The Moment of Inertia of a Conveyor

Posted in: , on 19. Apr. 2011 - 23:18

Calculating the moment of inertia of a conveyor

Hello all,

So I have this problem. I have to calculate the moment of inertia of a whole conveyor, and I don't know how to perform some parts.

I know the moment of inertia is integer(r^2*dm) . But here starts the questions.

-This distance is from the axis of the motor to each element?

-How can i do the calculation of each drum. I will calculate the moment of inertia of a long conveyor which have 7 drums and i don´t know how these drums will affect in the moment of inertia. And how the idlers will affect in the moment of inertia?


Lyle Brown
(not verified)

Re: The Moment Of Inertia Of A Conveyor

Posted on 19. Apr. 2011 - 10:50

1. Generally it is around the rotating axis of the element being evaluated, however we do not understand how you are completing your calculations, hence cannot comment with certainty.

2. Summation of individual components, the individual components can be calculated via:



Conveyor Moment Of Inertia

Posted on 19. Apr. 2011 - 11:14

Most often the moment of inertia is taken about the driving drums to reference their collective masses at the belt line. F=M x a (force = mass x acceration). The mass used here is the combined belt line mass plus all drive inertias. The Force is really the belt tension as a function incrementally calculated around the belt geometry (carry + return) at all reference locations.

Many questions can be asked. One is how long does it take to accelerate the conveyor of know mass and dF (differential or incremental force available to accelerate the conveyor to speed).

Another is that since you know the motor size(s), you also can estimate the additional motor capacity left to accelerate ( maybe about 40% of motor nameplate), and find the time given the added power.

Another is the available overstress allowance of the belt along its length usually stated as a limit of the belt breaking strength (ST-XXXX N/mm) ( maybe a safety factor of 5.5:1 during acceleration and 6:1 for steady-state, leaving a certain tension (0.5 of 6.0 or 8% of the ST value) available for starting and stopping that can be converted into a belt line tension that then can be converted into a motor torque and time to accelerate).

Why would you need to go in the opposite direction unless for academic interest? Academic interest are not the primary reason for this forum.

Lawrence Nordell Conveyor Dynamics, Inc. website, email & phone contacts: phone: USA 360-671-2200 fax: USA 360-671-8450

Moment Of Inertia Of A Conveyor

Posted on 6. Apr. 2012 - 05:22

As Larry mentioned all of your inertias must be resolved at a common refence and this is most typically the belt line. Thus all rotational inertias must be converted into belt line equivalents. All of the inertial data that you need can be obtained from the suppliers of the equipment and components. You must set up a table with all of the inertias, say with respect to their local reference in one column and their belt line equivalent mass in the next column. Once you complete the second column then you are ready to do accel time/tension calculations the Larry has outlined.

Joe Dos Santos

Dos Santos International 531 Roselane St NW Suite 810 Marietta, GA 30060 USA Tel: 1 770 423 9895 Fax 1 866 473 2252 Email: jds@ Web Site: [url][/url]

Re: The Moment Of Inertia Of A Conveyor

Posted on 1. Jun. 2013 - 05:48


The machines where all items in machine have rotary motion (like turbine, generator, motor driving centrifugal pump, etc.) are possibly less as compared to machines where some items have rotary motion and some items have linear motion.

The rotary motion formula is used for calculating acceleration, deceleration, etc. for machines which have rotary motion only. The rotary motion formulae are identical to linear motion formulae wherein moment of inertia ‘I’ represents mass M, angular velocity ‘w’ radian/second represent velocity m/s, angular acceleration radian/sq. second represents linear acceleration m/sq. second and angular displacement theta radian represents displacement m.

Many machines have some components having linear motion and some components having linear motion. Such machines acceleration - deceleration can be analysed as below:

Converting rotating mass moment of inertia ‘I’ effect to equivalent linear mass M and appropriately addition of it into linear mass. It will give total equivalent linear mass, for analysing as per linear motion formulae.

Conversely convert linear mass effect into equivalent moment of inertia and appropriately add into other moment of inertia ‘I’ of rotating items. Then the machine motion can be analysed by using rotary motion formulae.

Conveyor has linear moving masses as well as rotating masses. The linear moving mass are generally dominant. It is convenient for the conveyor designer to analyse the motion of conveyor as linear system.

But motor manufacturer or fluid coupling manufacturer will be more comfortable with rotary system. So for them conveyor moving mass are converted as an equivalent moment of inertia for which the formal engineering name is mk2. There is no engineering quantity by the name GD2 (as per engineering textbooks and engineering formulae). GD2 = 4 x mk2. Thus GD2 is mass x square of diameter of gyration instead of radius of gyration. The general formulae for converting linear mass to equivalent moment of inertia (mk2) or vice versa is as below:

M = I x [{( 2 x pi x N) / (60 x v)} raised to 2]

Where M = Mass in kg, I = Moment of inertia on kg-m2, N = rpm, v = Speed in mps.

Refer literature for better understanding of the subject.


Ishwar G. Mulani

Author of Book : Engineering Science And Application Design For Belt Conveyors (new print November, 2012)

Author of Book : Belt Feeder Design And Hopper Bin Silo

Advisor / Consultant for Bulk Material Handling System & Issues.

Pune, India.

Tel.: 0091 (0)20 25871916