Mechanics

This text covers topics such as circular motion, unbalance, balancing, primary and secondary forces, equation of motion, and pendulum equation in mechanics. It also mentions the availability of study material, solved assignments, essays, and dissertations on mechanics at Desklib.

Running head: MECHANICS 1
Mechanics
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Mechanics
Task 1.
In description of circular motion, we refer angular quantities as the analogous to linear
quantities. Consider the wheel that freely rotates about axle. Axle is the axis of
circular rotation for wheel. Angular position is the angle, which keeps on changing
from the reference point. Its SI unit is radian. θ < 0 is clockwise rotation θ > 0 is
anticlockwise rotation . A radian is angle in which the arc measurement, S, on a disk
of radius r is equal to the radius of circle.
Unbalance occur any time flywheel or a component fixed on a flywheel has a mass
center (center of gravity) which is not equivalent with axis rotation. When this
happens, a force is produced due to gyration of shaft that is final by the following
equation as shown below;
Fbalance=m× e × ω2
Where m=mass, e=eccentricity of part of rotor and ω=shaft speed
Magnitude of an oscillation can be stated as displacement between limits reached by
motion or distance from some principal point to the maximum eccentricity (the
highest value). Frequently, magnitude vibration is stated in terms of an average degree
of acceleration of circular motion, commonly known as, root-mean-square value
(m/s2).⃗
AU =¿ clockwise)⃗
AUB =¿ counter clockwise)
AB ¿
Initial imbalance AU
AB
×50 g= 0.165
0.2323 ×50=35.50 g
Angular position ¿ cos−1
[ AU
2 + AB
2− AU +B
2
2 AU AU + B ]=cos−1
[ 0.1652 +0.23232−0.2252
2 ×0.165 × 0.2323 ] =66.50°
Flywheel will balance if the weight of 35.50g will be added at 66.50 ° clockwise from the
position of the trial weight from the phase mark.

MECHANICS 3
Task 2
Here we let W to denote weights, r to denote lengths and θ to denotes angles.
Let Wc=2 Kg , WD =3 Kg ,W E =2 Kg
rc=50 mm , r D=75 mm ,r E =25 mm
θc=90θ ,θD=220θ , θE=−30θ Let W A , r A , θ A∧W G , rG , θG denote the weights added in
planes A and G respectively static balancing;
W c rc cos θ+W D rD cos θD + W E r E cos θE +W G rG cos θG +W A r A cos θA =0
W G rG cos θG +W A r A cos θA =6.2136 (1)
WG rG sin θG+ W A r A sin θA +W C rC sin θC+W E r E sin θE +W D rD sin θD =0
W G rG sin θG +W A r A sin θA =5.2136 (2)
Dynamic balancing (Rao, 2007), we take moments about the left bearing about plane B:
W C rC ∝C cos θC+W D r D ∝D cos θD+W E rE ∝E cos θE +W A r A ∝A cos θ A +W G rG cos θG
¿−16 W A r A cos θ A +88 W G rG cos θG=201.408 (3)
W C rC ∝C sin θC +W D r D ∝D sin θD +W E rE ∝E sin θE +W A +W G rG ∝G sinθG=0
=−16 W A r A sinθ A +88 W G rG sin θG =372.544 (4)
Equation (1) and (2) results
(−16 × 6.594 ) +104 × ( W G rG cos θG )=201.48
W G rG cos θG=2.9511 (5)
Equation (2) and (4) gives
(−16 ×5.2136 )+ 104 WG rG sin θG =372.544
W G rG sin θG =4.3842 (6)

Equation (5) and ( 6) give
( 2.95112 +4.38422 )
1
2 =5.2849 mm And θG =tan−1 ( 4.3842
2.9511 ¿)=56.054 ¿ ( 7)
Equation (1), (2) and (7) yields ;
W A r A cos θA =6.594−5.2849 cos 56.0549=3.649∧¿ W A r A sin θA =5.2136−5.2849sin 56.0549=0.9007 ¿
(8)
Equation (8) provide
W A r A= ( 3.64292+ 0.90072 )
2
=3.7526 Kg∧θA =tan−1 (¿ 0.9007
3.6429)=13.8877θ ¿
If balancing weight is placed at a radial distance of 50mm in place A and G.
We have rG =r A =50 mm∧thus W A =1.8763 Kg ,θA ¿13.8877θ W G =2.645 Kg ,θG=56.0549θ
Task 3
balanced Primary forces are given by the following equation
Fxp=∑
i=1
6
( Fx ) pi=∑ ( M p +M c )i r ω2 cos (ωt + αi) (1)
F yp=∑
i=1
6
( F y ) pi=∑ (−M c ) i r ω2 sin( ωt+ αi) (2)
unbalanced Secondary force is given by
F xs =∑
i=1
6
( Fx ) si=∑ ( M p )i
r 2 ω2
l cos( 2ωt +2 αi) (3)
Unbalanced primary and secondary moments (Barnacle, 2014) are given by
( Ms ) p=∑
i=1
6
( F x ) pi li (4)

MECHANICS 5
( Ms )s=∑
i=1
6
( Fx ) si li (5)
( Mx ) p=∑
i=1
6
( F y ) pi li (6)
( M p + M c ) r ω2
∑
i =1
6
cos ( αi ) = ( M p+ Mc ) r ω2 ¿
Equation (2) gives
−M c r ω2
∑
i=1
6
sin ( αi )=−M c r ω2 ( 2sin ( 0 )+ 2sin 120+2 sin 240 ) =0
Equation (3) results into
M p r2 ω2
l ∑
i=1
6
cos 2 αi= M p r 2 ω2
l ( 2 cos ( 0 ) +2 cos 240+2 cos 480 )=0
Equation (4) results into;
( M p + M c ) r ω2
∑
i =1
6
li cos ( αi )= ( M p +M c ) r ω2 α ¿
Equation (5) results into
M p r2 ω2
l ∑
i=1
6
li cos 2 αi= M p r2 ω2 α
l ( cos ( 240 ) +2 cos 480+ 3 cos 480+4 cos 240+5 cos 0 )=0
Equation (6) yields:
−M c r ω2
∑
i=1
6
li sin ( αi )=−M c r ω2 α ( sin (120 )+ 2sin 240+3 sin 240+4 sin 120+5 sin 0 )=0
Thus, engine is completely force and moment balanced.

Task 4
a) Equation of motion:
j0 ¨θ1 +2 k1 θ1−k 1 θ2=0
2 j0 ¨θ2−k1 θ1+ k1 θ1….. (1)
Rearranging and substituting in the harmonic solution
θi ( t )=φi cos (ωt +∅ ¿) ;i=1,2,3 ¿(2)
It brings frequency equation
2 ω4 j0
2+5 ω2 j0 kt +kt
2=0 (3)
The solution of equation (3) gives the natural frequency (the occurrence at which a
system fluctuates when not exposed to a constant or recurrent external force.
b) Mass of the car =m, radius of gyration (Hibbeler, 2001) is r = Jo mr2
Equation of the motion;
yf ( yr )= ground displacements of front wheels, downwards for motion along x
m ¨x +x ( k f +kr ) +θ ( kr l2−kf l1 )=kf
For motion along θ;
J0 ¨θ+ x ( l2 kr −l1 kf ) +θ ( kr l2
2 +k f l2
1 ) =(kr l2 yr −k f l1 yf )
Where ground motions can be expressed; yf (t)= ysinωt.
c) using newton’s second law of motion:
From Hooke’s law F = −kx(t) and second newton law F = mx′′(t) gives two free
equations for force acting on a system. Associating opposing forces implies that
signed displacement x(t) fulfills free vibration equation as illustrated below
mx′′(t) + kx(t) = 0
For mass m2

MECHANICS 7
m2 ¨x2 + c2 ¿ (1)
For mass m1
m1 ¨x1 + c2 ( ˙x1+ ˙y ) +k1 ( x1+ y )−c1 ( ˙x2 ˙−x1 )−k 2 ( x2−x1 ) =0 (2)
Equation (1) and (2) can be rewritten as;
m1 ¨x1 + ˙x1 ( l1 +l2 ) − ˙x2 c2 + x1 ( k1 +k2 ) −x2 k2=c1 ˙y +k1 y (3)
m2 ¨x2 + ˙x2 c2− ˙x1 c2 + x2 k2−x1 k2=0 (4)
Equations (3) and (4) can be expressed in matrix form as:
[ m1 0
0 m2 ] ( ¨x1
¨x2 ) + [ c1 +c2 −c2
−c2 c2 ] ( ˙x1
˙x2 ) + [ k1+ k2 −k2
−k2 k2 ] ( x1
x2 ) =
[ c1 ˙y+ k1 y
0 ] (5)
d) Using x1∧x2, equation of motion of piston
A pendulum is built from a shrill massless rope or bar of measurement L and a body of
mass m. Beside circular arc covered by mass, velocity is dS
dt where S = Lθ(t) is
arclength. Acceleration is Lθ′′(t). Newton’s second law motion for force alongside
this arc is F = mLθ′′(t). Additional relation for force can be established by
determining the vector gravitational force m-g into its common and tangential
components. From trigonometry, tangential component gives second force equation F
= −mg sin θ(t).
Equating opposing forces and annulling m results in the pendulum equation
m1 ¨x1 + ( k1 +k2 ) x1− psinθ=0 (1)
Equation of the motion of the pendulum bob
m2 ¨x2 + psinθ=0 (2)
In equation (1) and equation (2) sin θ can be expressed as sin θ ≈ θ= x2−x1
l (3)
This for a small angle θ.

For the vertical equilibrium of the mass m2
p=m2 g cos θ ≈ m2 g (4)
Substituting equation (3) and (4) in (1) and (2) yields:
m1 ¨x1 + ( k1 +k2 ) x1−m2 g ( x2−x1
l )=0 (5)
m2 ¨x2 +m2 g ( x2−x1
l )=0 (6)
Assuming harmonic solutions as
x1 ( t )=x1 cos(ωt +Φ) (7a)
x2 ( t ) =x2 cos( ωt+Φ ) (7b)
Equations (5) and (6) becomes
[−m1 ω2+ ( k1 + k2 ) +( m2 g
l ) y ]x1− m2 g
l x2 =0 (8)
−m2 g
l x1 + [−m1 ω2 + m2 g
l ] x2=0 (9)
By setting the determinants of the coefficient matrix of x1 and x2, we obtain the frequency
equation as
|−m1 ω2
( k1 + k2 ) +( m2 g
l )
−m2 g
l −m2 ω2 + m2 g
l |=0 (10)
i.e m1 m2 ω4− m1 m2 g ω2
l + m2 g
l ( k1 +k2 + m2 g
l ) =0 (11)
The roots of the equation (11) give the natural frequency (Singiresu, 1995) of the system.
e) Taking moments about 0 and mass m1
m1 l1
2 ¨θ1=−ω1 ( l1 sin θ1 ) +α sin θ2 ( l1 cos θ1 )−α cos θ2 ( l1 sin θ1 ) =−ω1 l1 θ1+ ω2 l1 (θ2−θ¿¿ 1)❑¿
(1)
Assuming that α ≈ ω2
m2 l2
2 ¨θ2+m2 l2 l1 ¨θ1=−ω2 ( l2 sin θ2 )=−ω2 l2 θ2 (2)
Using the relations θ1= x1
l1
and θ2= x2−x1
l2
equations (1) and (2) becomes

MECHANICS 9
m1 l1
2 ¨x1 +
( ω1 +ω2 ( l1 +l2
l2 ) ) x1− ω2 l1 x2
l2
=0 (3)
m2 l2 ¨x2 +ω2 x1 +ω2 x2=0 (4)
When m1=m2=m, l1=l2=l and ω1=ω2=mg equations (3) and (4) gives
ml ¨x1 +3 mg x1−mg x2=0 (5a)
ml ¨x2−mg x1+ mg x2=0 (5b)
For harmonic motion (Pain, 2005) ; xi ( t )=xi cos ωt; where i=1 , 2 ,3 , … equation (5) becomes
−ω2 ml x1 +3 mg x1−mg x2=0
−ω2 ml x2−mg x1+ mg x2=0 (6)
From which the frequency equation can be obtained as
ω4 m2 l2−(4 m2 lg)ω2 +2 m2 g2=0
Let R1 , R2∧R3 be the restoring forces in the spring. Equations of motion ot mass m in x
and y directions are:
m ¨x =∑
i=1
3
Ri cos α i (1)
m ¨y=∑
i=1
3
Ri sin αi (2)
Where Ri=−ki ( x cos α i+ y sin α i) (3)
Equations (1) to (3) gives
m ¨x +∑
i=1
3
ki ( x cos2 α i + y sin αi cos αi )=0 (4)
m ¨y+∑
i=1
3
ki ( x sin αi cos α i+ y sin2 α i )=0 (5)
For α 1=45 °, α 2=135 °, α 3=270 ° and k1 =k2=k3=k, equations (4) and (5)
m ¨x + kx=0 (6)
m ¨y+ 2ky =0 (7)

These two equations are uncoupled for harmonic motion.
x (t)=x cos (ωt +Φ)
y (t )= y cos(ωt+Φ )
And hence ω1= √ k
m for motion in x-direction.
ω2= √ 2 k
m for motion in the y-direction.
Natural modes are given by
x (t)=x cos ( √ k
m t +Φ1 )
y (t )= y cos( √ 2 k
m t+ Φ2 )
Where x , y , Φ1∧Φ2 can be determined by initial conditions.
Task 5.
(a) Equation of motion;
Assume that θ1, θ2 are small.
Moment equilibrium equations of the masses about P and Q:
ml2 ¨θ+ mgl θ1 +kd2 ¿) =0 ……..(1)
ml2 ¨θ−mgl2 −kd2 ( θ1+ θ2 )=0 … … … …(2)
(b)Natural frequencies and mode shapes.
Frequency or frequencies at which an object be likely to vibrate with when struck, hit
plucked, or sometime disturbed is referred to as the natural frequency of an object
.
A mode shape of a system is gotten when you determine its response due to initial conditions.
If you acquire a guitar rope, for instance, at relaxation and offer it initial conditions (then
discharge the string), it will pulsate in a precise way that depends on its Eigen frequencies,
and separately one of them is connected to a particular mode shape.
Assume: Harmonic motion with;
θ j (t)=φj cos (ωt−∅ ¿); j=1 , 2 ,3 … …..(3) ¿

MECHANICS
11
Whereφ1∧φ2 are the amplitudes of θ1∧θ2 respectively,
ω is the natural frequency ,∧∅ is the phase angle:
Using equation (3), equation (1) and equation (2), they can be expressed in matrix form as;
ω2 ml2
[ 1 0
0 1 ] φ1
φ2
+[ mgl +kd2 −k d2
−k d2 mgl +kd2 ] φ1
φ2
=0
0 ………….(4)
Frequency equation:
[ω2 ml2 +mgl +kd2 −k d2
−kd2 −ωml2 +mlg+ kd2 ]=0 ………..(5)
or
ω4−ω2 +
( 2 g
l
2 kd2
ml2 )+ g2
l2 + 2 gkd2
ml2 =0 ………(6)
By substituting for ω1
2∧ω2
2 in to equation (4), we obtain
φ2
φ1
(1)
=1∨φ1
φ2
(1)
=1
1 φ1
(1) and φ2
φ1
(2)
=−1∨φ1
φ2
(2 )
= 1
−1 φ1
(2)
Thus, the motion of the masses in the two modes is given by;⃗
θt
(1 ) ( t )=¿ 41
(1) = φ1
(1) 1
1 cos( ω1 t−∅1 ¿) ¿…… (7)⃗
θt
(2)=φ1
(2) 1
−1(cos ω2 +∅ 2)…………..(8)
c)Free vibration response.
Using linear superposition of natural modes, the free vibration response is given by;⃗
θ ( t )=C1⃗ θ ( 1 ) (t)+C2⃗ θ (2 ) (t)………….(9)
By using C1=C2 =1, without loss of originality (7) and (9) lead to ;
θ1 ( t )=φ(1)
1 cos ¿ + φ1
(2) cos ¿…..(10)
θ2 ( t ) =φ(1)
1 cos ¿ - φ1
(2) cos ¿…..(11)
Where φ1
(1), ∅ 1, φ1
2 and ∅ 2 are constants to be determined from initial conditions. Where
θ1 ( 0 ) =a ,θ2 ( 0 ) =0, ˙θ1 ( 0 ) =0∧ ˙θ2 ( 0 ) =0, eqn (10) and (11) yield
a=φ1
(1) cos ∅ 1+ φ1
(2) cos ∅2
0=φ1
(1)cos ∅1 −φ1
( 2 ) cos ∅2

σ =−ω1 φ1
(1 ) sin θ1−ω2 φ1
2 sin ∅ 2 (12)
σ =−ω1 φ1
(1 ) sin θ1 +ω2 φ1
2 sin ∅ 2
Equation (12) can be solved for φ1
(1) ,∅ 1 , φ1
2∧∅ 2 to obtain
∅ 1 ( t ) =a cos ω2−ω1
2 t . cos ω2 +ω1
2 t
(13)
∅ 2 ( t ) =a sin ω2−ω1
2 t . sin ω2+ω1
2 t
Task 6.
(a) The complete building can be modeled as shown below.
k 2( x2−x1 ) k3 ( x3 −x2 ) k 4 (x4 −x3 )
F1(t) F2( t) F3 (t) F4 (t)
c2 ( ˙x2− ˙x1) c3 ( ˙x3− ˙x2) c4 ( ˙x4− ˙x3)
(a) Equations of motion from the above diagram;
m1 ¨x1 +c1 ˙x1 +k1 x1 −c2 ( ˙x2− ˙x1 )−k2 ( x2−x1 )=F1 (t )
m2 ¨x +c2 ( ˙x2 − ˙x1 ) + k2 ( x2−x1 ) −c3 ( ˙x3−x2 )−k3 ( x3 −x2 ) =F2( t)
m3 ¨x3 +c3 ( ˙x3− ˙x2 ) + k3 ( x3 −x2 ) −c4 ( ˙x4 − ˙x3 ) −k 4 ( x4−x3 )=F3 (t)
m4 ¨x4 +c4 ( ˙x4 − ˙x3 )+ k4 ( x4 −x3 )=F4 (t)…………………(1)
(b) Langrage’s equations are;
Re m1 ¨x1
----
Gs m2 ¨x2
----
Jh m3 ¨x3
----
Bs m4 ¨x4
----

MECHANICS
13
d
dt ( ∂ T
∂ Qj )− ∂ K
∂Qj + ∂ P
∂ Qj + ∂ R
∂ Qj =Qj ………………(2)
Where K is the kinetic energy, P is the potential energy and R is the Rayleigh’s dissipation
function (Minguzzi, 2005) , Q j= jth, generalized coordinate.
K= 1
2 ( m1 ˙x1
2 +m2 ˙x2
2 +m3 ˙x3
2+ m4 ˙x4
2)
P= 1
2 ¿
R= 1
2 ¿
Q j=M j ; j=1,2,3 , … ….
Using Q j=x j ; j=1,2,3 , … … … .
The application equation (2) gives equation of motion in the equation (1).

MECHANICS
15
Reference:

Barnacle, H.E.,(2014) Mechanics of automobiles. Amsterdam. Elsevier.
Hibbeler, R.C.,(2001) Engineering mechanics. London. Pearson education.
Minguzzi, E., (2005) Rayleigh’s dissipation function at work. Roma. European Journal of
Physics, 36(3), 035014.
Pain, H.J & Pain, H.J., (2005) The Physics of Vibration and Waves (vol.2). London.
Chichester: Wiley.
Rao, S.S., (2007) Mechanics Vibration in SI units. London. Pearson Higher Ed.
Singiresu, S.R., (1995) Mechanical Vibration. Boston. Addison Wesley Publishing.