Calculating Maximum Kinetic Energy in a Flywheel

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In summary, the conversation is about calculating the maximum kinetic energy that can be stored in a flywheel with specific dimensions and mass to prevent structural failure. The formula for kinetic energy is discussed, as well as the moment of inertia and how it depends on the axis of rotation. The correct answer is provided and the use of parallel and perpendicular axis theorems is mentioned as being useful for other axes of rotation.
  • #1
makeAwish
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Homework Statement

Energy is to be stored in a flywheel in the shape of a uniform solid disk with a radius of = 1.30m and a mass of 66.0kg . To prevent structural failure of the flywheel, the maximum allowed radial acceleration of a point on its rim is 3600m/s² .
What is the maximum kinetic energy that can be stored in the flywheel?


The attempt at a solution

I took I = 3/2MR² (at rim)
where R = 2.6m
Then K = 1/2 * I * ω², where ω² = a/R

so K = (1/2)(3/2)(66kg)(2.6m)²(3600/2.6) = 463320J
which is wrong..
The ans is 7.72x10^4 J


can someone pls tell me where i gone wrong?
Thanks!
 
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  • #2
Moment of inertia of flywheel is 1/2*M*R^2 and radius is 1.3 m.
 
  • #3
Radial acceleration = [tex]m r^2 \omega[/tex]
Find out [tex]\omega[/tex]
Calc. K.E using the equation [tex]\frac{1}{2} I \omega^2[/tex]..
 
  • #4
janettaywx said:
Energy is to be stored in a flywheel in the shape of a uniform solid disk with a radius of = 1.30m and a mass of 66.0kg . To prevent structural failure of the flywheel, the maximum allowed radial acceleration of a point on its rim is 3600m/s² .
What is the maximum kinetic energy that can be stored in the flywheel?

I took I = 3/2MR² (at rim)
where R = 2.6m
Then K = 1/2 * I * ω², where ω² = a/R

Not quite. The equation is correct, but ω = v/r = 3600/1.3

As noted by rl.bhat a uniform disk has a moment of inertia of of 1/2mr².
 
  • #5
The Moment of Inertia isn't always 1/2 mr^2, is it?
It does depend upon the axis chosen, with M.I. of the new axis found out using parallel and perpendicular axis theorems.
 
  • #6
sArGe99 said:
The Moment of Inertia isn't always 1/2 mr^2, is it?
It does depend upon the axis chosen, with M.I. of the new axis found out using parallel and perpendicular axis theorems.

Rotating about its central axis it is.
http://hyperphysics.phy-astr.gsu.edu/hbase/icyl.html#icyl2

The || and ⊥ axis theorem are useful of course for other axes of rotation.
 
  • #7
i got it. thanks!
 

Related to Calculating Maximum Kinetic Energy in a Flywheel

What is rotational kinetic energy?

Rotational kinetic energy is the energy an object possesses due to its rotational motion. It is dependent on the object's angular velocity and moment of inertia.

How is rotational kinetic energy calculated?

The formula for rotational kinetic energy is KE = 1/2 * I * ω², where KE is the kinetic energy, I is the moment of inertia, and ω is the angular velocity.

What factors affect an object's rotational kinetic energy?

The rotational kinetic energy of an object is affected by its mass, moment of inertia, and angular velocity. As these values increase, so does the object's rotational kinetic energy.

What is the relationship between rotational and translational kinetic energy?

Rotational and translational kinetic energy are different types of kinetic energy, but they are interrelated. In a rolling motion, an object possesses both rotational and translational kinetic energy. The total kinetic energy is the sum of these two types of energy.

How is rotational kinetic energy used in practical applications?

Rotational kinetic energy is used in many practical applications such as power generation, transportation, and sports. For example, the rotational kinetic energy of wind turbines is converted into electrical energy, and rotational kinetic energy is used to power vehicles such as cars and bicycles. In sports, rotational kinetic energy is utilized in activities like throwing a ball or swinging a golf club.

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