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Rotational kinetic energy(flywheel) problem

  • #1
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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.
 

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
12,403
2,838
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?
 
  • #3
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,798
1,666

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 ...
 
  • #4
123
0
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?
 
  • #5
TSny
Homework Helper
Gold Member
12,403
2,838
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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