# Thread: Rotational inertia

1. ## Rotational inertia

Hello,

It is known that T=Ia is used for relating rotational torque, acceleration, and inertia.. I have a couple questions concerning this...

• Is the Torque from this formula ONLY the torque due to the system inertia? Or does it cover all torques? To help explain, picture an IC engine.. For the most part, cylinder pressure, inertias and frictions are what effect torque W.R.T. the crankshaft... So If I hook up a data logger and directly monitor torque and angular acceleration, can I simply solve for I to get my system inertia? Or is there more to it than that.

• Now say I turn the crank with a drill, there is obviously torque here, however angular acceleration is zero... So say I derived the inertia via 3D software and now I want to calc. torque using T=Ia... My result will be zero because a=0, but this obviously is not accurate because torque is required to overcome things like friction, inertias, gravity etc...

If somebody could shed some light on this for me, I would greatly appreciate it.  Reply With Quote

2. The 'a' is angular acceleration so there must be a change in rotational speed involved. Starting with the second part of the question, first:

If you spin the crank and remove the drill, the rotation will slow to a stop owing to friction. You can measure the torque required to turn the crank in the bearings with a torque wrench or other method. Spinning the crank up to drill speed and then removing the power and measuring the time required for it to come to a stop would permit you to calculate an angular de-acceleration.

You can also roll the crank down a couple of inclined ramps under the main journals and get a time to descend owing to the force of gravity. There are also formulas derived from the pendulum formula.

Perhaps the best compilation of methods is in 'A Handbook of Torsional Vibration" published by Cambridge University. This 1950's classic has been recently reprinted.

Which brings us to the first part of the question and data logging. The periodic inputs of the piston(s) through the connection rods do cause accelerations of the crankshaft ans subsequently slow the rotation during intake, compression and exhaust. The flywheel is designed to have sufficient rotational inertia to absorb and release energy so that the change in speed is kept to a small percentage of the average crankshaft speed.

The flexible nature of the crankshaft can result in some torsional (twisting) vibrations of the crankshaft. You may have noticed vibration dampers on the ends of some crankshafts.

Your data logger will pick up these vibrations along with the torque pulses. The friction and forces of compression act on the crankshaft and your data logger will reflect this slowing action. The inertia of the dynamometer or engine loads also act on the system.

The 3D software can produce a number that should be as accurate as any.  Reply With Quote

3. Thank you for the reply, but I am looking for specific answers to the questions I listed... I do not actually have a problem such as the one mentioned in my problem statement, I just made up an example to help others help me.

In a nutshell, I need to understand the limits of the equation T=Ia... Is this only torque due to inertia? Or des this solve for all moments, ect.. If I data log torque and angular acceleration, will this give me inertia, or is there more to the equation.

Thanks guys,
Chris  Reply With Quote

4. T=Ia is analogous to F=ma If you need to accelerate something in a straight line use F=ma. If you want to start something spinning or slow the rotation down, T=Ia fills the bill.

If you engine is running, The data logger gets all kinds of torque pulses that will interfere with your efforts. Do you need the crank inertia or the inertia of the whole collection of moving bits?  Reply With Quote

5. In the linear equation, F=ma, the force F causes m mass to linearly accelerate. In the rotary equation, T=Ia, torque causes a body of inertia I to angularly accelerate. Both are "laboratory" type equations that provide an understanding of basic physics. They do not account for "real world" affects such as friction or restrictive geometries. Those other affects will always vary in each application, but the basic physical laws of motion still apply. Force or torque can be consumed creating acceleration or overcoming friction, but not both. If you just have enough force or torque to accelerate your mass but not enough to overcome whatever real world friction may apply in your application, you will have no acceleration. Likewise if you only have enough force or torque to overcome friction but none left over for acceleration, you will still have no acceleration. If your drill can only overcome the friction in your system it will turn but it will not accelerate.  Reply With Quote

6. I need the inertia of the whole collection of moving bits with respect to crank centerline. The application is actually a type of press that I am trying to size a servo motor for, so I need to figure the load inertia on the motor that attaches to the crank.  Reply With Quote

7. Originally Posted by jboggs In the linear equation, F=ma, the force F causes m mass to linearly accelerate. In the rotary equation, T=Ia, torque causes a body of inertia I to angularly accelerate. Both are "laboratory" type equations that provide an understanding of basic physics. They do not account for "real world" affects such as friction or restrictive geometries. Those other affects will always vary in each application, but the basic physical laws of motion still apply. Force or torque can be consumed creating acceleration or overcoming friction, but not both. If you just have enough force or torque to accelerate your mass but not enough to overcome whatever real world friction may apply in your application, you will have no acceleration. Likewise if you only have enough force or torque to overcome friction but none left over for acceleration, you will still have no acceleration. If your drill can only overcome the friction in your system it will turn but it will not accelerate.
So if I understand this, if I wanted to calculate crankshaft torque here, I need more than simply T=Ia? You make it sound as if T=Ia is only torque due to the rotating assembly inertia. But say I wanted to calculate torque with respect to inertia, gravity acting on the rot. assy., and cylinder pressure... Would I need a longer equation? Again, I am not dealing with an engine, but my application is very similar. My aim is to understand how to model dynamic systems better here. And I don't see how T=Ia accounts for things like cylinder pressure and gravity. Is this accurate?  Reply With Quote

8. Yes, that is accurate. Gravity is not an issue for a purely rotating assembly. It will however be a factor on those components that are moving in some other motion, linear or combination linear-rotary, such as the crank link arm and the moving platen. Most mechanical presses have a large flywheel the only purpose of which it to smooth out the peaks and valleys of required torque during the cycle. You must also account for the effect of the work being done by the press, such as punching or forming a part. You keep mentioning a cylinder. Not sure how that fits in here. Yes, calculation of required torque in this case will be MUCH more complex than simply T=Ia.

Don't forget - the basic physical equation only refers to acceleration, not velocity. Once a rotating body is up to speed, the only torque needed will be that required to overcome friction and actually do the work.  Reply With Quote

9. The T=Ia will not account for cylinder pressure or gravity.
Asking questions related to engines when you are not dealing with engines is not getting you where you want to be.

You have a press question ask a question about presses. Many presses store energy in a flywheel so that a small motor can input modest amounts of energy constantly while the press mechanism extracts large amounts of energy intermittently. Are you trying to deal with a flywheel inertia problem?  Reply With Quote

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