- #1

andycampbell1

- 35

- 0

## 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[tex]\alpha[/tex]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?