Mechanical Design: Inertia

[Rich Dionne]

The last few weeks in our mechanical design class have had students wrestling with two concepts: the force required to overcome an object’s inertia, and the force required to overcome friction between and object and the world around it. In concept, neither of these topics are inherently complicated to understand; however, in practice—when applied to complex, real-world problems—it can be difficult to determine how and why to deploy particular equations.

 

Inertia

Inertia, as stated by Newton, represents the tendency of a stationery object to stay at rest, and a moving object to stay moving. Because of this tendency, it takes a certain amount of force to cause a stationery object to move and a moving object to stop moving (or otherwise change its velocity). This force is dependent on two things: the object’s mass, and the rate of acceleration desired. You remember mass: the property of an object that represents how much stuff makes it up. Mass is different from weight; weight is a force: the result of the pull of gravity on our mass. This is why we weigh differently when on the moon, say, or in orbit around the Earth, but need to provide the same amount of push to move around: we have the same mass, but are acted upon differently by different pulls of gravity. In order to determine the force required to overcome an object’s inertia, we need to know that object’s mass, as opposed to its weight.

 

The equation to determine the force required to overcome inertia is a simple one:

Where F_a is the force required (i.e., how hard you have to push on an object to get it moving if it is stationary) and, in the US, is usually in units of pounds; m is the mass of an object, and a is the rate of acceleration desired for the movement. An object’s weight is determined by the same formula: your weight is the force you feel when your mass is accelerated by gravity. To determine an object’s mass is as simple as substituting known quantities into the above equation (the weight of an object and the rate of acceleration due to gravity) and solving for mass. We can determine the mass of a 100 pound object (the acceleration due to gravity on Earth is 32.2 ft/sec²):

(Slugs is an imperial unit of measurement to represent the mass of an object; it’s a little easier to say and write than “pound-seconds-squared per foot.”)

After you’ve determined the mass of the object to be moved, you multiply that value by how quickly you intend to get the object moving (i.e., its rate of acceleration) to determine how much force you need to apply to make that acceleration happen. (Determining the rate of acceleration is part of the development of the motion profiles for a particular move, as discussed [last week].) For example, a flown unit which weighs 100 pounds must accelerate at about 3.2 ft/sec². We know from above that the mass of a 100 pound object is 3.11 slugs; substituting mass and acceleration in our formula, we get:

In order to get this 100# object moving, we’d need to apply about 10 pounds of force to overcome inertia. (There’s still friction to deal with, but let’s save that for next week!)

 

Fair enough. The application of the formula is pretty straightforward if you consider accelerating an object that is at rest. But in a scenic move, in what other situations might we need to apply this formula? There are two that immediately come to mind: decelerating an object (i.e., getting a planned movement to stop) and the deceleration that occurs in an emergency-stop situation.

 

Consider that during the move described above, our scenic unit reaches a maximum velocity of 8 ft/second. That’s really fast, but the motion profile I used to determine the properties of this move is triangular, meaning we accelerate to this speed over half the move, and decelerate for the other half of the move—in other words, we’re only travelling at this rate for a fraction of a second. Because the motion profile is triangular, and we’re accelerating and decelerating each for half the move, the rate of deceleration is the same as the rate of acceleration—and, consequently, the force required to decelerate the mass of the unit is the same as that to accelerate it.

 

But what if we had to stop this unit abruptly because of some unexpected, dangerous situation? Assuming that we’re using an integrated-brake-motor, we know that the application of the brake will stop the unit at a particular rate; in our case, we might determine that the application of the brake in an e-stop situation might bring the entire system to a halt in about half a second. In a worst-case scenario—the brake kicking on at our highest velocity—we’d be decelerating from 8 ft/sec to 0 ft/sec in 0.5 seconds: a rate of deceleration of 16 ft/sec². When the brake kicks on, the moving unit will pull on the system with a force equal to that required to decelerate the mass of the unit at 16 ft/sec²:

The force generated by the e-stop situation is nearly five times that generated by our relatively-gentle acceleration during the actual move! This is important information, as it means that all of the elements of our system need to be rated to handle this five-times-higher load, or when that brake kicks in, there’s a chance, for example, that the shackle connecting the unit to the aircraft cable on the winch drum might snap, and the wagon will keep moving. (Obviously, this is less than likely with a 50-pound load, but remember that our unit was fairly light-weight at 100 pounds; a typical scenic wagon might easily be as much as 10 times this weight.)

 

Next week: Friction