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
omega_{b} are the angular velocities of A and B, then
work rate in = work rate out so:
Torque_{a} omega_{a} = Torque_{b} omega_{b} so
omega_{b} / omega_{a} = n = Torque_{a} / Torque_{b}
If a torque M is applied to pinion A to accelerate the system then:
alpha_{a} and alpha_{b} are the accelerations of
A and B respectively.
The acceleration ratios: alpha_{b} / alpha_{a} = n
The torque on A to accelerate only A is: I_{a} alpha_{a}
The torque required on B to accelerate only B is: I_{b} alpha_{b}
= I_{b} n alpha_{a}
Therefore the torque on A to accelerate only B is: n^{2} I_{b} alpha_{a}
So total torque on A to accelerate A and B :
= I_{a} alpha_{a} + n^{2} I_{b} alpha_{a}
= ( I_{a} + n^{2} I_{b} ) alpha_{a}
The quantity:
I_{a} + n^{2} I_{b} 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 n^{2} I_{b} / eta so the equivalent inertia of A and
B becomes
I_{a} + (n^{2} I_{b})/eta
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: dalpha / dn =0. This is then solved for n.
David Grieve. Modified: 10th March 2011