Acceleration of a geared system

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 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: n² I_b alpha_a

So total torque on A to accelerate A and B :

= I_a alpha_a + n² I_b alpha_a
= ( I_a + n² I_b ) alpha_a
The quantity: I_a + n² 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 n² I_b / eta so the equivalent inertia of A and B becomes

I_a + (n² 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.

Example on a geared hoist

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David Grieve 13th November.