Acceleration of a Geared System

A pinion: A, moment of inertia Ia radius ra drives a gear B, moment of inertia Ib and radius rb.

If it is assumed that the system is 100% efficient, omegaa and omegab are the angular velocities of A and B, then
work rate in = work rate out so:

Torquea omegaa = Torqueb omegab so
omegab / omegaa = n = Torquea / Torqueb

If a torque M is applied to pinion A to accelerate the system then:
alphaa and alphab are the accelerations of A and B respectively.
The acceleration ratios: alphab / alphaa = n

The torque on A to accelerate only A is: Ia alphaa
The torque required on B to accelerate only B is: Ib alphab = Ib n alphaa
Therefore the torque on A to accelerate only B is: n2 Ib alphaa

So total torque on A to accelerate A and B :

= Ia alphaa + n2 Ib alphaa
= ( Ia + n2 Ib ) alphaa

The quantity: Ia + n2 Ib may be regarded as the equivalent moment of inertia of the gears referred to shaft A. This principle may be extended to any number of wheels geared together (and shafts, drums components fixed to them) the moment of inertia of each wheel/shaft in the train being multiplied by the square of it's speed ratio relative to the reference wheel. Thus hoists, etc can be reduced to an equivalent moment of inertia at the motor shaft or drum shaft.

Care must be taken when using equivalent inertias in problems where the efficiency (eta) of the gearing is given. If the efficiency of the gearing is eta then the torque required on A to accelerate B is n2 Ib / eta so the equivalent inertia of A and B becomes

Ia + (n2 Ib)/eta

To find the gear ratio which gives the maximum acceleration, an expression for alpha is found in terms of n, Ia, Ib, resisting and accelerating torques etc. This expression for alpha is then differentiated with respect to n and this is then equated to zero, ie: dalpha / dn =0. This is then solved for n.

David Grieve. Modified: 10th March 2011