Energy Method

 Energy techniques helps to simplify more complex problems

 Involves the rate of change of total energy (Potential and Kinetic)

in the system

 It is just an alternative way of arriving at the EoMs through a

different way of looking at Newton’s 2nd law.

 We must understand the concepts of Kinetic and Potential energy

Dr Asif Israr - Institute of Space Technology Week 3

 Energy Method
2

Kinetic Energy (K.E)

 Energy stored in a system by virtue of motion

 Must include both translational kinetic energy (1/2mv2 in the

simplest form) and rotational kinetic energy (1/2Iω2 in its simplest

form)
Potential Energy (P.E)
 Energy stored in a system by virtue of position

 Can only be measured as change from some datum

 Includes gravitational potential energy (mgh in its simplest form)

 Also includes elastic strain energy stored in springs (1/2kx2 in its

simples form)
Dr Asif Israr - Institute of Space Technology Week 3
 Energy Method
3

Potential Energy Datum

 Must choose a P.E datum at which P.E is considered to be zero
and from which P.E changes can be measured.

 Choice of datum not a problem but it should be somewhere which

makes the problem easier.

 In all our systems we will consider the un-stretched spring

position to be the P.E datum.

Dr Asif Israr - Institute of Space Technology Week 3

 Energy Method
4

 Let ‘T’ be the total kinetic energy in the system

 Let ‘V’ be the total potential energy in the system
 Then from work/energy relationships:
d (T  V )
 Rate of doing work by external forces
dt
 If the system is conservative (no external forces acting) then:
T  V  const
 Therefore, d (T  V )
0
dt
 We shall use examples to illustrate the use of this equation but
first some comments

Dr Asif Israr - Institute of Space Technology Week 3

 Co-ordinate choice
5

 Just as when we considered Newton’s 2nd law directly, we need to

choose a coordinate to define position.

 Can be either a translational co-ordinate ‘x’ or a rotational co-

ordinate ‘θ’.

 Choice depends on what makes problem easier. No “right answer”

to this

Dr Asif Israr - Institute of Space Technology Week 3

 Energy Method – Mass/Spring System
6

 A reinvestigation using Energy Method

 Translational displacement ‘x’ of mass to be used as co-ordinate
 ‘x’ measured from un-stretched spring position and therefore we
will measure Potential Energy changes from this position.
 Assume that mass has velocity x  at some position ‘x’
 Calculate system energies

Dr Asif Israr - Institute of Space Technology Week 3

 Energy Method – Mass/Spring System
7

Kinetic Energy (K.E)

 The mass possesses kinetic energy only by virtue of its translational velocity
since it is not rotating. Hence: 1
Potential Energy (P.E)
K .E  T  M 2
x
2
 The system potential energy is made up of two components

 The elastic P.E stored by virtue of deforming the spring

P.Eelastic   stiffness spring _ extension

1 2

2
in this case 1
P.Eelastic  kx2
2
 Gravitational potential energy gained or lost by virtue of the fact that the
mass has moved away from the P.E datum
P.Egrav  Mgx Loss of P.E

Dr Asif Israr - Institute of Space Technology Week 3

 Energy Method – Mass/Spring System
8

 Thus the total potential energy change is

1 2
P.Etotal  P.Eelastic  P.Egrav  kx  Mgx  V
2
 This gives a total system energy of
1 2 1 2
T V  Mx  kx  Mgx
2 2
 There are no external forces acting so we may say that this is a conservative
system i.e. T + V = const
d (T  V )
 Since this is a conservative system we have: 0
1 1  dt
d  Mx 2  kx 2  Mgx
 Therefore, d (T  V )   2 2 
dt dt
1 1
 M  2 xx  k  2 xx  Mgx  0
2 2
Dr Asif Israr - Institute of Space Technology Week 3
 Energy Method – Mass/Spring System
9

  is common to all the terms in the above equation.

We note that x
Thus:
Mx  kx  Mg  0
 If we ignore the gravitational effect then the above equation will
take the following form:
Mx  kx  0
 And by comparison with the standard form
x  n2 x  0
yields k as before
n 
M

Dr Asif Israr - Institute of Space Technology Week 3

 What about gravity?
10

kD mg  kD  0, from FBD,
and static equilibriu m
m
k
1
0
V
+x(t) spring  k ( D  x ) 2

2
mg Vgrav  mgx
m D
+x(t)
1 2
T  mx
2
Dr Asif Israr - Institute of Space Technology Week 3
 What about gravity?
11

d
(T  V )  0
dt
d  1 2 1 2
  mx  mgx  k (D  x)   0
dt  2 2 
 mxx  mgx  k (D  x) x  0
 x (mx  kx)  x (kD  mg )  0  kD  w  mg
0 from static
equilibriu m

 mx  kx  0
Gravity does not effect the EoM or the natural frequency of the
system for a linear system as shown previously with a force
balance.

Dr Asif Israr - Institute of Space Technology Week 3

 Problem 1
12

 Determine the EoM of the pendulum using Energy Method

1 2 1 2 2
T  J oq  ml q
2 2

V  mgl(1  cosq ) q
m
The change in elevation is l(1-cosθ)
d d  1 2 2  mg
(T  V )   ml q  mgl(1  cosq )   0
dt dt  2 
 J  m2

Dr Asif Israr - Institute of Space Technology Week 3

 Problem 1
13

m 2qq  mg (sin q )q  0

 q m q  mg (sin q )  0
2

 m 2q  mg (sin q )  0
g
 q (t )  sin q (t )  0

g g
 q (t )  q (t )  0  n 

Dr Asif Israr - Institute of Space Technology Week 3

 Problem 2
14

 Uniform beam of mass ‘M’

 Small oscillations about pivot ‘O’ in vertical plane
 Take the angular displacement ‘θ’ of the beam as co-
ordinate

Dr Asif Israr - Institute of Space Technology Week 3

 Problem 2
15

 The beam is rotating about support ‘O’, therefore there is no

whole body translation. The means that K.E arises only as a result
of rotational motion.
1 2 1 2
K .E  I  Iq
2 2
dq
 Since   q
dt

 The mass moment of inertia ‘I’ of a uniform beam of mass ‘M’

about one end is given by:
ML2
I
3
Dr Asif Israr - Institute of Space Technology Week 3
 Problem 2
16

 Therefore, the K.E of the system is:

1 ML2  2 ML2  2
K .E  T   q  q
2 3 6
 As before the P.E is made up of elastic and gravitational
components. We choose the potential energy datum to be the un-
stretched spring position. If the beam rotates through angle ‘θ’ as
shown:
 The spring at the tip of the beam will be compressed by:
L sin q  Lq By the small angle approximation
 And the spring at the centre of the beam will be compressed by:
L L
sin q  q By the small angle approximation
2 2
Dr Asif Israr - Institute of Space Technology Week 3
 Problem 2
17

 This gives the total Elastic Potential Energy as:

2
1 1 L  5 2 2
P.Eelastic  k ( Lq )  k  q   kL q
2

2 2 2  8

 Gravitational Potential Energy:

 Uniform beam, therefore C.G is distance L/2 from pivot
 When beam moves through small angle ‘θ’, C.G descend below P.E
datum by: L L
sin q  q
2 2
 Change in gravitational P.E is then:
L
P.Egrav  Mg q
2
Dr Asif Israr - Institute of Space Technology Week 3
 Problem 2
18

 From the above we get the total potential energy change:

5 2 2 L
P.E  P.Eelastic  P.Egav  V  kL q  Mg q
8 2
 There are no external forces acting so we can say the system is
conservative and
ML2 2 5 2 2 MgL
T  V  const  q  kL q  q
6 8 2
 Differentiation w.r.t time gives
d (T  V )
0
dt
Dr Asif Israr - Institute of Space Technology Week 3
 Problem 2
19

 From the above we obtain the EoM

d (T  V ) ML2 5kL 2
MgL 
  2qq   2qq  q 0
dt 6 8 2
 Which reduces to the EoM
ML2  5kL2 MgL
q q 0
3 4 2
 As we have a gravitational term in the above equation. We
suppose at this stage that the effect is negligible then the equation
will reduce to the following form:
ML2  5kL2
q q 0
3 4
Dr Asif Israr - Institute of Space Technology Week 3
 Example 2:
20

 Re-arranging so that we can compare with standard form

q
15k
q  0 comparing with q  n2q  0
4M

yields the natural frequency for this system

15k
n  rad / sec
4M

Dr Asif Israr - Institute of Space Technology Week 3

 Problem 3
21

 Find natural frequency of the system by the use of Energy Method

 Kinetic Energy
K .E.total  K .E.trans  K .E.rotat
1  2 1 2
 mx  J oq
2 2
1
where the moment of inertia of the cylinder is J0  mr 2
2
 Also, we know that x  rq and x  rq
2
 Thus, for the system at any time 1 2 11  x 
K .E.  mx   mr 2  
2 22  r 
3
 mx 2
4
Dr Asif Israr - Institute of Space Technology Week 3
 Problem 3
22

 Potential Energy
1 2
P.E  kx
2

 Total Energy T  K .E  P.E  const

d d
or, T  ( K .E  P.E )  0
dt dt
 Substituting the values d  3 2 1 2 
 mx  kx   0
dt 4 2 
 3  
 mx  kx  x  0
2 
Dr Asif Israr - Institute of Space Technology Week 3
 Problem 3
23

 3 
 As x  0 then mx  kx  0
2

 or, x  2k x  0
3m

2k
 and so, n  rad / sec
3m

Dr Asif Israr - Institute of Space Technology Week 3

 Problem 4
24

 Figure shows a uniform cylinder of radius ‘r’ and weight

‘W’ which rolls without slipping on a cylindrical surface
of radius ‘R’. Determine
(a) the differential equation of motion
(b) the natural frequency of oscillation

Dr Asif Israr - Institute of Space Technology Week 3

 Problem 4
25

 In moving from A to B through angle ‘θ’, the disc rotates through

angle ‘φ’
 Arc length AB = Rθ = rφ, hence φ = Rθ/r
 The K.E consists of a translational and a rotational components

K .Etrans 
1
2
 
m R  r q
2

1 2 1  2
K .Erotat  I  Iq
2 2
1 R 1 1 R 
 I q 2  Iq 2  I   1q 2
2 r 2 2 r 
Dr Asif Israr - Institute of Space Technology Week 3
 Problem 4
26

 This last term arises because the angular velocity of the disc
relative to its centre is  and the angular velocity of the disc
relative to an axis fixed at the centre of the track is q (motion A
to B is opposite direction to B to B΄) this enables us to calculate
the absolute angular velocity which is needed in the K.E
calculation of the disc.

 
2
1 1 2  R  2
K .E  m R  r q   mr   1 q
1  2

2 2 2 r 

Dr Asif Israr - Institute of Space Technology Week 3

 Problem 4
27

 The P.E only arises from the change in position of the disc C.G

P.E  mg ( R  r )(1  cosq )

d
 Then ( K .E  P.E )  0
dt
 Which in this case is:
2
mr  R 
mR  r   2qq 
2
1  
2
  1  2qq  mg ( R  r ) sin q  0
2 4  r 

mr 2 R  r  
2
mR  r  q 
2 
2
q  mg ( R  r )q  0
4 r Assuming small angle
Dr Asif Israr - Institute of Space Technology Week 3
 Problem 4
28

mR  r  q  mg ( R  r )q  0
3 2
Or,
2
 Comparing with standard form the natural frequency is revealed
as

2g
n  rad / sec
3( R  r )

Dr Asif Israr - Institute of Space Technology Week 3

 Problem 5
29

 Find the natural frequency of the following system

 Solution: K.E of the system is K .Etotal  K .Etrans  K .Erotat

1 2 1 2
 mx  J oq
2 2
Dr Asif Israr - Institute of Space Technology Week 3
 Problem 5
30

1 2 2 1  1 2  2 3 2 2
K .Etotal  mr q   mr q  mr q
2 22  4
where ‘m’ is the mass of cylinder and Jo its mass moment of
inertia
1 2 1
 P.E is P.E  kx  2  k (r  a) 2q 2
2 2
where x  (r  a)q and sinq  q then
d
K .E.  P.E.  0
dt
3
or mr 2qq  2k (r  a)2qq  0
2
 2 4k (r  a)2
or q 
4 k ( r a )
q 0 n  2
rad / sec
3mr 2 3mr
Dr Asif Israr - Institute of Space Technology Week 3
 Problem 6: Inverted Pendulum
31

 In this case we can’t remove the

gravitational component. The reason is
that the equilibrium position coincides
with the unstretched spring position.
 Choose angular displacement ‘θ’ as co-
ordinate
 Uniform bar mass ‘M’ vibrates as small
angle motion about pivot ‘O’.
 K.E. results from rotation about pivot (no
translational K.E.)

Dr Asif Israr - Institute of Space Technology Week 3

 Problem 6: System Energies
32

 K.E. is given by:

1 2 1 ML2 2
K .E.  T  Iq  q
2 2 3

 Elastic potential arises because from compression of one spring

by asinθ = aθ and the extension of the other by the same amount

   aq     aq   ka q
1 k 2 1 k 2 1 2 2
P.E.elastic
2 2 2 2 2

Dr Asif Israr - Institute of Space Technology Week 3

 Problem 6: Gravitational Potential Energy
33

 Must examine motion of the centre of gravity located at the centre

of the beam at G.
 This moves to G΄ when beam moves through ‘θ’
 From the diagram we can see that the
centre of gravity falls through a vertical
distance i.e. L L
 cosq
2 2
 Resulting loss of gravitational P.E. is

P.E.grav 
MgL
1  cosq 
2

Dr Asif Israr - Institute of Space Technology Week 3

 Problem 6: System energy
34

 The total P.E. Change in moving through angle ‘θ’ is:

1 2 2 MgL
P.E.  V  ka q  (1  cosq )
2 2
 The total system is then (conservative system):
ML2 2 1 2 2 MgL
T V  q  ka q  (1  cosq )  const
6 2 2
 Differentiation yields the equation of motion for small oscillations
of the beam
d T  V  ML2   1 2  MgL
  2qq  ka  2qq   sinq  q  0
dt 6 2 2
Dr Asif Israr - Institute of Space Technology Week 3
 Problem 6
35

 Eliminating the common factor this gives

ML2  MgL
q  ka q 
2
sinq  0
3 2
 Employing the small angle approximation this may be written:
ML2   2 MgL 
q   ka  q  0
3  2 
 Gravitational term is part of the coefficient of ‘θ’ in this case and
can’t be removed from the equation by considering vibration
about the equilibrium position.

Dr Asif Israr - Institute of Space Technology Week 3

 Problem 6
36

 The EoM may be re-written

 3  2 MgL 
q  ka 
2 
q  0
ML  2 

 Which gives the natural frequency

3  2 MgL 
n   ka  
ML2  2 

Dr Asif Israr - Institute of Space Technology Week 3

 Problem 6: Observations
37

 To oscillate with this frequency we must have MgL

ka 2   0 (why?)
2
MgL
 If ka 2  0
2
 Natural frequency would be zero – No vibration and beam
would remain in any position to which it was displaced.

MgL
 If ka2  0
2
 An imaginary natural frequency would be generated which is
not possible. Physically this represents a state of instability and
system would collapse.

Dr Asif Israr - Institute of Space Technology Week 3

 Problem 7
38

 Pulley mass ‘M’, radius of gyration about O ‘ko’ stiffness ‘k’,

radius of inner disc ‘r’. As pulley descends it rotates without slip
 Centre of pulley descends ‘x’. Therefore LHS of the string must
lengthen by distance ‘x’, and RHS must also lengthen by distance
‘x’
 Thus string total extension is ‘2x’
 K.E. consists of
1 2 1 2 1
K .E.trans  Mx and K .E.rotat  Iq  Mko2q2
2 2 2
Dr Asif Israr - Institute of Space Technology Week 3
 Problem 7
39

 Since the wheel rotates without slip we have the geometric

relationship between ‘x’ and ‘θ’

x  rq from which x  rq

 The K.E of rotation may be written 2

1 2 x

K .Erotat  Mk o  
2 r
 Total K.E is then 2
1 1  

x
K .E  T  K .Etrans  K .Erotat  Mx 2  Mk o2  
2 2 r
1  k 2

 Mx 1  2 
2 o
2  r 
Dr Asif Israr - Institute of Space Technology Week 3
 Problem 7
40

 Potential energy is made up of elastic strain energy in the string

and the change in gravitational P.E.
1
P.E.elastic  k (2 x)2  2kx 2 and P.E.grav  Mgx
2
 Giving P. E .  V  2 kx 2
 Mgx
 The system is conservative and so:
1  ko2 
T  V  const  M 1  2  2kx 2  Mgx
2  r 

d (T  V ) 1  ko2  
and  M 1  2  2 xx  2k  2 xx  Mgx  0
dt 2  r 
Dr Asif Israr - Institute of Space Technology Week 3
 Problem 7
41

 Giving the equation of motion for small oscillations

 k 2 
M 1  o2  x  4kx  Mg  0
 r 
 
 Neglect the gravitational effect the equation reduces in the
following form
 2
k 
M 1  2  x  4kx  0
o
 r 
 
4k
n  rad / sec
 k  2

 and the natural frequency is 

M 1 o
 r  2

Dr Asif Israr - Institute of Space Technology Week 3
 Problem 8
42

 The system consists of a Tee-piece of mass 20kg and radius of

gyration 0.15m about its centre of gravity ‘G’ pivoted at ‘O’.
 A light rigid rod attaches it to a uniform cylinder of mass 25kg.
 Restraining springs and dimensions are shown in the next slide.
 Find the equation of motion for small oscillations and the
natural frequency.

Dr Asif Israr - Institute of Space Technology Week 3

 Problem 8
43

We shall use this as

co-ordinate

Convince yourself that these are correctly derived from our choice of co-ordinate

Dr Asif Israr - Institute of Space Technology Week 3

 Problem 8 Geometry of the system
44

 If the tip of the Tee-piece descends by a small distance’x’ it will

rotate through a small angle x/0.4 rad and it will have angular
velocity x / 0.4 rad/sec

 The centre of gravity ‘G’ is located a distance OG  h  0.2 2 m

from the pivot ‘O’

 The mass moment of inertia of the Tee-piece about the pivot is,
by the parallel axis theorem

  
Io  IG  Mh  M KG  h  2.05 kgm
2 2 2 2

Dr Asif Israr - Institute of Space Technology Week 3
 Problem 8: Kinetic Energy
45

 For the Tee-piece

2
1 1  x 
K .E.Tee  piece  I o   2.05 
2
  6.41x 2

2 2  0.4 
 For the uniform cylinder K.E. comes from the translational and
rotational motions
1 1
K .E.cyl  M cyl xcyl  I cylcyl
2 2

2 2
 Remember that the mass moment of inertia of a uniform
cylinder about its centre is M cylr 2
I cyl 
2
x
 And its angular velocity is cyl 
r
Dr Asif Israr - Institute of Space Technology Week 3
 Problem 8: Kinetic Energy
46

 So cylinder K.E. is
2
1 1 25  0.125  x  2
K .E.cyl   25 x  
2
  

2 2 2  0.125 
 18.75x 2 joules

 From the above the total K.E. is:

T  K .E.Tee  piece  K .E.cyl  6.41x 2  18.75x 2

 25.16x 2 joules
Dr Asif Israr - Institute of Space Technology Week 3
 Problem 6: Elastic Potential Energy
47

 From the geometry the spring at the tip of the Tee-piece is

compressed by ‘x’ and the spring attached to the cylinder is
compressed by ‘2x’. The elastic P.E. is therefore:

1 1
P.E.elastic   3010  x  10103  (2 x)2
3 2

2 2
 35000x 2

Dr Asif Israr - Institute of Space Technology Week 3

 Problem 8: Gravitational Potential Energy
48

Dr Asif Israr - Institute of Space Technology Week 3

 Problem 8: Gravitational Potential Energy
49

 From the figure on the previous slide.

 When the tip of the tee-piece moves vertically through distance
‘x’, the C.G. moves from ‘G’ to ‘G΄’.

length _ OG  0.2 2  length _ OG

 Vertical distance C.G descends is ‘h’ where

 x 
h  0.2 2 cos    0.2
 4 0 .4 
  x  
 0.2 2 cos    1
  4 0.4  
Dr Asif Israr - Institute of Space Technology Week 3
 Problem 8: Gravitational Potential Energy
50

 We may simplify the trig. Term as follows using the identity

cos(A  B)  cos A cos B  sin A sin B

 Thus
 x    x  x 
2 cos    2  cos cos  sin sin 
 4 0.4   4 0.4 4 0.4 
x x
 cos  sin
0.4 0.4
  1
 since cos  sin 
4 4 2
Dr Asif Israr - Institute of Space Technology Week 3
 Problem 8: Gravitational Potential Energy
51

 Therefore the C.G falls:

 x x 
h  0.2 cos  sin  1
 0.4 0.4 
when Tee-piece tip descends ‘x’ vertically
 Therefore gravitational potential energy is
  x  
P.E.grav  20  9.81 0.2 2 cos    1
  4 0.4  
 x x 
 39.24 cos  sin  1
 0 .4 0.4 

Dr Asif Israr - Institute of Space Technology Week 3

 Problem 8: Potential Energy
52

 From this the total potential energy is

P.E.  V  P.E.elastic  P.E.grav

 x x 
 35000x  39.24 cos
2
 sin  1
 0.4 0.4 
 Making the small angle approximations
x x x
sin  and cos 1
0.4 0.4 0.4
 The total P.E. may be written as x
V  35000x  39.242

0.4
Dr Asif Israr - Institute of Space Technology Week 3
 Problem 8: System Energy and EoM
53

x
 The total system energy is T  V  25.15x 2  35000x2  39.24
0.4
 And since this is a conservative system
d (T  V )
 25.15 2 xx  35000 2 xx  9.81x  0
dt
 This leads to the system equation of motion
50.3x  70000x  9.81  0
 Since the gravitational P.E. is zero therefore
50.3x  70000x  0
70000 n
and n  rad / sec giving fn   6Hz
50.3 2
Dr Asif Israr - Institute of Space Technology Week 3
 Problem 8: Summary of the problem
54

 Complicated systems can be shown to be single degree of

freedom systems.
 Using techniques developed earlier these systems can be shown
to have very simple equations of motion.
 Careful consideration of geometry and trigonometry is required.
 Energy techniques are quicker and easier than direct application
of Newton’s 2nd law.

Dr Asif Israr - Institute of Space Technology Week 3