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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/m^3 , in the shape of a uniform disk with a thickness of 11.3 cm.

Required:
a. What would the diameter of such a disk need to be if it is to store an amount of kinetic energy of 14.1 MJ when spinning at an angular velocity of 93.0 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 :

Answer:

Explanation:

kinetic energy = 14.1 MJ = 14.1 x 10ā¶ J

Let radius of flywheel be r .

volume of flywheel = Ļ€ rĀ² x t where t is thickness

= 3.14 x rĀ² x .113 mĀ³

= .04 rĀ² mĀ³

mass = volume x density

= .04 rĀ² x 7800 = 312.73 rĀ²kg

moment of inertia I = 1 / 2 mass x radiusĀ²

= .5 x 312.73 rĀ² x rĀ²

= 156.37 rā“ kg mĀ²

angular velocity Ļ‰ = 2Ļ€ x 93/60

= 9.734 rad /s

kinetic energy = 1/2 IĻ‰Ā² where Ļ‰ is angular velocity

= .5 x 156.37 rā“ x 9.734Ā²

= 7408.08 rā“

Given

7408.08 rā“ =  14.1 x 10ā¶

rā“ = .19 x 10ā“

r = .66 x 10

= 6.60 m .

Diameter = 13.2 m

b )

centripetal acceleration of a point on its rim = Ļ‰Ā² r

= 9.734Ā² x 6.6

= 625.35 m /sĀ²

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