LECTURE 26- MASS MOMENT OF
INERTIA OF PULLEY SYSTEMS
Course Name : DESIGN OF MACHINE ELEMENTS
Course Number: MET 214
Moment of Inertia combinations
Power transmission systems often include a variety of power transmission components mounted to
multiple drive shafts. Example system is shown below
To assist in the determination of the acceleration, torque, power, time to stop, time to start and/or speed
performance associated with the operation of a power transmission system including conveyor systems, the
mass moment of inertia of the entire system needs to be identified.
The mass moment of inertia of a pulley can be determine from adjustments to the mass moment of a short
cylinder. Envision the formation of a pulley by removing the center portion from a solid cylinder.
where
J mp  J mC 2  J mC 1
J mP   t ( J C 2  J C 1 )
Where t is thickness of cylinders
J mP 
J mC 1 
Mass moment of inertia of pulley
Mass moment of inertia of cylinder #1 (or just mass
moment)
J mC 2  Mass moment of inertia of cylinder #2
J C1 
J C2 
Area moment of inertia of cylinder #1 (or just mass
moment)
Area moment of inertia of cylinder #2
Alternatively, the mass moment of inertia Jm can be expressed in terms of the radius of
gyration of mass km .
Km 
Jm
m
  Km 
2
J m  mK m   J m 
2
w
g
Jm
m
2
Km
Recall Torque is related to angular acceleration by the following relationship
T  J m
For constant α
T  Jm

t
For customized units:
T  . 105  J m
n
t
 n  n f  ni
where
nf 
ni 
t 
final speed after accelerating, rpm
initial speed prior to accelerating ,rpm
change in time during which α is experienced
Alternatively,
w 2
T  . 105  K m
 g
2
wK m  n
T 
g
t
. 105
2
wK m  n
T 
308  t
 n

 t
It will be assumed that the Jm values for pulleys, including the effects of thru hole and/or
a V groove are available from the suppliers of pulleys.
Mass moment if inertia for solid shafts have already been identified.
J MS 
1
2
MTR 
2
1 W
2
( )R
2 g
where J MS  mass moment of inertia of solid shaft
M T  mass of solid shaft
W  weight of solid shaft
R  radius of shaft
To determine the mass moment of inertia of a combination involving a pulley mounted
on a shaft, simply add the mass moment of inertias of the components.
J mT  J mP  J mS
where
J mS 
mass moment of inertia of shaft
J mP  mass moment of inertia of pulley
J mT  total mass moment of inertia of shaft and pulley
Combinations involving two or more shafts with power transfer between shafts using pulleys.
When two shafts are interconnected by a pulley arrangement, the mass moment of inertia of the two
pulleys and the two shafts can be combined to form a single equivalent mass moment of inertia to simplify
calculations involving the torque required from a motor to achieve a desired acceleration, time to stop,
time to start, etc.
Reflected Impedances:
A technique for combining mass moments of inertias and other mechanical aspects effecting angular
acceleration to form a single, equivalent alternative representation useful in simplifying calculations
involving angular accelerations, torque, etc.
Consider the single pulley labeled as pulley #1 connected to a shaft labeled as shaft #1. The motor
supplies a torque to shaft #1
The amount of work associated with rotating pulley #1 and shaft #1 due to an external torque T1 applied
to the shaft by the motor can be determined as follows.
where
W 1  T1 1  [ J mT 1  1 ] 1
T1  Torque applied to shaft #1 by motor to rotate pulley #1 and shaft #1
 1  Angular rotation of pulley #1 due to T1
J mT  Total mass moment of inertia of shaft #1 and pulley #1
1
 1  Angular acceleration associated with motion of pulley #1
𝑊1 = 𝑤𝑜𝑟𝑘 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟𝑟𝑒𝑑 𝑓𝑟𝑜𝑚 𝑚𝑜𝑡𝑜𝑟 𝑡𝑜 𝑠ℎ𝑎𝑓𝑡 #1 𝑡𝑜 𝑟𝑜𝑡𝑎𝑡𝑒 𝑠ℎ𝑎𝑓𝑡 #1 𝑎𝑛𝑑 𝑝𝑢𝑙𝑙𝑒𝑦 #1
Assume a pulley system is formed by using a belt to link pulley #1 to a second pulley labeled as pulley #2
mounted to a second shaft labeled as shaft #2.
Note:
W T1  W 1  W 2 / 1
where
W T1  Total work performed by motor to rotate shaft #1 and pulley #1 and shaft #2 and pulley #2
W 1  work performed to rotate pulley #1 and shaft #1
W 2 / 1  work performed by pulley #1 to rotate shaft #2 and pulley #2
W T1  T1 1  T 2 2
W T1  [ J mT 1  1 ] 1  [ J mT 2  2 ] 2
Where
T 2  Torque required to rotate pulley #2 and shaft #2
 2  Angular rotation of pulley #2 due to T2
 2  Angular acceleration associated with motion of pulley #2
J mT  Mass moment of inertia of pulley #2 and shaft #2
2
Recall the following relationships exist for belt drives
r1 1  r2 2   2 
r1 1  r2 2   2 
r1
r2
r1
r2
1
r1 1  r2  2   2 
r1
r2
1
1
Substituting for θ2 and α2 in terms of θ1 and α1, the second term can be rewritten as follows:
r 
r
J mT 2 2 2  J mT 2  1  1 1  1
 r2  r2
2
r 
 J mT 2  1   1 1
 r2 
Using the above enables the expression for WT1 to be reconfigured in terms of α1θ1 as shown below
2
W T1  [ J m 1T1  J m 2 T 2
 r1 
  ] 1 1
 r2 
The reconfigured expression for 𝑊𝑇1 represents the total amount of work to be performed by the motor
to rotate pulley #1 and shaft #1 and pulley #2 and shaft #2 with the requirements for pulley #2 and shaft
#2 expressed in terms associated with pulley #1. 𝑊𝑇1 can be expressed as the product of the total
amount of torque that must be applied to pulley #1 to rotate both pulleys.
W T1  TT1  1
TT1  total amount of torque applied to shaft #1 so pulley #1 and shaft #1 and pulley #2 and shaft
#2can rotate with their respective requirements.
Equating the above expressions for WT1 and cancelling the θ1 terms enables the following expression to
be derived for the total torque necessary to perform WT1.
2
TT 1  [ J mT 1  J mT 2
TT1  J eq T  1
 r1 
  ] 1
 r2 
2
J eq  [ J mT 1  J mT 2
where
 r1 
  ]
 r2 
Given a two pulley system with the pulley shaft combinations having mass moment of inertias J mT and
J mT respectively, the amount of torque TT1 that must be applied to shaft #1 so pulley #1 and shaft #1 will
accelerate with α1 and pulley #2 and shaft #2 will accelerate with     r1 
can be determined
2
1

from the product of Jeq with α1.
 r2 
1
2
2
r 
The term involving J mT  1  is referred to as the reflected impedance. In this example, the reflected
2
 r2 
2
impedance is given a special designation
r 
J 2 R 1  J mT 2  1 
 r2 
The reflected impedance represents the resistance to the acceleration of pulley #1 due to pulley #2 with
the effects of pulley #2 being described in terms of pulley #1.
The reflected impedance consists of two terms. The JmT2 represents the resistance (impedance) to motion due to mass
moment of inertia effects. The term (r1/r2)2 represents how the impedance due to mass moment JMT2 must be adjusted
so its overall effect can be consolidated with shaft #1 and pulley #1
When determining reflected impedances in pulley systems, so the overall effects can be combined into the context of a
single perspective, the following statement is helpful for generalizing the procedure for determining reflected
impedances.
Reflected Impedances:
Rotational mechanical impedances can be reflected (transferred) through a pulley system by multiplying the mechanical
impedance by the following scale factor.
reflected impedance scale factor = [radius of destination pulley]2
[radius of source pulley]2
In the example analyzed previously, the term involving JmT2 has been modified by the impedance scaling factor (r1/r2)2
J 2 R1  J MT
 r1 
2 

 r2 
2
Where
J 2 R1  the reflected impedance of JMT2
r1  radius of destination pulley
r2  radius of source pulley
Pulley system example : Air compressor, wood working and metal working tools, rudiment perspective for
conveyor systems.
Typical arrangement involving a belt drive.
Total torque required from motor drive can
be resolved into components to facilitate
establishing drive requirements for system
 r 
T S 1  J meq  1  T L 2  1 
 r2 
Where
Angular effects
reflected through
system
TT1 
Total torque applied to shaft #1
J meq 
Equivalent mass moment of inertia of system about shaft #1
1 
Angular acceleration of shaft #1
T L2 
Torque load on shaft #2 due to load force FN
r1 
Radius of pulley #1
r2 
Radius of pulley #2
Linear effect due
to load forces
propagated
through system
Total torque TT1 is determined by resolving TT1 into two components
J meq  1  Components due to angular effects propagated to drive shaft #1
r 
T L 2  1   Component due to load forces propagated through system.
 r2 
As shown in example below each component is determined separately, and the effects combined to
determine the total torque the motor must supply to the system.
Angular effects propagated through system
Note:
J mT  J mP 2  J mS 2  J MG
2
Where
J mT 2  Total mass moment of inertia about shaft #2
J mP 2 
Mass moment of inertia of pulley #2
J mG 
Mass moment of inertia for grinding wheel
J mS 2 
Mass moment of inertia of shaft #2
Effects of FN are considered with linear
effects and is not considered as part of
angular effects.
J mT  J mP1  J mS 1
1
Where
J mT 1 
Total mass moment of inertia for shaft #1
J mP1  Mass moment of inertia for pulley #1
J mS 1 
Mass moment of inertia for shaft #1
T S 1  J eq 1  1
Where
J eq 1  J mT 1
r 
 J mT 2  1 
 r2 
2
2
T S 1  [ J mT 1
T S1 
r 
 J mT 2  1  ] 1
 r2 
Torque to be applied to shaft #1 by motor to acceleration equivalent inertia
J eq 1  Equivalent inertia of system about shaft# 1
1 
Angular acceleration of shaft #1
Linear effects due to load forces propagated through system
Consider a load force applied to grinding wheel.
FT   N F N
Where
F N  Normal force applied to part being ground
FT  Tangential force due to grinding operation
N 
Co-efficient of dynamic friction of grinding wheel
T L 2  rG FT
Where
T L 2  Torque about axis #2 due to grinding operation
rG 
Radius of grinding wheel
r 
T L1  T L 2  1 
 r2 
Where
T L1  Torque about axis #1 due to grinding operation on axis #2
r1 
Radius of pulley #1
r2 
Radius of pulley #2
TT1  T S 1  T L1
Where
TT1  Total torque required from motor to perform grinding operation.
2
TT1  [ J mT 1
Note:
 r1 
 r1
 J mT 2   ] 1  T L 2 
 r2 
 r2
r1  r2 
r1




1
r2
Comments about J mT and J mT
If speed of grinding wheel is constant during grinding operation, then
 r 
TT1  T L1  T L 2  1 
 r2 
1
2
PT1  TT1  1  T L 2  2
Where
PT1 
Power required from motor to operate grinder
1 
Speed of axis if during grinding operations presumed steady
2 
Speed of axis #2 during grinding operations presumed steady
 2  0  1   2
r2
r1
0
Example: Consider the configuration shown below
T m  24 ft  lbs
2
J mS 1  . 0068 ft  lb sec
2
J mP 1  . 0038 ft  lb sec
2
J mS 2  . 00726 ft  lb sec
2
J mP 2  . 218 ft  lb sec
2
J mload  2 . 474 ft  lb sec
Find time to accelerate shaft #1 from n1=0 to n2=550 rpm
J mload is attached to shaft 2 and is not shown in the figure.