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.

J mp J mP



J mC

2 

J mC

1  

t

(

J C

2 

J C

1 ) Where t is thickness of cylinders where

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 C

1  Area moment of inertia of cylinder #1 (or just mass moment)

J C

2  Area moment of inertia of cylinder #2

Alternatively, the mass moment of inertia J m gyration of mass k m . can be expressed in terms of the radius of

K m



J m

 

K m

2 

m J m m J m



mK m

2  

J m



w K m

2

g

Recall Torque is related to angular acceleration by the following relationship

T



J m

 For constant α

T



J m

  

t

For customized units:

T

 

n

 .

105 

J



n f m



n



t



n i

where

n f



n i



t

  final speed after accelerating, rpm initial speed prior to accelerating ,rpm change in time during which α is experienced

Alternatively,

T T T

    .

105

wK g



t

.

105

wK

2

m



n

308 2

m

  

w g

 

t n K m

2   

n



t

It will be assumed that the J m 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

M T R

2  1 2 (

W g

)

R

2 where

J MS



M T W R

   mass moment of inertia of solid shaft mass of solid shaft weight of solid shaft 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 J J mP mT

   mass moment of inertia of shaft mass moment of inertia of pulley 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.

W

1 

T

1  1  [

J mT

1  1 ]  1 where

T

1  1

J mT

1  1 𝑊 1     Torque applied to shaft #1 by motor to rotate pulley #1 and shaft #1 Angular rotation of pulley #1 due to T 1 Total mass moment of inertia of shaft #1 and pulley #1 Angular acceleration associated with motion of pulley #1 = 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 T

1 

W

1 

W

2 / 1 where

W T

1

W

1

W

2 / 1    Total work performed by motor to rotate shaft #1 and pulley #1 and shaft #2 and pulley #2 work performed to rotate pulley #1 and shaft #1 work performed by pulley #1 to rotate shaft #2 and pulley #2

W T

1

W T

1 

T

1  1 

T

2  2  [

J mT

1  1 ]  1  [

J mT

2  2 ]  2 Where

T

2  2   Torque required to rotate pulley #2 and shaft #2 Angular rotation of pulley #2 due to T 2  2

J mT

2   Angular acceleration associated with motion of pulley #2 Mass moment of inertia of pulley #2 and shaft #2 Recall the following relationships exist for belt drives

r

1  1  2  2 

r r

2 1  1

r

1  1 

r

2  2   2 

r

1

r

2  1

r

1  1 

r

2  2   2 

r

1

r

2  1 Substituting for θ 2 and α 2 in terms of θ 1 and α 1 , the second term can be rewritten as follows:

J mT

2  2  2 

J mT

2  

r

1

r

2    1

r

1

r

2  1 

J mT

2  

r r

2 1   2  1  1 Using the above enables the expression for W T1 to be reconfigured in terms of α 1 θ 1 as shown below

W T

1  [

J m

1

T

1 

J m

2

T

2  

r

1

r

2   2 ]  1  1

The reconfigured expression for 𝑊 𝑇1 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.

𝑊 𝑇1 can be expressed as the product of the total amount of torque that must be applied to pulley #1 to rotate both pulleys.

W T T

1

T

1  

T T

1  1 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.

T T

1 Equating the above expressions for W T1 and cancelling the θ 1 terms enables the following expression to be derived for the total torque necessary to perform W T1 .

 [

J mT

1 

J mT

2  

r r

2 1   2 ]  1

T T

1 

J eq T

 1 where

J eq

 [

J mT

1 

J mT

2  

r r

2 1   2 ] Given a two pulley system with the pulley shaft combinations having mass moment of inertias

J

and

mT

1

J mT

2 respectively, the amount of torque T accelerate with α 1 and pulley #2 and shaft #2 will accelerate with from the product of J eq with α 1.

T1 that must be applied to shaft #1 so pulley #1 and shaft #1 will  2   1  

r

1

r

2   can be determined The term involving

J mT

2  

r r

2 1   2 is referred to as the reflected impedance. In this example, the reflected impedance is given a special designation

J

2

R

1 

J mT

2  

r r

2 1   2 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 J mT2 represents the resistance (impedance) to motion due to mass moment of inertia effects. The term (r 1 /r 2 ) 2 represents how the impedance due to mass moment J MT2 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 J mT2 has been modified by the impedance scaling factor (r 1 /r 2 ) 2

J

2

R

1 

J MT

2  

r

1   2

r

2 Where

J r

1 2

R

1 

r

2   the reflected impedance of J MT2 radius of destination pulley 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 Where

T T

1  Total torque applied to shaft #1

T S

1 

J meq

 1 

T L

2  

r r

2 1   Angular effects reflected through system

J meq

  1  Equivalent mass moment of inertia of system about shaft #1 Angular acceleration of shaft #1

T L

2  Torque load on shaft #2 due to load force F N

r

1  Radius of pulley #1

r

2  Radius of pulley #2 Linear effect due to load forces propagated through system

Total torque T T1 is determined by resolving T T1 into two components

J meq

 1  Components due to angular effects propagated to drive shaft #1

T L

2  

r r

2 1    Component due to load forces propagated through system.

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: Effects of FN are considered with linear effects and is not considered as part of angular effects.

J mT

2 

J mP

2 

J mS

2 

J MG

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

J mT

1 

J mP

1 

J mS

1

J

Where 

mT

1

J mP

1  Total mass moment of inertia for shaft #1 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 

J mT

2  

r r

2 1   2

T T S

1

S

1   [

J mT

1 

J mT

2  

r r

2 1   2 ]  1 Torque to be applied to shaft #1 by motor to acceleration equivalent inertia

J eq

1   1  Equivalent inertia of system about shaft# 1 Angular acceleration of shaft #1

Linear effects due to load forces propagated through system

Consider a load force applied to grinding wheel.

F T

 

N F N

Where

F N

 Normal force applied to part being ground

F T

 Tangential force due to grinding operation 

N

 Co-efficient of dynamic friction of grinding wheel

T L

2 

r G F T

Where

T L

2  Torque about axis #2 due to grinding operation

r G

 Radius of grinding wheel

T L

1 

T L

2  

r r

2 1 Where  

T L

1  Torque about axis #1 due to grinding operation on axis #2

r

1  Radius of pulley #1

r

2  Radius of pulley #2

T T

1 

T S

1 

T L

1 Where

T T

1  Total torque required from motor to perform grinding operation.

T T

1  [

J mT

1 

J mT

2   

r r

2 1    2 ]  1 

T L

2   

r r

2 1    Note:

r

1 

r

2 

r r

2 1  1 Comments about

J mT

1 and

J mT

2 If speed of grinding wheel is constant during grinding operation, then

T T

1 

T L

1 

T L

2  

r r

1 2  

P T

1 

T T

1  1 

T L

2  2 Where  2  0   1   2

r

2

r

1  0 

P T

1  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

Example: Consider the configuration shown below

T m J J J J J mS mS mP

 1

mP

1 2 2

mload

24

    

.

ft



.

0068 0038 .

00726 .

218 2 .

lbs ft ft ft

474



ft

 

ft lb lb



lb



sec sec

lb

sec

lb

2 2 2

sec sec

2 2 Find time to accelerate shaft #1 from n 1 =0 to n 2 =550 rpm

J mload

is attached to shaft 2 and is not shown in the figure.