Matching moments of inertia

Actuators have to develop forces to overcome both dissipative forces (friction) and inertial forces. That is, to move anything from position A to position B, you have to first apply force to accelerate the thing from a state of rest to some state of motion, then apply (usually smaller) force to maintain it in the accelerated-motion state against friction, and finally apply a force to decelerate...

December 1, 2006

Actuators have to develop forces to overcome both dissipative forces (friction) and inertial forces. That is, to move anything from position A to position B, you have to first apply force to accelerate the thing from a state of rest to some state of motion, then apply (usually smaller) force to maintain it in the accelerated-motion state against friction, and finally apply a force to decelerate it back to rest.

Moment of inertia is the rotational equivalent of mass. Just as mass quantifies an object’s tendency to resist changes in its translational velocity, moment of inertia quantifies its resistance to changes in rotational speed. The moment of inertia of any component increases directly with its mass and with the square of the distance that mass is from the rotation axis. That is, if you double the mass for the same size and shape, you double the moment of inertia, but if you keep the mass constant while doubling the size, the moment of inertia goes up by a factor of four.

The formula

So, in general the formula for moment of inertia is

I = Kmr2 ,

where I is the moment of inertia, m is the mass, and r is the object’s radius. The additional parameter K is a numeric value that depends on how its mass is distributed. When most of the mass is concentrated near the axis, such as for a solid ball, K is low. If it’s mostly far from the center, such as for a mass on the end of an arm, K is much larger.

Electric motor armatures are generally cylindrical with more or less uniform density, so K is approximately 0.5. Values for other shapes are given in most undergraduate physics textbooks and can be found by searching the Internet. Of course, you should be able to get the moment of inertia for the armature of any specific electric motor from its manufacturer.

Understanding the moment of inertia that the load presents is a little more complicated because loads come in all shapes and sizes, and perform complicated motions. Generally, start with the rotational equivalent of Newton’s second law:

M = Iu ,

where M is the moment or torque driving the rotation, and u is the angular acceleration.

Suppose that we want to, for example, accelerate a 100 gm weight to a linear velocity of 10 cm/sec in 0.1 sec. In a mechanism with a lever arm, if the arm length is 20 cm, then the angular acceleration is 10 cm/sec÷ 0.1 sec ÷ 20 cm = 5 radians/sec2.

The force applied to the mass would have to be (from Newton’s second law) 100 gm × 10 cm/sec÷ 0.1 sec = 10,000 dynes. Applying that force through the 20 cm lever arm gives a moment of 10,000 dynes × 20 cm = 2×105dyne-cm. Solving the rotational version of Newton’s second law for the moment of inertia gives 40,000 gm-cm2.

The motor’s maximum rotational speed will be (in radians per second) 10 cm/sec÷ 20 cm = 0.5 radians/sec. To convert from rad/sec to rpm, multiply by 9.55 to get just under 5 rpm.

Very few electric motors run that slowly. It’s a pretty typical output speed for a gearmotor, however. Suppose we find a gearmotor whose armature runs at 2,500 rpm while the output shaft runs at 5 rpm. That means the gear head has a ratio of 500:1.

Motor designers’ rule of thumb is that the armature moment of inertia should match the load moment of inertia, after accounting for the gear ratio. The way to do this is to divide the load moment of inertia by the square of the gear ratio to get the appropriate armature moment of inertia, which in this case turns out to be 0.16 gm-cm2.

Why match the armature moment of inertia to that of the load? If the load is too heavy, the motor won’t be able to control it. If it’s too light, most power will go into accelerating and decelerating the armature, rather than the load. Not only is this a waste of power, it could lead to overheating the motor.

How good does the match have to be? It turns out to be not all that critical. A mismatch by a factor of 2-3 will not cause problems, but it could be as large as 10:1 depending on the application. If the mismatch is much larger, however, serious problems may develop.

Based on a conversation with Mike Anselmo, application engineering manager, Alpha Gear Drives, Bartless, IL.

Author Information

Charlie Masi is Senior Editor for Control Engineering. He can be reached at charlie.masi@reedbusiness.com