Rotational kinetic energy(flywheel) problem

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The discussion revolves around designing a flywheel to store the kinetic energy of a 600 kg car moving at 100 km/hr, with a maximum rotational speed of 12,000 rpm. Key calculations include determining the angular velocity (ω) as 1256.6 rad/s and the translational kinetic energy (TKE) as 231,481 J. The moment of inertia (Ig) for the flywheel was calculated, but confusion arose regarding its relation to mass and diameter. Participants suggested using the density of steel (approximately 7830 kg/m^3) to establish a relationship between mass and dimensions, leading to a calculated radius of 0.1476 m. The discussion emphasizes the importance of correctly applying formulas for rotational dynamics and material properties in engineering design.
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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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