Acceleration of a Geared System |

A pinion: A, moment of inertia I_{a} radius r_{a} drives a gear B,
moment of inertia I_{b} and radius r_{b}.

If it is assumed that the system is 100% efficient, *omega _{a}* and

work rate in = work rate out so:

If a torque M is applied to pinion A to accelerate the system then:

The acceleration ratios:

The torque on A to accelerate only A is: I

The torque required on B to accelerate only B is: I

Therefore the torque on A to accelerate only B is: n

So total torque on A to accelerate A and B :

= ( I

The quantity:

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 n^{2} I_{b} / *eta* . so the equivalent inertia of A and
B becomes

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

David Grieve 31st October 2001.