POWER EQUATIONS
(EQUATION OF MOTION)
1
Power equations (Equation of Motion)
➢ In kinetostatic analysis, the input speed is typically assumed constant.
(330-5-7) (See Cleghorn Chap. 9)
➢ The exact input velocity of a mechanism is an unknown variable and it
varies during a motion cycle with load, inertia force, friction force, etc.
(330-5-8)
➢ In time-response analysis, it is assumed that the mechanism
dimensions, geometry, mass properties, driving force or torque, and
the external loads are known, then we can solve for actual motion
response of the input member(s) by solving dynamic differential
equations of the mechanism. (330-5-b6)
➢ Consequently, motions of all other links can be solved by kinematic
analysis, and bearing forces, input torque required, etc., could be
determined by kinetostatic force analysis. (332-1-1)
2
Power Equation
1. Power equations (Equation of Motion) (5.1)
➢ An energy balance for a machine can be written as:
W = T + U + Wf
where
W = net work input to the machine;
= (work in) – (work out).
T = change in kinetic energy of the moving parts.
U = change in potential energy stored in the machine.
Wf = energy dissipated through friction.
3
Power Equation
1. Power Equations
➢ An energy balance on a machine can be written as
W = T + U + Wf
➢ In terms of time rates we have the power equation. (5.1)
P=
dT dU
+
+ Pf
dt
dt
where
P = the net power into the machine
dT/dt = rate of change of stored kinetic energy of the moving parts
dU/dt = rate of change of potential energy stored in the machine
Pf = power dissipated through friction
4
Power Equation
2. Kinetic Energy
➢ The kinetic energy, T, of a rigid body in plane motion is (5.2)
T=
1
(mV 2 + I 2 )
2
where
m = mass of body,
V = velocity of the center of mass,
 = angular velocity of the body,
I = moment of inertia.
5
Power Equation
2. Kinetic Energy
➢ Kinematic coefficients are used for derivatives of motion variables
with respect to the input variable.
d k / dSi = hk
d 2 k / dSi2 = hk
drk / dSi = fk
d 2rk / dSi2 = fk
where Si is the input variable, which can be either angle or length.
k =
d k dSi
= hk Si
dSi dt
 k = hk Si +
rk = fk Si
dhk dSi
Si = hk Si + hkSi2
dSi dt
rk = fk Si + fkSi2
6
Power Equation
2. Kinetic Energy
fx =
dX
,
dSi
fy =
dY
,
dSi
h=
d
dSi
where
X and Y are the coordinates of the body
center of mass,
 is the angular position of the body.
(
( )
d fx2
dt
)
1
1
(mV 2 + I 2 ) = m fx2 + fy2 + Ih 2  Si2
2
2
d fx2 dSi
dh2 dh2 dSi
dh
dfx
=
=
2
h
Si = 2hhSi
=
= 2fx
Si = 2fx fxSi
dt
dSi dt
dSi
dSi dt
dSi
T =
( )
(
)
dT 
= m fx2 + fy2 + Ih 2  Si Si + m ( fx fx + fy fy ) + Ihh Si3
dt
A
= ASi Si + BSi3
B
(A and B are functions of Si) (5.3)
7
Power Equation
2. Kinetic Energy
(
)
dT 
= m fx2 + fy2 + Ih 2  Si Si + m ( fx fx + fy fy ) + Ihh Si3
dt
= ASi Si + BSi3
(A and B are functions of Si)
➢ For a machine, the time rate of change of kinetic energy would be (5.3)
dT
= ( A ) Si Si + ( B ) Si3
dt
where A and B are the sums of A and B for all moving parts
of the machine.
8
3. Equivalent Mass or Moment of Inertia (5.4)
(
Power Equation
)
 A =  m fx2 + fy2 + Ih2 
T =
(
)
1
1 
1
2
2
2
2
2
2
2

(
mV
+
I

)
=
m
f
+
f
+
I
h
S
=

AS


x
y
i
 i
2
2 
2
A = me (equivalent mass) if Si is a length, ri. (Unit of mass)
A = Ie
(equivalent moment of inertia) if Si is an angle, i.
(Unit of mass moment of inertia)
➢ It is the mass or moment of inertia which, if attached to the input link,
would have the same kinetic energy as the total machine. (5.4)
1
1
2
T = ( A ) Si = me ri 2
2
2
or
1 2
Ie i
2
➢ Ie and me generally change with the configuration of the machine.
Geartrains have constant Ie because the first-order kinematic
coefficients are constants.
9
Power Equation
3. Equivalent Mass or Moment of Inertia
(
)
dT 
= m fx2 + fy2 + Ih 2  Si Si + m ( fx fx + fy fy ) + Ihh Si3
dt
= ASi Si + BSi3
1 d ( A )





 B =  m ( fx fx + fy fy ) + Ihh  =
2 dSi
1 dme
1 dIe
=
or
2 dri
2 d i
➢ ΣB is one-half the rate of change with respect to Si of the system’s
equivalent inertia or mass.
➢ For geartrains, ΣB is always zero.
T =
dT
1 dIe 3
= Ie i i +
i
dt
2 d i
1
1
2

A
S
=
me ri 2
( ) i
2
2
for angular
input i
or
1 2
Ie i
2
dT
1 dme 3 for linear
= me ri ri +
ri
dt
2 dri
input ri
10
4. Example (5.4)
Power Equation
r2 − r3 − r1 = 0
 2 = 30
r2 cos  2 − r3 cos 3 = 0

r2 sin 2 − r3 sin3 + r1 = 0
r3 = 40.32 cm
3 = 70.8934
r3 sin 3 h3 − cos  3f3 = r2 sin 2
−r3 cos3 h3 − sin3f3 = −r2 cos 2
 h3 = 0.286, f3 = 9.97 cm
r3 S3 h3 − C3f3 = −r3 C3 h32 − 2f3h3 S3 + r2 C 2
−r3 C3 h3 − S3f3 = −r3 S3 h32 + 2f3 h3 C3 + r2 S 2

OA = r2 = 15.24 cm,
h3 = 0.10605,
f3 = −8.23 cm
OC = r1 = 30.48 cm
11
Power Equation
4. Example
➢ Coordinate components of mass centers
Link 2
X g 2 = r4 cos  2 = 6.60 cm
Yg 2 = r4 sin 2 = 3.81cm
Link 3
X g 3 = r2 cos  2 = 13.20 cm
Yg 3 = r2 sin 2 = 7.62 cm
Link 4
X g 4 = r5 cos 3 = 7.48 cm
Yg 4 = − r1 + r5 sin3 = −8.88 cm
OA = 15.24 cm, OC = 30.48 cm, OG2 = 7.62 cm, CG4 = 22.86 cm,
m2 = 1.36 kg, m3 = 0.9072 kg, m4 = 3.629 kg,
I2 = 35.12 kg  cm2 , I3 = 1.46 kg  cm2, I4 = 585.58 kg  cm 2
12
Power Equation
4. Example First-order kinematic coefficients of mass centers
fg 2 x = dX g 2 / d 2
= − r4 sin 2 = −3.81 cm
fg 2 y = dYg 2 / d 2
= r4 cos  2 = 6.60 cm
fg 3 x = dX g 3 / d 2
= − r2 sin 2 = −7.62 cm
fg 3 y = dYg 3 / d 2
= r2 cos  2 = 13.20 cm
fg 4 x = dX g 4 / d 2
= − r5 h3 sin 3 = −6.17 cm
fg 4 y = dYg 4 / d 2
= r5 h3 cos  3 = 2.14 cm
(
)
A2 = m2 fg22 x + fg22 y + I2h22
= 1.36[(-3.81)2 + 6.602] + 35.12
= 114.09 kg∙cm2
(
)
A3 = m3 fg23 x + fg23 y + I3h32
= 0.9072[(-7.62)2 + 13.202] + 1.46(0.286)2
= 210.82 kg∙cm2
(
)
A4 = m4 fg24 x + fg24 y + I4h42
= 3.629[(-6.17)2 + 2.142] + 585.28(0.286)2
= 202.61 kg∙cm2
A = A2 + A3 + A4 = 527.53 kg∙cm2
➢ The contribution from Link 3 is the highest.
13
Power Equation
4. Example
Second-order kinematic coefficients of mass centers
fg2 x = d 2 X g 2 / d 22
= − r4 cos  2 = −6.60 cm
fg2 y = d Yg 2 / d
2
2
2
= − r4 sin 2 = −3.81 cm
fg3 x = d 2 X g 3 / d 22
B2 = m2 ( fg 2 x fg2 x + fg 2 y fg2 y ) + I2h2h2
=1.36[(-3.81)(-6.60)+(6.60)(-3.81)]
+ 35.12(1)(0) = 0
B3 = m3 ( fg 3 x fg3 x + fg 3 y fg3 y ) + I3h3h3
= − r2 cos  2 = −13.20 cm
= 0.9072[(-7.62)(-13.20)+(13.20)(-7.62)]
fg3 y = d 2Yg 3 / d 22
+ 1.46(0.286)(0.106) = 0.0442 kg∙cm2
= − r2 sin 2 = −7.62 cm
fg4 x = d X g 4 / d
2
2
2
B4 = m4 ( fg 4 x fg4 x + fg 4 y fg4 y ) + I4h4h4
= − r5h3 sin 3 − r h cos 3
= 3.629[(-6.17)(-2.90)+2.14(-0.970)]
= −2.90 cm
+ 585.28(0.286)(0.106)
= 75.20 kg∙cm2
2
5 3
fg4 y = d 2Yg 4 / d 22
= r5h3 cos  3 − r h sin3
2
5 3
= −0.970 cm
B = B2 + B3 + B4 = 75.25 kg∙cm2
➢ The rate of change of Link 4 is the highest.
14
New15
Power Equation
4. Example
➢ The variation of the equivalent moment of inertia is mainly contributed
by Link 4.
➢ The largest equivalent moment of inertia of Link 4 occurs when θ2 =
270°. At this position, the effect of fg4x is the largest.
θ2 = 270°
θ2 = 30°
θ2 = 30°
16
4. Example
Power Equation
dT dU
P=
+
+ Pf
dt
dt
A = 0.052753 kg∙m2
B = 0.007525 kg∙m2
➢ The time rate of change of kinetic energy for
2
this machine, in this position of the driving
crank 2, is (power equation) (5.7-b)
dT
= ( A )  2 2 + ( B )  23
dt
T2
= 0.052753 2 2 + 0.007525 23
( 2
in rad/sec,  2
Nm/sec
in rad/ sec 2 )
➢ If the only external loading was T2 2 , (5.8-t)
T2 2 = 0.052753 2 2 + 0.007525 23
Equation of motion
T2 = 0.052753 2 + 0.007525 22
17
Power Equation
4. Example
Equation of motion
T2 = 0.052753 2 + 0.007525 22
The equation holds regardless of the specifications on T2, 2 and  2 .
➢ If it was also specified that
 2 = 40 rad/sec
and  2 = 0
T2 = 0.007525(40)2 = 12.039 Nm.
It is the kineto-static analysis (see Cleghorn Chap. 9).
➢ If it was specified that
T2 = 27.12 Nm and  2 = 80 rad/sec,
 2 = (T2 − 0.007525 22 ) / 0.052753 = −398 rad/ sec 2
It is the time-response analysis.
18
Power Equation
4. Example
➢ Suppose that we specify unknown couple T2 acting on crank 2,
a couple T4 = 203.37 Nm counterclockwise acting on link 4,  2 = 30
and  2 = −2 rad/sec (cw ),  2 = 0, the power equation is
T2 2 + T4 4 = (T2 + T4h4 ) 2 = (A) 2 2 + (B) 23
The equation of motion
2
T2 + T4h4 = (A) 2 + (B) 22
In kinetostatic analysis
T2
T2 = (A) 2 + (B) 22 − T4h4
= 0.052753(0) + 0.007525(-6.28)2 − (203.37)(0.286)
= −57.87 Nm
T4
What is T4:
https://blog.orientalmotor.com/motor-sizing-basics-part-1-load-torque
19
Power Equation
4. Example
In kinetostatic analysis
T2 = (A) 2 + (B) 22 − T4h4
= 0.052753(0) + 0.007525(-6.28)2 − (203.37)(0.286)
= −57.87 Nm
2
T2
T4
New
20
Power Equation
5. Potential Energy, Elevation (5.9)
Potential energy due to elevation, H, is
where
P=
dT dU
+
+ Pf
dt
dt
Ue = m g H
m = mass of the body,
g = gravitational acceleration magnitude,
H = vertical distance from reference
level to the mass center.
The time rate of change of it is
dUe
= mg (dH / dt ) = mgVe
dt
where
Ve =

dH dH dSi
=
= feSi
dt
dSi dt
dUe
= (mgfe )Si
dt
fe is the kinematic coefficient of elevation.
Check notes for another way of calculation Sup.
21
Power Equation
6. Potential Energy, Linear Spring (5.11)
P=
dT dU
+
+ Pf
dt
dt
➢ The spring force Fs is linearly proportional to the change in length of the
spring
Fs = Fo + k (rs − rso )
Fo = preload in the spring,
k = spring stiffness,
rs = the spring length,
rso = the length of the spring
when Fs = Fo
➢ The potential energy stored in the spring
1
U = A1 + A2 = Fo (rs − rso ) + (Fs − Fo )(rs − rso )
2
Fs
Spring force
where
A2
Fo
A1
rso
Spring deflection
rs
Sup.
22
Power Equation
6. Potential Energy, Linear Spring (5.11)
P=
dT dU
+
+ Pf
dt
dt
➢ The potential energy in the spring
1
U = A1 + A2 = Fo (rs − rso ) + (Fs − Fo )(rs − rso )
2
= k(rs - rso)
1
2
= Fo (rs − rso ) + k (rs − rso )
2
➢ Differentiating the above equation with respect to time gives our result
𝑑𝑈
= 𝐹𝑜 𝑟𝑠ሶ + 𝑘(𝑟𝑠 − 𝑟𝑠𝑜 )𝑟𝑠ሶ
𝑑𝑡
➢ When expressed in terms of the kinematic coefficient of the spring fs
𝑑𝑈
= 𝐹𝑠 𝑓𝑠 𝑆𝑖ሶ
𝑑𝑡
where
fs = drs dSi
Sup.
23
Power Equation
7. Energy Dissipated, Viscous Damper (5.11)
P=
dT dU
+
+ Pf
dt
dt
➢ The force magnitude required to move the elements of a viscous
damper relative to each other is
Fc = crc = cfcSi
where fc = drc dSi is the kinematic coefficient of the damper;
the definition is the same as the kinematic coefficient of the spring.
➢ The work done to move the damper elements a small distance,
dW = crc drc
➢ The rate of energy dissipated in the damper is
Pf = dW / dt = crc (drc / dt ) = crc2 = cfc2Si2
Sup.
24
Power Equation
8. Example
➢ Find the kinematic coefficients of the spring
for a given value of the input angle θ2
➢ For the first vector loop, h3
and h4 can be solved.
➢ The vector loop equation for
the 2nd loop is r7 + r6 − r3 − r2 = 0
where the X and Y components are
r7 cos 7 + r6 cos  6 − r3 cos( 3 + 3 ) − r2 cos  2 = 0
r7 sin7 + r6 sin 6 − r3 sin( 3 + 3 ) − r2 sin 2 = 0
Solve 6 and r6
➢ Differentiating the above equations with respect to 2:
cos  6
 sin
6

− r6 sin 6   f6   − r3 sin( 3 + 3 )h3 − r2 sin 2  Solve f6 and h6 using
=



r6 cos  6   h6   r3 cos( 3 + 3 )h3 + r2 cos  2  Cramer’s rule
➢ If the spring is replaced by a damper, the above equations can be
used to obtain the kinematic coefficients of the damper.
Sup.
25
Power Equation
9. Equation of Motion (5.14)
Without U and Pf : T2 + T4h4 = (A) 2 + (B) 22
With U and Pf :
(T2 + T4h4 ) 2 = (A) 2 2 + (B) 23 +
dU
+ Pf
dt
➢ If the state of motion (input position, velocity, and acceleration) is
completely known, the equation can be used to calculate the
relationship between the external forces or couple. (Kinetostatic
analysis) (5.14-1-5)
➢ If T2, T4 and the state of motion at time t = 0 are given, the equation
can be used to determine the state of motion at any later time t.
(Dynamic analysis, or time-response analysis) (5.14-2-2)
➢ It is non-linear and the coefficients A and B are not constant but are
function of 2. Hence the most feasible way to solve the problem is a
numerical (such as Euler’s method or Runge-Kutta method), step-bystep procedure.(5.14-2-7)
26
Power Equation
9. Equation of Motion: Euler’s Method
Without U and Pf : T2 + T4h4 = (A) 2 + (B) 22
With U and Pf :
(T2 + T4h4 ) 2 = (A) 2 2 + (B) 23 +
dU
+ Pf
dt
2
➢ Given a set of initial conditions at some time t = t0
 20
 2
 2 (t 0 ) =  20 and  2 (t 0 ) =  20
We can solve the equation of motion and compute  2 (t 0 ) =  20
t
∆t
➢ The change of position  2 and velocity  2 during the time step ∆t
using constant acceleration assumption (Euler’s method)
 2 =  20 t
 2 =  20 t + 21  20 t 2
from which the position and velocity at t = t0 + ∆t are
 2 =  20 +  2
 2 =  20 +  2
2
 20 +  20 t
 20
 2
∆t
t
➢ Repeat the process and obtain the position & velocity at t = t0 + 2∆t so on
27
9. Equation of Motion: Step Size of Euler’s Method
Power Equation
Without U and Pf : T2 + T4h4 = (A) 2 + (B) 22
With U and Pf :
(T2 + T4h4 ) 2 = (A) 2 2 + (B) 23 +
dU
+ Pf
dt
➢ The smaller ∆t is, the more valid the assumption of constant acceleration
is, but the computation time increases as sell.
➢ A general rule is that ∆t should be small enough that ∆θ2 is reasonable
(5° or less). If this value is exceeded, ∆t needs to be reduced; otherwise
the computation would lead to divergence.
➢ As a rule of thumb, set ∆t to 0.01 second would ensure convergence of
most cases.
28
Power Equation
10. Example: Input Motor
T2 + T4h4 = (A) 2 + (B) 22
(1) If the input is from an electric motor or
internal combustion engine, the input torque
depends on the input rotational speed. The
torque-speed curve (Tm-ωm) is nonlinear and
described by a torque-speed curve.
2
Full load
operating
torque and
speed
T2
T4
(2) An alternating current (AC) induction motor without load has
a rotational speed (ωsync) that is an integer multiple of 60. Any
amount of load causes the motor to slip and lag behind ωsync
29
Power Equation
10. Example: Input Motor
T2 + T4h4 = (A) 2 + (B) 22
(1) The torque-speed curve of driving motor
(Baldor VL350) can be approximated as
piecewise linear function.
2
Tf = 1.36 Nm
Tp = 3.46 Nm
T2
Tb = 4.00 Nm
Ts = 4.88 Nm
T4
ωp = 450 rpm ωb = 1440 rpm ωf = 1730 rpm ωsync = 1800 rpm
(2) ωm in rpm, motor rotation is ccw.
30
Power Equation
10. Example: Gearbox
T2 + T4h4 = (A) 2 + (B) 22
(1) The motor gearbox has a ratio of 30
T2 = −30Tm
2
Nm
(2) The motor speed is computed from  2
m = −30 2 rad/sec
(3) The motor mass center is stationary and has
zero first and second order kinematic coefficients.
Kinematic coefficient of the rotational inertia
must be included
 = fym
 =0
fxm = fym = 0, fxm
T2
hm = −30, hm = 0, Im = 204.85 kg  cm2
T4
(4) Equivalent moment of inertia of the motor
Am = Im hm2 = 18.44 kg  m2
OA = 15.24 cm, OC = 30.48 cm, OG2 = 7.62 cm, CG4 = 22.86 cm,
m2 = 1.36 kg, m3 = 0.9072 kg, m4 = 3.629 kg,
I2 = 35.12 kg  cm2 , I3 = 1.46 kg  cm2, I4 = 585.58 kg  cm 2
31
10. Example: Load & Initial Conditions
Power Equation
T2 + T4h4 = (A) 2 + (B) 22
(1) The load torque T4 appears when link 4
has an indexing operation in the clockwise
direction
T4 = 203.5 (ccw) Nm if  4  0
2
T4 = 0
(ccw) Nm if  4  0
(2) Initial conditions:  2 = 0,  2 = 0
T2
at
t =0
h2 = 1, h3 = h4 = 0.2
Tm = 4.88 Nm, T2 = −146.4 Nm
A = A2 + A3 + A4 + Am = 18.48 kg∙m2
T4
B = B2 + B3 + B4 = 0.0119 kg∙m2
 20 = −5.72 rad/sec
OA = 15.24 cm, OC = 30.48 cm, OG2 = 7.62 cm, CG4 = 22.86 cm,
m2 = 1.36 kg, m3 = 0.9072 kg, m4 = 3.629 kg,
I2 = 35.12 kg  cm2 , I3 = 1.46 kg  cm2, I4 = 585.58 kg  cm 2
32
Power Equation
10. Example: Remarks
T2 + T4h4 = (A) 2 + (B) 22
1. Enter all geometric and physical data and the initial conditions.
Also state the time step size, t, to be used.
2. Place all the computations needed to establish numerical values for
A, B, and h4. Then write the differential equation.
3. Use a numerical integration method, such as Euler’s method or
Runge-Kutta method, to solve the differential equation, and
calculate the values of 2 and  2 at the end of the time interval.
4. Increment t and resolve the differential equation. Repeat until some
previously established limit in t, or 2 , or  2 is reached.
5. Be very sure that a consistent set of units is used throughout.
33
Power Equation
10. Example: Results
➢ Steady state is reached at about t = 2 sec. Maximum and minimum Link
2 speeds are 59.99 and 55.46 rpm. The average speed is 57.73 rpm.
➢ Speed fluctuation is undesirable. It leads to higher bearing force,
stress on the links, and loading on the motor.
➢ Define coefficient of fluctuation: Cf =
 + min
where ave = max
2
➢ A general rule of thumb is that Cf
should never exceed 10% and a
maximum of 5% for AC motors.
max − min 59.99 − 55.46
=
= 7.85%
ave
57.73
ωmin = 55.46 rpm
ωave
∆θ2 less than 5° for ∆t = 0.01 s
ωmax = 59.99 rpm
34
Power Equation
10. Example: Flywheels
➢ A flywheel is a rotating cylindrical mass
attached to the input crank or output shaft
of the motor. A flywheel stores and
releases kinetic energy in order to resist
changes in motion. A flywheel can reduce
steady state speed fluctuation.
➢ A flywheel is balanced so that its mass
center is stationary, and the kinematic
coefficients of the mass center are zero.
ω2
T2
Gearbox
ωm
Tm
➢ If Ifly = 35.12 kg∙m2 is added to I2 when
computing A2 and B2, the maximum and
minimum Link 2 speeds are 59.12 and
56.85 rpm., which results in Cf = 3.91%.
The time to reach steady state has
increased to about t = 5 sec.
35
Power Equation
10. Example: Motor Torque versus Speed
➢ The Baldor VL3501 motor has a full load operating speed of ωf =
1730 rpm. Accounting for the gearbox ratio of 30:1, this corresponds
to Link 2 speed of 57.6 rpm. This is very close to the average speed
from the forward dynamic simulation.
➢ It should be able to avoid overheating the motor.
ωf = 1730 rpm
36
Power Equation
=0
Example: Inverse Dynamic Problem T + T h = (A) + (B) 2
2
4 4
2
2
➢ Link 2 is driven at a constant speed. Link 4
performs an indexing operation during its slow cw
rotation. During its quick ccw rotation, Link 4
returns to begin another indexing operation
2
T4 = 203.5 (ccw) Nm if  4  0
T4 = 0
(ccw) Nm if  4  0
°
➢ Initial conditions:  2 = 0,  2 = 60 rpm cw at t = 0
T2
h2 = 1, h3 = h4 = 0.2
B = B2 + B3 + B4
= 0.0119 kg∙m2
T4
T2 = −40.23 Nm
➢ Link 2 needs to provide
input torque of at least
67.84 Nm at ω2 = 60 rpm.
T2 = −67.84 Nm
θ2 = 30°
37