Moment of Inertia of Rotating Load... The moment of inertia, indicated with symbol J in the formula in the opposite column, must be calculated before accelerating torque can be figured. Examples for three common shapes are shown here. Many other shapes are shown in machinery handbooks.
HOLLOW CYLINDER (Pipe)... J (moment of inertia) of a pipe about an axis running lengthwise is: J = W x (R^{2} + r^{2}) ÷ 2g lnchLbsSecs^{2} , in which: W is total weight of pipe in pounds R is outside radius of pipe in inches r is inside radius of pipe in inches g is acceleration of gravity, always 32.16
SOLID CYLINDER... J (moment of inertia) of a solid cylinder about an axis running lengthwise is: J = W x R^{2} ÷ 2g lnchLbsSecs^{2} , in which: W is total weight of the cylinder in pounds R is outside radius of the cylinder in inches g is acceleration of gravity, always 32.16
PRISM... J (moment of inertia) of a prism of uniform cross section about the axis shown is: J = W x (A^{2} + 8^{2}) ÷ 12g lnchLbsSecs^{2}, in which: W is total weight of prism in pounds A and B are cross section dimensions in inches g is acceleration of gravity, always 32.16

Rotating Loads  Driven With Hydraulic Motors Centrifugal pipe casting in which a high mass must be rapidly accelerated to a high rotational speed, is a typical application where a significant amount of extra torque, in addition to that required to keep the pipe spinning at a constant speed, must be supplied by the hydraulic motor. Incidentally, most rotating loads driven at high speeds do require extra torque to get them up to speed in a reasonable time because of the very high kinetic energy contained in almost any rotating load driven at high speed. The extra torque for acceleration can be calculated with this basic formula:
(c) T = J x 7 π x RPM ÷ 360 x t inchlbs., in which: T is the extra torque in inchlbs. J is the moment of inertia of rotating load which must be calculated. See formula in box in opposite column. π is always 3.14. RPM is the change in velocity, revolutions per minute, from standstill to final velocity, or from a lower to a higher speed. t is the time, in seconds, allowed for the acceleration. After finding torque, T, the additional PSI on the hydraulic motor to produce this torque can be determined from catalog information on the particular motor used.
Example of Pipe Spinning... Find the torque required to accelerate a 500 lb. pipe from a standstill to 700 RPM in a time of 5 seconds. Pipe diameter is 10" O.D. x 7" I.D.
First, solve for the moment of inertia using the formula:
J = W x (R^{2} + r^{2}) ÷ 2g lnchLbsSecs^{2}
J (moment of inertia) = 500 lbs x (52 + 3.52) ÷ 2 x 32.16 J = 500 X (25 + 12.25) ÷ 64.32 = 289.6
Next, use this calculated value of J in Formula (c) above to determine torque required just for acceleration.
T = 289.6 X 3.14 X 700 ÷ 360 X 5 = 353.6 inch lbs.
Remember, this calculated value of torque is in addition to steady torque required to keep the pipe spinning at a constant RPM. Also remember that the GPM oil flow to the motor must be sufficient to produce the desired 700 RPM.
Acceleration From a Lower to a Higher Speed When accelerating from a lower to a higher speed, subtract the difference between the two speeds and use this as RPM in the basic formula (c) at the top of this page.
