Gearbox and flywheel question, How to calculate input torque

[David144]

Homework Statement

A flywheel is attached on the output of the gearbox. The output shaft rotates in the opposite direction to the input shaft at 5 times its speed.
The gearbox has an efficiency of 92%. If the flywheel is solid, has a mass of 50kg, a diameter of 1.5m and is to accelerate from rest to 300 revs min-1 in 1min:

a) Calculate the torque required at input T1.

Homework Equations

T=I*α

I=0.5*m*r2

α=I(ω2-ω1)/t

The Attempt at a Solution

Angular velocity of output shaft 300 revs min = 31.41593 rads/s (10π)
Angular velocity of input shaft 60 revs min = 6.283 rads/s (2π)

moment of inertia of flywheel = 0.5*50*0.75^2
=14.0625 Kg m^2

Change in angular momentum =I(ω2-ω1)
=14.0625(10π-2π)
=112π kg m^2 s

Torque = change of angular momentum / time
T=112π/60
=5.86NmI have seen a similar thread but with a completely different method and answer. Any advice please?

 

[Dr.D]

Looks like you did not take the gear box efficiency into your calculations. Probably more important (but beyond the scope of your problem), you did not take the mass moment of inertia of the gears into account.

 

[CWatters]

Change in angular momentum =I(ω2-ω1)
=14.0625(10π-2π)
=112π kg m^2 s

Torque = change of angular momentum / time
T=112π/60
=5.86Nm

That looks wrong to me.

The change in angular momentum I(ω₂-ω₁) is due to the acceleration, not the gearbox ratio.

The general equation for the torque required to accelerate a flywheel is..

T_f= Iα

where
I is the moment of inertia and
α is the angular acceleration

(Aside: There is a similarity with Newton's law F = ma)

To calculate the torque required at the flywheel..
T_f= Iα
α = (ω_f - ω_i)/t
so
T_f = I(ω_f - ω_i)/t

The initial angular momentum ω_i is zero (because it starts from rest).

I get a figure for the torque at the flywheel of just over 7Nm. The torque at the input to the gearbox would be 5 times that (if you ignore the losses in the gearbox). So I would expect the input torque to be at least 35Nm.

(Sorry I had to edit a few mistakes in this post)

 

[David144]

Hi

Thanks for the reply. I see from using the equation that I get an answer of 7.33 Nm for the torque at the flywheel , however is it as simple as multiplying by 5 as the shaft rotates at 5 times the speed?

If I use the same equations for the input shaft I get a lower figure

T= I(ωf - ωi)/t

as ωi=0

T=I*(ωf)/t

therefore P=I(ωf-ωi)/t

as I=14.062 kgm-2
ωf=2π rads-1
T=14.062*2π/60
T=1.466Nm

 

[CWatters]

Yes multiply by 5 to get the input torque.

The gearbox multiplies the input rpm by 5 to get the output rpm. This increase in speed doesn't come for free. The cost is a five times increase in torque required to turn the input.

Another way of looking at this is to say the gearbox increases the moment of inertia of the flywheel by a factor of five. Eg it's five times harder to turn.

 

[CWatters]

Experiment with a bicycle. If you select a fast gear the wheels turn faster but it's harder to pedal.

 

[David144]

Many thanks! I think i am there now.

7.33Nm x 5 = 36.65Nm

36.65 /0.92 to compensate for frictional losses = 39.84Nm