Rotational kinetic energy(flywheel) problem

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Homework Help Overview

The discussion revolves around a problem involving the design of a flywheel intended to store the translational kinetic energy of a car. The flywheel is specified to be a uniform cylindrical disk made of steel, with constraints on its thickness and maximum rotational speed.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore calculations related to angular velocity, translational kinetic energy, and moment of inertia. There are attempts to relate the diameter and mass of the flywheel using the density of steel. Questions arise about the correct interpretation of units and how to incorporate the moment of inertia into the calculations.

Discussion Status

Participants are actively engaging with the problem, sharing calculations and questioning assumptions. Some guidance has been provided regarding the correct expressions for moment of inertia and the variability of steel density. However, there is no explicit consensus on the next steps or final values.

Contextual Notes

There are mentions of potential unit errors and the need for clarification on the density of steel, indicating that assumptions about material properties are still under discussion.

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Homework Statement


You have been assigned the task of designing a flywheel capable of storing
the translational kinetic energy of a 600 kg car traveling at 100 km/hr. The
flywheel is to be manufactured out of steel and the maximum rotational
speed is 12,000 rpm.
a. Assuming that the flywheel is a uniform cylindrical disk 50 mm in
thickness, calculate the mass and diameter of the flywheel.

Homework Equations


ω=2π/60
TKE=1/2mv^2
Ig=1/2mr^2
RKE=1/2Ig*ω^2

The Attempt at a Solution


Solving for ω[/B]
12000rpm*2pi/60=1256.6 rad/s

Solving for energy TKE
100*1000/3600=27.8m/s
1/2*600*27.8m/s=231481J
Solving for Ig(moment of inertia flywheel)
RKE=1/2Ig*ω^2

231481=1/2Ig*1256.6^2
Ig=0.293KG

At this point I'm not sure how to go further.
 
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Can you think of a way to relate the diameter and mass of the wheel using the fact that the wheel is made of steel?
 
orangeincup said:

Homework Statement


You have been assigned the task of designing a flywheel capable of storing
the translational kinetic energy of a 600 kg car traveling at 100 km/hr. The
flywheel is to be manufactured out of steel and the maximum rotational
speed is 12,000 rpm.
a. Assuming that the flywheel is a uniform cylindrical disk 50 mm in
thickness, calculate the mass and diameter of the flywheel.

Homework Equations


ω=2π/60
TKE=1/2mv^2
Ig=1/2mr^2
RKE=1/2Ig*ω^2

The Attempt at a Solution


Solving for ω[/B]
12000rpm*2pi/60=1256.6 rad/s

Solving for energy TKE
100*1000/3600=27.8m/s
1/2*600*27.8m/s=231481J
Solving for Ig(moment of inertia flywheel)
RKE=1/2Ig*ω^2

231481=1/2Ig*1256.6^2
Ig=0.293KG

At this point I'm not sure how to go further.
You've made a small mistake in the units for Ig, which should be ...
 
TSny said:
Can you think of a way to relate the diameter and mass of the wheel using the fact that the wheel is made of steel?
Density and volume?
So I googled the density of steel is 7830 kg/m^3

So would it be m=d*(h*π*r^2)?

I'm confused on what to do with the Ig value mostly, I know it's not the mass but I don't know how to put it into an equation to solve for mass.
I=mr^2?
I=1/2mr^2?

So if I say 0.293=1/2m*r^2
m=7830*(.05m*π*r^2)

0.293=1/2*7830*(.05m*π*r^2)*r^2

r=0.1476m?
 
That looks good. You used the correct expression for I for a cylinder. (As SteamKing pointed out, the unit for I is not kg.)

The density of steel varies, so I'm not sure what value you should use. If you are using a textbook, then you can see if there is a table of densities there that you can use.

Anyway, your answer looks correct to me.
 

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