Calculating Braking Torque for a Flywheel

[andycampbell1]

Homework Statement

A flywheel, together with its shaft, has a total mass of 300 kg with a radius of gyration of 900 mm. Friction in the bearing produces the equivalent of a torque 70 Nm. Calculate the braking Torque required to bring the the flywheel to rest from a speed of 12 rev/s in 8 seconds.

Homework Equations

ΣF =ma
• F – Force (N)
• m –mass (kg)
• a=(v‐u)/t (m/s2)
• a=(v2‐u2)/2s


ΣT =Iα
• T‐ Torque (Nm)
• I =mk2(flywheel
moment)
• α=(ω2‐ω1)/t
• α=(ω2
2‐ ω1
2)/2Θ

F=T/r
m =I/k2
a=α r
v=ω r
s=Θ r

T = F x r
I = mk2

α = a/ r
ω = v / r
Θ = s / r

The Attempt at a Solution

I have worked out the moment of inertia as I=mk^2 = 300x900^2= 243kgm^2
I also converted the rev/s to rad/s so that would be 12x2pi = 75.4 rad/s
I devided the rads/s by the time to get angular acceleration = 9.42
Then I used the formula T= m$$\alpha$$k^2 which gave me 300x9.42x900^2 this gave me the answer 2289. The solution to the answer gives me T=2221 Nm where have I went wrong?

 

[tiny-tim]

hi andy!

A flywheel, together with its shaft, has a total mass of 300 kg with a radius of gyration of 900 mm. Friction in the bearing produces the equivalent of a torque 70 Nm. Calculate the braking Torque required to bring the the flywheel to rest from a speed of 12 rev/s in 8 seconds.
…
… this gave me the answer 2289. The solution to the answer gives me T=2221 Nm where have I went wrong?

if you subtract the friction torque, isn't that roughly correct?

 

[andycampbell1]

Yep ha, it was right under my nose and I missed it, thanks for the help anyway.