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A flywheel is a solid disk that rotates about an axis that is perpendicular to t

ID: 2210062 • Letter: A

Question

A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provide a means for storing energy in the form of rotational kinetic energy and are being considered as a possible alternative to batteries in electric cars. The gasoline burned in a 324-mile trip in a typical midsize car produces about 4.65 x 109 J of energy. How fast would a 28.4-kg flywheel with a radius of 0.495 m have to rotate to store this much energy? Give your answer in rev/min.

Explanation / Answer

A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provide a means for storing energy in the form of rotational kinetic energy and are being considered as a possible alternative to batteries in electric cars. The gasoline burned in a 499-mile trip in a typical midsize car produces about 1.60 x 109 J of energy. How fast would a 40.5-kg flywheel with a radius of 0.438 m have to rotate to store this much energy? Give your answer in rev/min.
  • ANS
A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provide a means for storing energy in the form of rotational kinetic energy and are being considered as a possible alternative to batteries in electric cars. The gasoline burned in a 499-mile trip in a typical midsize car produces about 1.60 x 109 J of energy. How fast would a 40.5-kg flywheel with a radius of 0.438 m have to rotate to store this much energy? Give your answer in rev/min.
  • ANS
A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provide a means for storing energy in the form of rotational kinetic energy and are being considered as a possible alternative to batteries in electric cars. The gasoline burned in a 499-mile trip in a typical midsize car produces about 1.60 x 109 J of energy. How fast would a 40.5-kg flywheel with a radius of 0.438 m have to rotate to store this much energy? Give your answer in rev/min. R U asking us to do your homework for you? It is a word problem. First you have to extract what is given. Then you have to locate the formulas you need to convert what is given to the answer you are seeking.

Given:

The trip distance is irrelevant as we are given the amount of energy expended

= 1.6 x 109 J of energy. (I assume that the 109 is an exponent)

Flywheel mass 40.5kg
Flywheel radius .438 meters

Solve for rev/min

Rotational formula = http://hyperphysics.phy-astr.gsu.edu/hba…

Source(s):

Actually the best flywheels are not solid disks but will have most of their mass on the outer ring. Hint: But calling it a disk makes it easier to work the formula. R U asking us to do your homework for you? It is a word problem. First you have to extract what is given. Then you have to locate the formulas you need to convert what is given to the answer you are seeking.

Given:

The trip distance is irrelevant as we are given the amount of energy expended

= 1.6 x 109 J of energy. (I assume that the 109 is an exponent)

Flywheel mass 40.5kg
Flywheel radius .438 meters

Solve for rev/min

Rotational formula = http://hyperphysics.phy-astr.gsu.edu/hba…

Source(s):

Actually the best flywheels are not solid disks but will have most of their mass on the outer ring. Hint: But calling it a disk makes it easier to work the formula. R U asking us to do your homework for you? It is a word problem. First you have to extract what is given. Then you have to locate the formulas you need to convert what is given to the answer you are seeking.

Given:

The trip distance is irrelevant as we are given the amount of energy expended

= 1.6 x 109 J of energy. (I assume that the 109 is an exponent)

Flywheel mass 40.5kg
Flywheel radius .438 meters

Solve for rev/min

Rotational formula = http://hyperphysics.phy-astr.gsu.edu/hba… Actually the best flywheels are not solid disks but will have most of their mass on the outer ring. Hint: But calling it a disk makes it easier to work the formula.
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