# Torque required to overcome inertia and friction on a flywheel

1. Mar 23, 2013

### Aldebo

∂hi all. I sort of understand this question but not fully so here it goes:

A shaft carrying a flywheel of diameter 565mm and mass 210 kg is required to run at a speed of 710 rev/min. If the working speed is to be reached in 16 seconds from rest and the coefficient of friction between shaft and bearing is 0.3, determine,

The torque required to overcome inertia and friction .

so far i've worked out T=IA (Torque = Inertia x Angular acceleration), however where is the coefficient of 0.3 involved and how.

any help will be much appreciated

Relevant equations

T=Ia ∴ I=(mr²)/2

w2=w1+at ∴ a=(w2-w1)/t

However i'm not sure where the coefficient of friction is involved in all this, maybe a seperate equation I don't know

2. Mar 23, 2013

### Aldebo

any help would be much appreciated.

3. Mar 23, 2013

### rude man

Friction torque is defined by τ = μω, analogous to daming force F = cv. So μ has units of N-m-s/radian.

Knowing that, what is the 2nd order differential equation relating angular acceleration θ'' to net torque, and can you simplify it by making the independent variable ω instead of θ? Then solve the eq. for ω(t), set t = 16s, and solve for required applied torque.