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Question

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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 12.1 cm.

1) What would the diameter of such a disk need to be if it is to store an amount of kinetic energy of 15.0 MJ when spinning at an angular velocity of 90.0 rpm about an axis perpendicular to the disk .

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

thank you so much in advance and i will choose a best answer

Answer 1

Kinetic energy store is
Ke= 0.5 I w^2
where moment of inertia
I=0.5 mR^2
then
Ke= 0.25 m (Rw)^2

Using density mass can be found as
m= pV= 7800 x .121 x pi R^2

Ke= 0.25 [7800 x .121 x pi R^2] (Rw)^2
Ke= 741 R^4 w^2


R= (Ke/(741 w^2))^1/4
R= (15.0 E6/(741 (90/60)^2))^1/4
R=9.74 m

2) We know that centripetal acceleration is
ac= V^2/R
and V is related to angular velocity as
V= wR actually V= 2piR (since given in rpm)
we have

ac=(2piR)^2/R= (2pi)^2 R
ac=(2pi)^2 (9.74 )=385 m/s^2

Answer 2

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