 / 60 rps

= 1.5 rps

= 1.5 x 2π rad /s

angular velocity of flywheel

ω = 3π rad /s

Let I be the moment of inertia of flywheel

kinetic energy = (1/2) I ω²

(1/2) I ω² = 10⁷ J

I = 2 x 10⁷ / ω²

=2 x 10⁷ / (3π)²

= 2.2538 x 10⁵ kg m²

Let radius of wheel be R

I = 1/2 M R² , M is mass of flywheel

= 1/2 πR² x t x d x R² , t is thickness , d is density of wheel .

1/2 πR⁴ x t x d = 2.2538 x 10⁵

R⁴ = 2 x 2.2538 x 10⁵ / πt d

= 4.5076 x 10⁵ / 3.14 x .1 x 7800

= 184

R= 3.683 m .

diameter = 7.366 m .

b ) centripetal accn required

= ω² R

= 9π² x 3.683

= 326.816 m /s²