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MECHANICS Available at: http://www.mathcity.org
[email protected] 0316-7017457
By Muzammil Tanveer
Lecture # 1
Mechanics: Mechanics is the branch of science which studies the state of rest and motion of objects and laws governing rest, equilibrium and motion. Since material objects exist in the form of liquids gases and solids there are corresponding types of mechanics to deal with them. (i) (ii) (iii)
Kinematics Dynamics Statics
Kinematics: Kinematics is the branch of mechanics which describe the motion of objects without consideration of their masses and force acting on them.
Dynamics: Dynamics is the branch of mechanics concerned with the motion of objects under the action of force.
Statics: Statics is the branch of mechanics concerned with objects at rest or in equilibrium under the action of forces.
Ch# Rigid Body Motion: Particle: An extremely small part of matter or an infinitesimal part of matter having negligible dimension is called a particle.
Rigid Body: A collection of particles such that distance between any two particles remains same, irrespective to the motion of the body or force acting upon. e.g.
metallic balls, wood, stones etc.
Note:
The concept of rigid body is an idealization.
Displacement: Directed change in position of particles of a rigid body is called displacement.
1 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Translational Motion: The motion in which all particles of a rigid body are displaced such that the line joining initial and final position of particles are parallel to each other.
Rotational Motion: If during the motion particles of the rigid body move in curved path or circular path or are displaced through some angle about (an imaginary) line called axis of rotation. Then that motion is called Rotational motion.
Chasles’ Theorem: The most general displacement of a rigid body is composed of pure translation followed by a rotation about some point (base point). ⃗ = (displacement)
+ (displacement)
⃗ = dt ⃗ + ( ω⃗ × r⃗) ⃗
= ⃗ + ( ω⃗ × r⃗)
where ω⃗ is angular velocity of body and r⃗ is position vector of P with respect to A (base point). ⃗ =
⃗ + ( ω⃗ × r⃗)
since
⃗
= ⃗
2 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Lecture # 2
Angular Equation motion: We know that = ω = dt By integrating ω = t+c (i) ; where c is constant of integration and can be found by using initial condition t = 0 then ω = ω ω = (0)+c
ω = t+ω
Put in (i)
c=ω (ii)
called first angular equation of motion. Now from (ii)
= t+ω
since ω =
d = (t+ω )dt = + ω
+c
(iii)
When t = 0 , = 0 put in (iii) 0 = a (0) + ω (0) + c c =0
Put in (iii)
= + ω
+0
= + ω
(iv)
Called Second Angular equation of motion. Now from (ii) t = ω − ω (
Put in (iv) =( =(
a
a
a
= ) )
a .
a
)
+ ω (
a
+ ω =
a
) (
a
t=
a
)
+ ω
or 2 = ω − ω
Called the third Angular equation of motion. 3 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Note: These equations are only useful when a rigid body is rotating with angular velocity about a point ‘o’.
Question: A rigid body is rotating about at a point say ‘0’. Find ⃗ (i) (ii)
ω⃗ = 2 ̂+3 ̂ − ω⃗ = ̂ −
r⃗ = ̂+ ̂ + r⃗ = 2 ̂ − ̂ − 2
, ,
Solution: (i)
ω⃗ = 2 ̂+3 ̂ −
,
r⃗ = ̂+ ̂ +
⃗ = ω⃗ × r⃗ ̂ ̂ ⃗ = 2 3 −1 1 1 1 ⃗ =4 ̂−3 ̂− (ii)
ω⃗ = ̂ − ̂ ̂ ⃗ = 1 0 −1 2 −1 −2 ⃗ =− ̂−0 ̂−
̂(3 + 1) − ̂(2+1)+ (2 − 3)
r⃗ = 2 ̂ − ̂ − 2
,
̂(0 − 1) − ̂(-2+2)+ (−1 − 0)
Screw Motion: The motion which consist of translation and rotation about a line along the translation is called screw motion. In this motion linear velocity of each particle on the axis of rotation is parallel ( or anti parallel) to the angular velocity.
Theorem: Find the equation of axis of rotation in vector form in case of screw motion Or Show that equation of axis of rotation in case of screw motion is r⃗ = a⃗ + ω⃗ where r⃗ is position vector of any point on the axis of rotation and a⃗ is any vector, is scalar and ω⃗ is the angular velocity Or Show that instantaneous general motion is screw motion. Proof: Consider a rigid body in general motion. Let A be the base point of the body with linear velocity ⃗. Let B be any other particle then Linear velocity of B is 4 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
⃗ =
⃗ + ( ω⃗ × r⃗)
(i) ; where r⃗ is position vector of B from base point .
In general, ⃗ and ω⃗ are not parallel (or antiparallel) but we can choose B such that Linear velocity ⃗ of B is parallel to angular velocity ω⃗ of the rigid body. Taking cross product of (i) with ω⃗ ω⃗ × ⃗ = ( ω⃗ × ⃗) + ω⃗ × ( ω⃗ × r⃗ ) Since ω⃗ ‖ ⃗ so ω⃗ × ⃗ = 0 0 = ( ω⃗ × ⃗) + ( ω⃗ . r⃗ ) ω⃗ −( ω⃗ . ω⃗ )r⃗ = ( ω⃗ × ⃗) + ( ω⃗ . r⃗ ) ω⃗ − ω r⃗ Since ω⃗ . ω⃗ = | ω⃗|| ω⃗|
0=ω
ω r⃗ = (ω⃗ × ⃗) + ( ω⃗ . r⃗ ) ω⃗ ⃗ × ⃗
r⃗ =
+
⃗ .⃗
ω⃗
r⃗ = a⃗ + ω⃗ ⃗ × ⃗
where a⃗ =
and =
⃗ .⃗
Question: Show that ω⃗ = curl ⃗ (for rotation). Solution: ∇=
̂+
̂+
⃗=
̂+
̂+
̂
̂
∇× ⃗=
̂+ (ω
−
̂+
̂+
̂ = ω
−ω
) ̂+ (ω
−ω ) + ̂ (ω
−ω
=ω
−ω
,
=ω
,
=ω ,
−
̂+
−
… (i)
̂ ω
= −ω ,
ω
−ω
= −ω ,
) =ω
=ω ,
−ω
= −ω ,
=ω
5 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Putting above values in eq. (i) ∇ × ⃗ = (ω − (−ω )) ̂ +(ω − (−ω )) ̂+(ω − (−ω )) = (ω + ω ) ̂ +(ω + ω ) + ̂ (ω +ω ) = 2ω ̂ +2ω + ̂ 2ω
or 2[ω ̂ +ω ̂+ω
]
(∇ × ⃗ ) = ω⃗ Or ω⃗ = curl ⃗ proved.
Question: A rigid body is rotating about a fix point ‘o’. The points A(0,-1,2) and B(2,0,0) are moving with velocities ⃗ = [7,-2,-1] and ⃗ = [ 0,6,-4] respectively. Find the angular velocity of the body. Solution: Position vector of ‘A’ ⃗ = [0,-1,2] ⃗ = ω⃗ × ⃗ ̂ [7,-2,-1] = ω 0
̂
̂ (−ω − 0) ω ω = (2ω + ω ) ̂ − (2ω − 0) + −1 2
[7,-2,-1] = (2ω + ω ) ̂ − 2ω ̂−ω 7 = 2ω + ω
-2 = −2ω
,
ω = 1
Position vector of ‘B’ ⃗ = [2,0,0] ⃗ = ω⃗ × ⃗ ̂ [0,6,-4] = ω 2
̂ ω 0
̂ (0 − 2ω ) ω = (0 − 0) ̂ + (2ω − 0) + 0
[0,6,-4] = 0 ̂+2ω ̂-2ω 0=0
,
6 = 2ω
−2ω = -4
ω =2
Angular Velocity ω⃗ = ω ̂ +ω ̂+ω
ω =3
ω⃗ = ̂ +2 ̂+ 3
6 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Lecture # 3
Question: The instantaneous linear velocity of three particles A(a,0,0) , B(0,
√
, 0) and
⃗ = (u,0,0), ⃗( , 0, ) and ⃗(u+v,−√3 , ) respectively w.r.t the cartesian coordinate system. Find |ω⃗| of rigid body. C(0,0,2a) are Solution: Let A(a,0,0) be a base point (fix point). We know that linear velocity of any particle ‘P’ of a rigid body in case of general motion ⃗ = ⃗ + ( ω ⃗ × ⃗ )
⃗ is position vector of P w.r.t A
Now ⃗ = (− , ,0) and ⃗ = (− , 0,2 ) √
,ω ,ω )
⃗ − ⃗ = ( ω ⃗ × ⃗ )
ω ⃗ = ( ω
⃗ = ⃗ + ( ω ⃗ × ⃗ )
Then
̂ ω −
( 0,0,V) = = ̂ (0− ( 0,0,V) =
̂ (−
̂ ω
ω 0
√
) + ̂(− ω − 0) + (
√
√
) + ̂(− ω ) + (
√
√
+ ω ) + ω )
On comparig −
√
=0 , − ω =0 ,
√
+ ω =V
ω =0 Now
⃗ = ⃗ + ( ω ⃗ × ⃗ ) ⃗ − ⃗ = ( ω ⃗ × ⃗ )
̂ (V , −√3V , ) = ω −
̂ ω 0
ω 2
= ̂ (2aω − 0)+ ̂(− ω − 2 ω ) + (0+aω ) (V , −√3V , )
= ̂ (2aω )+ ̂(− ω − 2 ω ) + (aω )
On comparing 7 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
2aω = V , ω =
− ω − 2 ω = −√3V
,
− (0) − 2 ω = −√3V
,
aω = ω =0
−2 ω = −√3V ω = √
ω ⃗ = (
,
, 0) √
Now |ω ⃗| =
√
+ (0)
+
|ω ⃗| =
=
+
|ω ⃗| =
Ans
Question: The instantaneous linear velocity of three particles A(a, 2a, − ) , B(− , − , ) and C(a,a,a) are
⃗ = (√ , 0 , √ ), ⃗ −
√
, 0, −
√
and ⃗(0, −
, )
√ √
respectively. Find direction of cosine of axis of rotation. Solution: Let A(a, 2a, − ) be a base point. We know that Linear velocity of any particle ‘P’ of a rigid body in general motion is ⃗ = ⃗ + ( ω ⃗ × ⃗ ) Now ⃗ = (−2 ,−3 ,2 ) and ⃗ = (0, − , 2 ) ,ω ,ω )
⃗ − ⃗ = ( ω ⃗ × ⃗ )
−
ω ⃗ = ( ω
⃗ = ⃗ + ( ω ⃗ × ⃗ )
Then
√
, 0, −
√
̂ = ω −2
̂ ω −3
ω 2
= ̂ (2aω + 3aω ) + ̂(−2 ω −2aω ) + (−3 ω −2aω ) On Comparing 2aω + 3aω = −
√
,
−2 ω −2aω = 0 , −3 ω −2aω = −
√
−ω = ω Now
⃗ = ⃗ + ( ω ⃗ × ⃗ ) 8
Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
⃗ − ⃗ = ( ω ⃗ × ⃗ ) (
√
,
√
,
√
̂ )= ω 0
̂ ω −
ω 2
= ̂ (2aω + aω ) + (̂ 0 −2aω ) + (− ω −0) = ̂ (2aω + aω ) + (̂ −2aω ) + (−aω ) On Comparing √
2aω + aω =
, −2aω = ω =
ω =
And 2aω + a( ω = ω ⃗ = (
√ √
√
,
√
√
√
)=
√
, −aω =
√
√
Now |ω ⃗| =
,
√
)
+
√
|ω ⃗| =
√
+
=
+
√
|ω ⃗| =
+ is the
magnitude of Angular velocity. ⃗ | ⃗|
=(
√
,
√
,
√
)
(
√
,
√
,
√
)
We know that direction of cosine of the axis of rotation are the component of unit vector along the angular velocity and are (
√
,
√
,
√
).
Inertia: Tendency of a rigid body to remain at rest or of a body in motion to stay in motion unless an external force acted upon. Or Measure of resistance in Linear acceleration.
Moment of Inertia (Rotational Inertia) Measure of resistance of a rigid body in angular acceleration. 9 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Mathematically I = m Where r is perpendicular distance of rigid body from a line about which it is rotating and r =
is radius of Gyration (Latin word means rotation or rapid
motion in circular path). In case of system of particles +
I=
+……..+
I=∑ Radius of Gyration =
∑
∑
=
Where M = ∑
For Continuous or Uniform mass distribution: I=∫
dm
Z
= m ( distance from y-axis) =m(
mass(m)
)
+
P(x,y,z)
Momentum: P⃗ = mV⃗
z X
Angular Momentum: L⃗ = r⃗ ´ P⃗
x
For system of particles L⃗ = ∑
⃗ × ⃗
=∑
Y
⃗)
⃗ × (
⃗ × ⃗
=∑ =∑
( ⃗ × ( ω⃗ × ⃗ ))
=∑
[( ⃗ . ⃗ )ω⃗ − ( ⃗ . ω⃗) r⃗)] If ⃗ = ( ,
=∑
∑
, ) and ω⃗ = (ω , ω , ω )
[( ⃗) ω , ω , ω [(
+
V⃗ = ω⃗ × r⃗
+
− {( ,
) ω ,ω ,ω
, ). ω , ω , ω }( ,
, )]
−( ω + ω + ω ) ( ,
, )]
10 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
[(
=∑
+
ω +
(
ω +
=∑
[(
[
=∑
ω ), ( ω +
= [{∑
ω +
ω +
ω ),
−( ω +
ω +
ω ),
ω
ω +
ω +
ω )−(
ω +
ω +
ω +
ω −
ω −
ω −
ω ,
ω +
ω +
ω −
ω −
ω −
ω ,
ω +
ω +
ω − −ω
) ω − ω
−ω
) ω − ω
−ω
) ω + (− ∑
+
(
ω −
ω −
, ] )ω + (− ∑
)ω }, , )ω } ,
(
+
) ω + (− ∑
)ω ) + (− ∑
{∑
(
+
) ω + (− ∑
)ω + (− ∑
ω +
ω +
ω ,
=
ω +
ω +
ω
=
ω +
ω +
ω
=
ω +
ω +
ω
Or
ω +
ω ]
,
{∑
=[
ω )]
ω )]
ω +
) ω − ω
)ω −
+ ω +
ω )−(
ω ), (
+ ω +
ω +
+
(
ω +
ω +
+
(
)ω , (
+
ω +
+
[(
+
ω +
( =∑
)ω ,(
+
ω +
ω ,
)ω }] ω +
ω +
ω ]
ω ω ω
=
Diagonal element (
,
,
) called Moment of Inertia about coordinate axis.
Upper and lower triangular matrix element called product of inertia about coordinate axis. L = I ω⃗ 11 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Where
=∑
+
(
)
=∑
(
+
)
=∑
(
+
)
= (− ∑
)
= (− ∑
)
= (− ∑
)
For continous (uniform mass distribution = ∫(
+
= −∫
)
, ,
= ∫( = −∫
)
+
,
,
= ∫(
= −∫
+
)
12 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Lecture # 4
Question: Find moment of inertia of a uniform rod about an axis passing through one of its extremities or passing through its end points and perpendicular to the length rod. Solution: Let the mass of the rod is ‘m’ and length ‘a’. We take small particle of mass dm Y and length dx and coordinate (x,0,0). mass ‘m’ dx dm Moment of Inertia of rod about Y-axis = ∫(
+
=∫
dm
) dm
O
z=0
(x,0,0)
As mass dm dm dx length
dx
a
a
0
0
I yy x 2 dx x 2 dx 3 a
a3 x 0 3 0 3
m a
m a 3 ma 2 a 3 3 mass lengthof rod I yy 3
2
If length of the rod is 2a mass (length of rod ) 2 m(2a ) 2 I yy 3 3 4ma 2 3 4ma 2 I yye ...........( A) 3
I yye , e end po int
13 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
a
X
Question:Find M.I of uniform rod about an axis passing through centre of rod and to the length of rod. Solution: Let the mass of rod in ‘m’ and length ‘2a’. Take a small particle of mass ‘dm’ having length dx and coordiantte (x,0,0) Y M.I about Y-axis = ∫(
+
=∫
dm
) dm
(x,0,0)
z = 0 -a o
a
As mass dm dm dx length
X
dm
dx
dx
a
a
a
a
I yy x 2 dx x 2 dx 3 a
a3 a3 3 3 a
x 3
m a
m 2a 3 ma 2 2a 3 3 ma 2 ........( B ) 3 From Aand B
I yyc , c centre
I yyc
I yyc 4 I yye
Question: Find M.I of rectangular lemina about an axis in the plane of lemina and passing through any of its edges. Soluiton: Let ‘a’ be the length and ‘b’ be the width of rectangular lemina. Take a point having mass dm and area dxdy and coordinates (x,y,0). Moment of inertia about width I yy ( x 2 z 2 ) dm
Y z 0
x 2 dm
b
0
a
0
b
dm (x,y,0)
O
a
dm dxdy
2
x dxdy
Area = dxdy
14 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
X
a
b
0
x3 dy 3 0
3 a a I yy 0 dy 0 3 3 3 a b a y 0 b 0 3 3 m a 3b ma 2 ab 3 3
m ab
ma 2 mass(lengthof Re c tan gular le min a) 2 3 3 Moment of inertia about length
I yyc
I xx ( y 2 z 2 ) dm z 0
y 2 dm
b
0
a
0
dm dxdy
y 2 dxdy
b
a
y 2 x 0 dy 0
b
y 2 (a 0)dy 0
b
y 2 dy 0
3
a
y 3
b
0
3
b a 0 3 3 m ab . ab 3 mb 2 mass( widthof rec tan gular le min a) I xx 3 3
m ab
15 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Question:Find M.I of square lemina about an axis in the plane of lemina and passing through any of its edges. Solution: Let the length and the width of square lemina is ‘a’. Take a small particle of mass dm and area dxdy and coordinates (x,y,0). Y Moment of inertia about width I yy ( x 2 z 2 ) dm
a z 0
x 2 dm
a
0
a
0
a
0
O
a
dm dxdy
2
x dxdy
x3 3
dm (dxdy) (x,y,0)
a
0
a3 I yy 0 dy 0 3 a3 a a3 y 0 a 0 3 3 4 m a ma 2 2 a 3 3 a
m m a.a a 2
Similarly, ma 2 I xx 3
Question: Find M.I of a rectagular lemina about an axis in the plane of lemina parallel to its edges and passing through its centre. Solution: Let the length of rectangular lemina is ‘a’ and width ‘b’ . Take a small particle of mass dm and area dx,dy and coordiante (x,y,0) M.I about width I yy ( x 2 z 2 ) dm z 0
x 2 dm 16 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
X
b
a 2 2 b a 2 2
3
b
b2
2
x 3
a
dm dxdy
x2 dxdy
Length = a b/2
2
a
Width = ab
dm (x,y,0)
a/2
-a/2
2
-b/2
a 3 a 3 b 2 2 I yy b2 dy 3 2 3 3 3 b 2a 2a b b y b2 2 24 24 2 2
m ab
m 2a 3 ma 2 .b ab 24 12 ma 2 mass (length of rec tan gular le min a ) I yy 12 12
M.I about length I xx ( y 2 z 2 ) dm z 0
y 2 dm b
b2
a
2 a 2 2
b a
x a2
2
2
b
y 2dy
2
3 a a y 2 2 3 3
dm dxdy
y 2 dxdy
b
2
b
2
3
b b a 24 24 am 2b3 mb 2 ab 24 12
m ab
mb 2 mass(lengthof rec tan gular le min a) I xx 12 12 17 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
For MCQ I xxe
mb 2 3
e edge , end poin t
I xxc
mb 2 12
c centre
mb2 2b3 I 4 4I xxc 3 12 e xx
I xxe 4I xxc
Question: Find M.I of square lemina about an axis in the plane of lemina parallel to its edges and passing through its center. Solution: Let the length and width of square lemina is ‘a’.Take a small particle of mass dm area dxdy and coordinates (x,y,0). M.I about width
a/2 dm
I yy ( x 2 z 2 ) dm z 0
x dm a
a 2 2 a a 2 2
a
a2
2
x3 3
(x,y,0) a/2
-a/2
2
a
dm dxdy
x2 dxdy
-a/2 2
a
2
a 3 a 3 a 2 2 2 I yy a dy 3 2 3 2a 3 b 2 a3 a a y b 2 24 12 2 2
m a2
ma 3 a ma 2 2 a 12 12 I yy
ma 2 ma 2 Similarly, I xx 12 12 18
Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Lecture # 5
Wallis Formula’s: (i)
If ‘m’ is even then =
(ii)
=
(
− 1)( − 3) … … … 1 . ( − 2)( − 4) … . .2 2
If ‘n’ is odd then =
(iii)
=
( − 1)( − 3) … … … 1 ( − 2)( − 4) … . .2
If both ‘m’ and ‘n’ are even then =
(iv)
(
− 1)( − 3) … … … 1. ( − 1)( − 3) … … 1 . ( + )( + − 2)( + − 4) … . .2 2
If both ‘m’ is even and ‘n’ is odd then =
− 1)( − 3) … … … 1. ( − 1)( − 3) … … 2 ( + )( + − 2)( + − 4) … . .2 If both ‘m’ and ‘n’ are odd then =
(v)
=
(
(
− 1)( − 3) … … … 2. ( − 1)( − 3) … … 2 ( + )( + − 2)( + − 4) … . .2
Theorem: (Only statement used) If f(x) is even function then ∫
l l
( )
= 2 ∫
l
( )
If f(x) is odd function then ∫
l l
( )
=0
Question: Find M.I of a circular lemina about an axis in the place of lemina and passing through its centre. Solution: Let ‘m’ be the mass of circular lemina and radius ‘a’. Take a small area element having dm and coordinates (x,y). 19 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Y a
= ∫(
+
) dm
= ∫
dm
(x,y) -a
a
2 2 a y a x
=
X
-a
y 2 dydx
Area dm
a y a 2 x 2
3
y = 3 a
y a2 x2
dm = dxdy dx
x changes from −a a
a
=
=
a 3 a
2
a x
=
2
3
2
a x
2
3
dx
a
3 3 2 2 2 2 2 2 a x a x dx 3 a a
=
+
y a 2 x 2
2 2a x 3 a
3 2 2
= −
= y=√
−
y changes from −√
−
√
−
dx
a 3 2 2 2 2 a x dx 3 a
Put x = asin , dx = acosd x = −a
=
(−1) = −
x=a
=
(1) =
2 = 3
3 2 2 2 2 .a cos d a a sin 2
2
=
2 3
3 2 2 a cos 2 .a cos d 2
2
20 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
2 = 3
2
a
4
cos4 d
2
2 a 3
=
4 2 4
cos d
2
By theorem f(x) is even
2 a4 2 = .2 cos 4 d 3 0
∫
l l
( )
=
= 2 ∫
l
( )
=
By Wallis cosine formula 4 a 4 3.1 . . 3 4.2 2
=
a 4
= =
4 m a 4 . a2 4
=
ma2 4
Similarly
=
ma2 4
Question: Find M.T of a elliptic lemina about an axis in the plane of lemina through its major and minor axis. Y Solution: dm √ − = ∫(
+
-a
) dm
dm
= ∫ a
=
y
a z=0
b 2 2 a x a
(x,y)
√
−
2
y dydx
a y b a2 x2 a
21 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
X
x changes from −a a
a
=
a
y
3
y 3
+
b 2 2 a x a
y= √
dx y
b 2 2 a x a
3
b a x a2 x2 = 3 a a a =
2
−
y changes from 3
a b
=1
2
− √ dx
−
√
a b3
3 3 b3 2 2 2 2 2 2 a x a x a3 dx 3 a a 3
a 2 b3 2 2 32 . = a x dx 3 a3 a
Put x = asin , dx = acosd x = −a
=
(−1) = −
x=a
=
(1) =
3 2 2 2 a a sin 2 .a cos d
3 2
=
2 b . 3 a3
2
3 2 2 a cos 2 .a cos d
3 2
2 b . = 3 a3
2
3 2
=
2 b 3 a3
a
4
cos4 d
2
3
2 b a = 3
2 4
cos d
2
By theorem f(x) is even
=
2 b 3a 2 .2 cos 4 d 3 0
∫
l l
( )
=
= 2 ∫
l
( )
=
By Wallis cosine formula 22 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
−
4 b 3a 3.1 = . . 3 4.2 2
ab3
= =
4 m ab 3 . ab 4
mb2 4
=
Similarly
=
ma2 4
Question: Find M.I of rectangular parallelepiped with respect to its edges with one corner at the origin. Z (0,0,c) Solution: = ∫(
+
) dm (0,0,0)
= dm = dxdydz
Y
c b a
(0,b,0)
(a,0,0) X
= ( y 2 z 2 )dxdydz 0 0 0
c b
= x( y
2
a
2
z )
dydz 0
0 0
c b
= (a 0)( y
2
2
z )
a
dydz 0
0 0
c b
= a ( y 2 z 2 )dydz 0 0
c
3
b
y yz 2 dz 3
= a 0
0
23 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
c b3 = a bz 2 0 dz 3 0
c
= a ( 0
b3 bz 2 )dz 3 3
= a
3
b z z b 3 3
c
0
cb3 bc3 = a 3 3 =
m abc 2 2 . b c abc 3
=
m 2 2 b c 3
=
=
Similarly, m 2 a c2 3
= = Find
,
m 2 a b2 3
, = xydm c b a
= xy dxdydz 0 0 0
x2 = y 2
a
c b
0 0
a2 = 2
dydz 0
c b
y dydz 0 0
24 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
2 c
=
a 2
0
2
y 2
b
dz 0
c
a 2 b2 = . dz 2 2 0 a2 b2 = . 2 2
z
c 0
m a 2b 2 c = . abc 4
=
=
=
mab 4
Similarly, =
mac 4
=
mbc 4
Find matrix
I xx I yx I zx
I xy I yy I yz
⎡ ⎢ I xz ⎢ I yz ⎢ ⎢ I zz ⎢ ⎢ ⎣
m 2 2 b c 3
mab 4 mac 4
mab 4
mac 4 mbc 4
⎤ ⎥ ⎥ ⎥ ⎥ ⎥ m 2 2 ⎥ a b ⎦ 3
m 2 a c2 3
mbc 4
Question: Find M.I of a cube w.r.t its edges with one corner at the origin. Solution: = ∫(
Z +
(0,0,a)
) dm
(0,0,0)
= dm = dxdydz
Y
(0,a,0)
(a,0,0) X
25 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
=
a
a
a
0
0
y
0
a
a
= 0 0 =
a
0
z 2 dxdydz
2
x( y z )
a
0
dydz
(a 0)( y 2 z 2 )dydz a
3
= a
a
2
0
a
0
2
y yz 2 3
dz 0
3 a a = a az 2 (0) dz 0 3
3
= a
a z az 3 3
3
a
0
a 3 .a a.a 3 = a 3 3 a4 a4 = a 3 3 2a 4 = a 3 ma 2a4 = 3 a 3 =
2ma 2 3
Similarly, =
2ma 2 , 3
=
2ma 2 3
= xy dm
a
0
a
a
0
0
dxdydz
26 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
=
x2 a y dydz 2 0
a a
0 0
a
a2 = 2
a
a
0
0
ydydz
=
a4 a dz 4 0
=
a4 a z 4 0
=
a2 a y 2 dz = 2 0 2 0
m a4 (a ) a3 4
ma 2 = 4
Similrarly, =
ma 2 4
,
=
ma 2 4
Matrix form 2
I xx I yx I zx
I xy I yy I yz
⎡ 2ma ⎢ 3 I xz ⎢ ma 2 I yz ⎢ ⎢ 4 I zz ⎢ 2 ⎢ ma ⎣ 4
ma 2 4 2ma 2 3
ma 2 4
ma 2 ⎤ 4 ⎥ ⎥ ma 2 ⎥ 4 ⎥ ⎥ 2ma 2 ⎥ 3 ⎦
27 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Lecture # 6
Question: Find M.I of a circular ring about a line passing through its Centre and perpendicular to its circumference. Solution: Consider the circular ring of radius ‘a’ in xy-plane. Consider a mass particle dm with arc length dl with central angle d Y =
dl
d =
=
dm
O a
X
dm = a d a M.I about z-axis =
=∫ ∫
=
0
d
O
=
Since l = r
2
l = a
=
(2 − 0)
=
2
=
dm = ∫
dl = ad
.2
=
= m
Perpendicular axis theorem: Statement: The M.I of Lemina (or a plane rigid body) about a normal axis (about an axis to its plane) is equal to the sum of M.Is about two mutually axis in the plane of lemina and passing through the intersection of lemina with the normal axis. Proof: Consider place Lemina in XY-plane. X-axis and Y-axis are axis. Z-axis is to X-axis and Y-axis. Let
be the ith particle of mass
with
28 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
dl
Position vector ⃗ M.I about z-axis n
I zz ( xi2 yi2 )dm i 1 n
n
i 1
i 1
I zz xi2 dm yi2 dm I zz I yy I xx I zz I xx I yy Pr oved 2
Or I zz ( x 2 y 2 ) dm x 2 dm y dm I yy I xx
K.E in general motion: Consider a rigid body in general motion with base point at the center of mass of the rigid body. Then linear velocity of the ith particle of the rigid body is ⃗
= ⃗ + ( ⃗ × ⃗) And its K.E is
⃗
= ∑
1 n m i V ( ´ ri ) . V ´ ri 2 i 1 1 n i 1 mi V 2 V .( ´ ri ) V . ´ ri 2
´ ri
2
2 1 n 2 i 1 mi V 2V .( ´ ri ) ´ ri 2
T
2
n 1 n 1 n mi V 2 mi ´ ri V . ´ mi ri 2 i 1 2 i 1 i 1
(1) n
If center of mass (base point) is taken at origin then
mi ri 0 i 1
Then from (1)
T=
1 n 1 n mi V 2 mi ´ ri 2 i 1 2 i 1
2
29 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
1 1 n = M V 2 mi ´ ri 2 2 i 1
2
n
where M mi i 1
T = Ttranslational K .E Trotaional K .E
Now Let ´ ri a 2
´ ri ´ ri . ´ ri ´ ri . a
.( ri ´ a )
.[ ri ´ ( ´ ri )]
Now
=
=
1 n m i ´ ri 2 i 1
2
1 n 1 n m .[ r ´ ( ´ r )] m [ = i i ri ´( ´ ri )] i i 2 i 1 2 i 1
1 .L = 2
where L
Now
=
mi [ ri ´( ´ ri )] i 1
1 MV 2 = 2
Hence
n
And
1 .L 2
=
1 [x y z ].[Lx Ly Lz ] 2
1 = [x Lx yLy z Lz ] 2
1 [x (xIxx y Ixy z Ixz ) y (xI yx y I yy z I yz ) z (xIzx y Izy z Izz ] 2 1 [(x2 )Ixx xy Ixy xz Ixz yxI yx (y2 )I yy yz I yz zxIzx zyIzy (z2 )Izz ] 2
Ixy Iyx , Ixz Izx , Iyz Izy 1 [(x )2 Ixx (y )2 I yy (z2 )Izz 2xy Ixy 2xz Ixz 2yz I yz ] 2 30 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Lecture # 7
Question: Find M.I of a rigid body about line having direction cosine , and v Solution: Consider a line ‘l’ through origin having direction cosine ,
and v. Let
̂ be a unit vector along line ‘l’. Then we can write ̂ =̂+ Let
̂ +
( , , ) be the particle of the rigid body and di be the distance of
From the line ‘l’ then from fig. di = | ⃗| sin where ⃗ =
̂+
̂+
Z
= | ̂ ´ ⃗| l
Now M.I of the rigid body about line ‘l’
Y
n
I
mi di
2
i 1
X
2 mi e ´ ri n
......(1)
i 1
Now i e ´ ri l xi
j
k
yi
zi
e ´ ri zi vyi i vxi l zi j l y i xi k
Put in (1) n
2 2 2 I mi zi vyi vxi l zi l yi xi i 1 n
I mi 2 zi2 v2 yi2 2vzi yi v2 xi2 l 2 zi2 2vl xi zi l 2 yi2 2 xi2 2l xi yi i 1
31 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
n
I mi ( yi2 zi2 )l 2 ( xi2 zi2 ) 2 ( xi2 yi2 )v2 2l xi yi 2vyi zi 2vl xi zi i 1
n n n n n n I l2 mi yi2 zi2 2 xi2 zi2 v2 xi2 yi2 2lmx y 2 v my z 2 v l mx iii 1i i i i zi i1 i1 i1 i1 i1 i1
I l 2 I xx 2 I yy v 2 I zz 2l I xy 2 vI yz 2vl I xz
Which is M.I of rigid body about a line having direction cosine ,
and v
Momental Ellipsoid: We know that M.I of a rigid body about a line having direction cosine , v is given as
and
I l 2 I xx 2 I yy v 2 I zz 2 l I xy 2 vI yz 2 vl I xz
Let
̂ √
⃗= Also
be vector along ‘l’ and Q(x,y,z) be the point on line ‘l’ such that ̂ √
and ⃗=
⃗ = ̂+
̂+
Now direction cosine of ⃗
i.e.
,
⃗
,I0
√
,
⃗ are
⃗
√ , √ , √
Since line ‘l’ and
⃗ have same direction cosines therefore
= √ , = √ and v = √ Put in eq (1) I x 2 I I xx y 2 I I yy z 2 I I zz 2 xyI I xy 2 yzI I yz 2 zxI I xz
Divide both side by I x 2 I xx y 2 I yy z 2 I zz 2 xyI xy 2 yzI yz 2 zx I xz 1
Which represents equation of ellipsoid thus ellipsoid is called Momental ellipsoid or ellipsoid of inertia (Ass , , are +ve) 32 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Question Show that Momental ellipsoid at the center of elliptic disc is
x2 y2 1 1 2 z cons tan t a2 b2 a2 b2
√
−
-a
a
Solution: We know that =
mb 4
√
2
, = = =
ma 4
= +
2
By axis theorem
mb2 ma2 + 4 4
m(a 2 b2 ) 4 y
x a
= xy dm
xa
y2 x 2 x a
b 2 2 a x a
x a
y
xy dydx
b 2 2 a x a
b 2 2 a x a
dx b 2 2 a x a
xa
b2 2 b2 2 2 x 2 a x 2 a x 2 dx a a x a xa
x (0) dx 0
x a
Similarly,
=
=0
Now by Momental ellipsoid formula x 2 I xx y 2 I yy z 2 I zz 2 xyI xy 2 yzI yz 2 zx I xz 1
By putting the above value
33 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
−
m a2 b2 2 2 mb ma 2 2 x2 y z 4 4 4
2xy(0) 2yz(0) 2zx(0) 1
2 ma2 2 ma2 mb2 x y z 1 4 4 4 4 2 mb
2
Divide by
x2
ma 2 b 2 4
y2
1 4 2 1 z a2 b2 a2 b2 ma2b2 x2 y2 1 1 2 z cons tan t a2 b2 a2 b2
Question: Write inertia matrix if equation of Momental ellipsoid is 3x2 3 y2 5z2 xy 2 yz 5xz 3
Solution: Given 2x2 3y2 5z2 xy 2 yz 5xz 3
Divide by 3 2 2 2 5 2 1 2 5 x y z xy yz xz 1 3 3 3 3 3
..................(1)
As we know x 2 I xx y 2 I yy z 2 I zz 2 xyI xy 2 yzI yz 2 zx I xz 1
……….(2)
Comparing (1) and (2) =
,
=1,
= ,
I xx Now Inertia Matrix is I yx I zx
= − , I xy I yy I yz
=
I xz I yz I zz
2 3 1 6 5 6
,
= 1 6 1 1 3
5 6 1 3 5 3
34 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Lecture # 8
Question: Find equation of momental ellipsoid of a uniform rectangular parallelepiped. Solution: x 2 I xx y 2 I yy z 2 I zz 2 xyI xy 2 yzI yz 2 zx I xz 1
As
…..(1)
We know that for a parallelopiped (
=
)
,
(
=
)
,
(
=
)
,
=−
,
=
,
=− Put these in (1) (
)
+
(
)
(
+
)
)xy+2(
+2(
)yz = 1
)xz +2(
Multiplying by 6/m both side 2(
+
)
+ 2(
+
)
2(
+
)
+ 2(
+
)
+( +(
+
)
+
)
-3abxy-3acxz-3bcyz = 6/m -3abxy-3acxz-3bcyz = c
c = 6/m Which is required equation of momental ellipsoid of a uniform parallelopiped.
Question: Find M.I of solid cylinder about (i) (ii)
its axis of symmetry or an axis passing through the Centre of cylinder and parallel to its length. its central diameter or an axis passing through its Centre and perpendicular to the axis of symmetry.
Solution: Let h be the height of cylinder ‘O’ be the Centre and z-axis be its axis of symmetry and x-axis & y-axis are the lines perpendicular to its axis of symmetry through its centre in cylindrical coordinate (r,,z) x = r cos
;
0ra
V=
y = rsin
;
0 2
=
z=z
;
-h/2 2 h/2
h
Z h/2
O
X
-h/2
35 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
h
Y
M.I about the axis of symmetry = ∫(
+
=
) dm
x
2
y2 dV
h /2 2 a
=
(r
2
cos2 r 2 sin 2 ) r drd dz
h /2 0 0
h /2 2 a
h/2 2 a 2
=
r .r drd dz
h /2 0 0
=
4
r 4
h / 2 2
r
3
drd dz
h/2 0 0
a
d dz
h /2 0
0
h /2 2
a4 = 0 d dz 4 h /2 0 a4 h/2 2 = d dz 4 h/2 0 a 4 h /2 = 4 h /2
2
dz 0
a 4 h/2 = (2 0)dz 4 h/2 a4 h /2 = .2 z h/2 4 a4 h h .2 ( ) 2 2 4
= =
m a4 a 2 h 4 =
.2 h
m a2 h
ma2 2
36 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
(ii). M.I about central diameter = = ∫( =
+
) dm
2 2 x z dV =
h /2 2 a
(r
2
cos 2 z 2 ) r drd dz
h /2 0 0
h /2 2 a
=
(r
3
cos 2 rz 2 )drd dz
h /2 0 0 h /2 2
=
a4 a2 2 2 cos z d dz 4 2 h /2 0 h /2 2
=
a 4 1 cos 2 a 2 2 z d dz 4 2 2 h /2 0
sin 2 h /2 a4 2 = 4 2 h /2
2
a2 z 2 d dz 2 0
sin 4 a 4 2 2 = 4 2 h /2 h /2
a2 z 2 .2 0 dz 2
h /2
=
a 4 2 0 ) a 2 z 2 dz ( 4 2 h /2 h /2
a 4 a 2 z 2 dz = 4 h /2 a 4 = 4
h /2
h /2 2
dz a
h /2
h /2
a4 h/2 z3 2 z a = 4 h/2 3 =
z 2 dz h/2
h/2
3 a 4 h h h3 2 h a 4 2 2 24 24
37 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
=
3 a 4 2 h h a 4 12
=
m 2 a 2 h2 .a h a2 h 4 12
=
ma 2 mh2 4 12 =
m a2 h
ma 2 mh 2 4 12
Question: Find M.I of a sphere about its diameter or about axis passing through the centre of sphere. Y Solution: = ∫( =
+
) dm
x
2
a
y dV ….(1) 2
x = rsincos
Z
y = r sinsin z = rcos Put in Eq (1) 2 a
I zz r 2 sin 2 cos 2 r 2 sin 2 sin 2 r 2 sin drd d 0 0 0
2 a
I zz r 2 sin 2 cos2 sin 2 r 2 sin drd d 0 0 0
2 a
I zz r 4 sin 3 drd d 0 0 0
2
a
r5 I zz sin3 d d 50 0 0
38 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
X
2
a5 0 sin 3 d d 5 0 0 a5 5
2
a5 5
2
a5 5
2
3
sin d d 0 0
2
sin sin d d 0 0
2
sin (1 cos )d d 0 0
Put
cos = u -sin d = du
Where = 0 u = 1 = u = -1
a5 5
2 1
a5 5
2
a5 5
2
2
(1 u )(du)d
a5 5
0 1
u3 u 0 3
2 1
(1 u
2
)dud
0 1
1
d 1 2
1 1 1 1 0 3 3 d
m a5 4 2 4 3 5 3 0 a 3
a5 4 d 5 3 0
m a5 (2 0) a3 5
2ma 2 I zz 5
39 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Lecture # 9
Theorem: Show that in matrix notation L I ´ L where '' mean derivative w.r.t ' t ' Proof: We know that for a rotating rigid body L⃗ = ∑
⃗ × ⃗
=∑
⃗)
⃗ × (
⃗ × ⃗
=∑ =∑
………(1)
V⃗ = ω⃗ × r⃗
( ⃗ × ( ω⃗ × ⃗ )) ……..(2)
In components form ω +
ω +
ω
=
=
ω +
ω +
ω
=
ω +
ω +
ω
Or
ω ω ω
=
[L] = [I] [ω]
Now from (1) Diff. w.r.t ‘t’ n L mi [ ri ´Vi ri ´Vi ] i 1
ri Vi and Vi ´ ri
mi [Vi ´Vi ri ´ i ´ ri ] n
i 1
mi [ ri ´ ´ ri ´ ri ] n
i 1
mi [ ri ´ ´ ri ] n
i 1
ri Vi
m [ r i i ´ ´ ri ] n
i 1
40 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
m [ r i i ´ ´ ri ] n
i 1
m [ r i i ´ ´ ( ´ vi ) ] n
i 1
From (2) and (3) we note that = ∑ matrix notation as [I] [ω] .In
m [ r i i ´ ´ ri ] n
i 1
...( A)
( ⃗ × ( ω⃗ × ⃗ )) can be written in the same way we can write
as
[I] [ ⃗∙ ] .
Now ri ´ ´ ( ´ ri )
= ri ´ (.ri ) ( ´ ) ri
r ´ ( . r ) () r
2 = ri ´ ( . ri ) () ri
=
2
i
i
i
= ( . ri )(ri ´ ) ()2 (ri ´ ri )
( . ri )( ´ vi ) =
...(4)
ri ´ ri 0 and ri ´ ( ´vi )
Similarly (or replace ⃗ × ω⃗ )
´ r ´ ( ´ r ) ´ ri ´ (ri ´ ) i
i
= ( . ri ) (ri ´ ) = ( . ri ) ( ´ ri )
Comparing (4) and (5)
ri ´ ´ ( ´ ri )
Now Apply ∑
ri ´ ´ ri
both side
m ( r i i ´ ´ ( ´ ri ) ) n
= ´ ri ´ ( ´ ri )
..(5)
i 1
m ( = i ´ ri ´ ( ´ ri ) ) n
i 1
n
= ´( mi ri ´ ( ´ ri ) ) i 1
n L mi (ri ´ ´ ri
= ´L
i 1
41 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
m ( r i i ´ ´ ( ´ ri ) ) n
So
i 1
can be written in matrix form as [ ⃗´ ⃗]
Hence from (A) L [ I ][ ] ´ L proved
Question:
X
Find M.I of a tringular lemina about one of its edges. A
Solution: p
M.I about y-axis )
(
=∫
x P R p-x
Q
B
∫( − ) dA
A
dm dA
…..(1)
PQRS become a rectangle
x
p
Area = length ´ width dA = |
Y
C
a
= ∫( − ) dm =
S
Q
P
|dx ⃗
⃗
=
⃗ ⃗
⃗
Area = ap
=
⃗ =
⃗ dx =
dA =
=
I yy
Put in (1)
( p x)2
x 0
a p
a p
2
x( p x 2 px)dx
x0
2
3
xp x 2 px
a p2 p 2
. p2
=
ax dx p
x p 2
=
dx x p
a
B
Since the APQ and ABC are similar so
2 p 0
a p
a p2
x p
( xp 2 x3 2 px 2 )dx
x0
p4 2 p4 .p (0) p 2 4 3
2
p4 2 p4 a 6 p4 3 p4 8 p4 4 3 p 12
I yy
mp 2 6
42 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
C
Lecture # 10
Equimomental System: Two systems are said to be Equimomental if they have same M.I about any line space.
Principle axis: Three mutually perpendicular axis are called Principle axis if product of inertia realtive to these axis if product of inertia relative to these axis is zero and corresponding M.I’s are called Principle moment of inertias i.e. A set of three mutaully axis relative to which Inertia matrix is diagonal.
Theorem: Two systems are said to be equimomental iff (i) (ii)
They have same masses They have same centroid (center of mass of an object with uniform density) They have same Principle axis and same Principle M.I at cenroid
(iii) Proof:
Suppose that condition (i) , (ii) and (iii) holds. Let and be two systems each having same masses ‘m’ same centroid ‘C’ and same Principal axis at centroid. Consider a line ‘l’ having direction cosines , and v passing throug centroid C. The M.I of system
about line ‘l’ is
I l l 2 I xx 2 I yy v 2 I zz 2l I xy 2 vI yz 2l vI xz 1
Since given axis are Principle axis
=
=
.....(1)
=0)
Therefore from (1) we have I l1 l 2 I xx 2 I yy v 2 I zz
Similarly M.I of system
....(2)
about line ‘l’ is
I l l 2 I xx 2 I yy v 2 I zz 2
....(3)
From (2) and (3) I l1 I l 2 I 43 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Let
be the line parallel to ‘l’ at a distance ‘d’. then M.I of
Il(1) I md 2 ' Similary M.I of
about line ‘ ’ is
..(4) ( By Parallel Axistheorem)
about line ‘ ’ is
Il(2) I md 2 '
..(5) ( By Parallel Axistheorem)
From (4) and (5) Hence both system
and
are equimomental
Conversely: (i)
Let
and
are equimomental
Consider a line ‘ ’ passing through centroids and of system respectively. Since system and are equimomental therefore
and
I l1 I l 2 I
Let ‘ ’ be the line parallel to ‘ ’ at a distance ‘ ’. Then M.I of ‘ ’ is
I l(1) I m1d12 2 Similarly M.I of
..(6) ( By Parallel Axistheorem)
about line ‘ ’ is
Il(2) I m2 d12 2 As
and
about line
..(7) ( By Parallel Axistheorem)
are equimomental so
I l(1) 2
=
I m1d12
I l(2) 2 = I m2 d12
m1 = m2
They have same masses. (ii)
Now consider a line ‘ ’ passing through centroid and to ‘ ’. Also consider a line ‘ ’ passing through and to ‘ ’. This means ‘ ’ and ‘ ’ are parllel
Now M.I of
about line ‘ ’ is
Il(1) Il(1) md 2 2 4 3 Similarly M.I of
Il(2) I l(2) md2 2 3 4
( By Parallel Axistheorem) about line ‘ ’ is
( By Parallel Axistheorem) 44
Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
As
and
are equimomental. So = I l(2) I l(1) 4 4
I l(1) md 2 2 3
2md 2 2 0
d2 0
and
I l(2) md 2 2 4
coincides.
=
same centroid (iii)
Now the two systems have same centroid and same masses if we take any line through the common centroid say G i.e. G =
=
The two systems have the same M.I about that line. It means that we have same Principle axis and same M.I at point G. Hence all the conditions are satisfied which complete the proof.
Question: Show that a circular plate of mass ‘m’ and radius ‘a’ is equimomental with a loop of same mass ‘m’ and radius √
Solution: We know M.I of circular plate about an axis passing through its Circular plate
centroid and to its plane =
loop
M.I of loop about same axis =
√
=
Hence circular plate and loop are equimomental.
√2
a
45 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Lecture # 11
Theorem (Existence of Principle axis theorem): Statement: For a rigid body a set of three mutually perpendicular axis relative to which product of inertia are zero and angular velocity ⃗ and a linear momentum ⃗ are in same direction. Proof: Consider an axis through a point ’0’ such that ⃗ and ⃗ are parallel to it. Then we can write ⃗= ⃗
; where is contant of proportionality
= = =
………(1)
Next we know that (In general )
L1 I111 I122 I133 L2 I 211 I 222 I 233 ....(2) L3 I 311 I322 I 333 From (1) and (2) we can write
( I11 l )1 I122 I133 0 I 211 ( I 22 l )2 I 233 0 ....(3) I311 I 322 ( I 33 l )3 0 This is homogenous system of equation which unknown , , . It has nontrivial soltution when determinant of the matrix made by the coefficient in set of equation (3) is given i.e.
=0 I11 l I 21
I12 I 22 l
I13 I 23
I 31
I 32
I 33 l
0
Which is cubic equation in which has three relation = , , . These relation are actually eigen value of the inertia matrix and these values are
46 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
principal moment of inertia of the rigid body. The corresponding eigen vectors give the direction of principal axis.
Working rule of finding Principle M.I and Principle Axis: (i) (ii)
(iii)
Find inertia matrix at given point (say origin) If inertia matrix is diagonal then corresponding axis are Principle. If inertia matrix is not diagonal the eigen values matrix is not diagonal the eigenvalues of this matrix give values of Principle moment of Inertia. Eigenvectors corresponding to each eigenvalue gives the direction of Principle axis.
Question: Consider a system of particle having masses m,2m,3m,4m located at points (a,a,a) . (a,-a,-a),(-a,a,-a),(-a,-a,a) respectively. Find Principle moment of inertia of the system about origin. Solution: n
I xx mi ( yi2 zi2 ) i 1
m (a 2 a 2 ) 2m (a 2 a 2 ) 3m (a 2 a 2 ) 4m (a 2 a 2 ) 2ma 2 4ma 2 6ma 2 8ma 2 20ma 2 n
I yy mi ( xi2 zi2 ) i 1
m (a 2 a 2 ) 2m (a 2 a 2 ) 3m (a 2 a 2 ) 4m (a 2 a 2 ) 2ma 2 4ma 2 6ma 2 8ma 2 20ma 2 n
I zz mi ( xi2 yi2 ) i 1
m (a 2 a 2 ) 2m (a 2 a 2 ) 3m (a 2 a 2 ) 4m (a 2 a 2 ) 2ma 2 4ma 2 6ma 2 8ma 2 20ma 2
47 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
n
I xy mi xi yi i 1
[m (a 2 ) 2m ( a 2 ) 3m ( a 2 ) 4m (a 2 )] [ m a 2 2m a 2 3m a 2 4m a 2 ] I xy 0 I yx n
I yz mi yi zi i 1
[m (a 2 ) 2m (a 2 ) 3m (a 2 ) 4m ( a 2 )] [ m a 2 2m a 2 3m a 2 4m a 2 ] I yz [4ma 2 ] 4ma 2 I zy n
I xz mi xi zi i 1
[m (a 2 ) 2m ( a 2 ) 3m (a 2 ) 4m ( a 2 )] [ m a 2 2m a 2 3m a 2 4m a 2 ] I xz [2ma 2 ] 2ma 2 I zx I xx I yx I zx
I xz 20ma 2 I yz 0 I zz 2ma 2
I xy I yy I zy
0 20ma 2 4ma 2
2ma 2 4ma 2 20ma 2
For Eigenvalue I xx l
I xy
I xz
I yx
I yy l
I yz
I zx
I zy
I zz l
0
20ma 2 l
0
2ma 2
0
20ma 2 l
4ma 2
2ma
2
4ma
2
0
2
20ma l
By expanding we get
20ma
2
l 384m 2 a 4 l 2 40ma 2 l 4m 2 a 4 0
48 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
20ma
2
l 0 and 380m 2 a 4 l 2 40ma 2 l 0
l 20ma 2
,
l
l 2 40ma 2 l 380m2 a 4 0
( 40ma 2 ) 1600m 2 a 4 1520m 2 a 4 2(1)
40ma 2 80m2 a 4 l 2(1)
l
40ma 2 4 5ma 2 2
l 20ma 2 2 5ma 2 l 20ma 2 2 5ma 2
,
l 20ma 2 2 5ma 2
l 2(10 5)ma 2
,
l 2(10 5)ma 2
l 20ma 2
,
l 20ma 2 2 5ma 2
are the Principle M.I about origin.
49 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Lecture # 12
Spherical Top: A rigid body is said to be spherical top if its all Principle moment of inertias are equal. i.e. = = . A sphere is an example of spherical top with axis passing through the centre of sphere.
Symmetrical Top: A rigid body is said to be symmetircal top if
= . A cylinder is an example of symmetrical top (taking third axis along the axis of cylinder).
Asymmetrical top: A rigid body is called Asymmetrical top if
. A rigid body in general is example of Asymmetircal top.
Rotor: A rigid body is called rotor if its two Principle M.I’s are equal and third M.I is zero i.e. = and = 0 or molecule is an example of rotor.
= 0 and
=
. A diatomic l
Question:Find equimomental system of particle for a rod of mass M.
m A
-a
M-2m m aB 0
Solution: Consider a uniform rod of length 2a. If ‘0’ be the centre of mass. Let the masses m, M-2m , m are located at point A,O,B respectively. The system of particle will be equimomental with the rod. If its M.I about any line is equal to M.I of the rod about the same line. Let ‘l’ be the line through ‘0’ perpendicular to this rod. Then M.I of rod about y-axis (axis passing through cenroid of rod) = = (Mass) (ℎ
ℎ) = M
M.I of system of particle about y-axis = m(− ) +(M-2m) (0) +m( ) = 2m If both system are equimomental then
=
Hence if we take two particles each having mass
M
= 2m
m=
at end points of rod and a
particle mass M at the centre of rod then this system of three particles is equimomental with the given rod of mass M 50 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Lecture # 13
Euler’s Dynamical equations: Consider a rigid body is rotating in OXYZ (Cartesian) coordiante system such that OX, OY & OZ are Principle axes. Then we know that Or in matrix form LI L1 I11 L I 2 21 L3 I 31 L1 I11 L 0 2 L3 0
I12 I 22 I 32 0 I 22 0
OX, OY and OZ are Principle axes. So
L1 I11 L 0 2 L3 0
0 I 22 0
I13 I 23 I 33
1 2 3
0 0 I 33
1 2 3
=
=
0 0 I 33
=0
I111 I 22 2 I 333
In vector form angular momentum is given below L I111 i I 222 i I 333 k We know that rate of change of vector function in fix and rotating coordiante system is related as follow dV dV ´V dt f dt r
Putting ⃗ = ⃗ Where V is any vector function dL dL ´ L dt f dt r dL I111 i I 222 i I 333 k ´ L dt f
51 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
dL Here is external torque acting on the system dt f d L d r ´ P d P V ´P r dt dt dt = V ´ P r ´ F = V ´ mV r ´ F = 0 + r´ F = So I111 i I 222 i I 333 k ´ L ….(A)
i
j
´ L 1 2 I111 I 222
k
3 I 333
´ L I3323 I 2223 i I 3313 I1113 j I 2212 I1112 k
´ L I 22 I 33 23 i I 33 I11 13 j I11 I 22 12 k
Put in (A)
I111 i I 222 i I 333 k I22 I33 23 i I33 I11 13 j I11 I22 12 k
1 I111 ( I 22 I 33 )23 2 I 222 ( I33 I11 )13 3 I 333 ( I11 I 33 )12 Are Eulers Dynamic equations Results: If there is no external torque then I111 ( I 22 I 33 )23 0 I 222 ( I 33 I11 )13 0
(ii)
I 333 ( I11 I 33 )12 0
Multiplying (i) by
, (ii) by
(i)
and (iii) by
(iii) and adding
I1111 I 22 I33 123 I 2222 I33 I11 123 I 3333 I11 I 22 123 0
52 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
I11 1 1 I22 1 23 I33 1 23 I2222 I33 1 23 I11 1 23 I33 3 3 I11 1 23 I22 1 23 0
I 1111 I 22 2 2 I 33 3 3 0
1 I11 .211 I 22 .2 2 2 I 33 .2 3 3 0 2
1 d I1112 I 22 22 I 33 32 0 I1112 I 2222 I 3332 = constant 2 dt
Now we know that Rotational K.E is T =
1 L 2
=
1 1 L1 2 L2 3 L3 2
=
1 1 I111 2 I 222 3 I333 2
=
1 I1112 I 2222 I 3332 = constant 2
In the absence of external torque the rotational K.E of system is constant. Now multiplying (i) by
, (ii) by
and (iii) by
and adding
2 2 I112 1 1 I11 I22 I33 123 I2222 I22 I33 I11 1 23 I3333 I33 I11 I22 1 23 0
2 I112 11 I11I22123 I11I33123 I222 22 I22 I33 1 23 I11 I22 1 23 I3333 I11 I33123 I22 I33 1 23 0
2 I 112 11 I 22 2 2 I 332 3 3 0
1 2 2 I11 .211 I 22 .2 2 2 I 332 .2 3 3 0 2
1 d 2 2 2 I111 I 22 22 I 332 32 0 2 dt
I
12 I 222 22 I 332 32 = constant
2 11
Now we know L I111 i I 222 i I 333 k L I112 12 I 222 22 I 332 32 = constant
In the absence of external torque magnitude of angular momentum is constant. 53 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Lecture # 14
Question: Find M.I of a solid right circular cone. (i) (ii)
About its axis of symmetry About an axis passing through its base and also through its center OR About the diameter of its base.
Solution: (i) Let M be the mass ‘a’ the radius and ‘h’ be the height of right circular cone. We choose the z-axis along the axis of symmetry and consider a typical disc of radius r and width dz at a distance z from the base. Z Since ∆OAC and ∆BCD are similar triangles. C
So r=
= (
)
h
B
=
=
r
D
dz
Now mass of disk = dm = dv = dz Also
Vertex
=
O a
z A
We know that Y
M.I of disk about z-axis (as shown in fig) =
dm
Now M.I of cone about z-axis (as shown in fig) h
1 1 r 2 dm r 2 r 2 dz 2 20 h
1 r 4 dz 2 0 h
h
1 a 4 (h z ) 4 1 a 4 dz ( ( z h)) 4 dz 4 4 2 h 2 h 0 0
1 a 4 ( z h)5 2 h4 5
h
0
54 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
X
1 3M a 4 ( h h)5 (0 h)5 . 2 a 2 h h 4 5 5
3Ma 2 2h 5
(h)5 3Ma 2 (h)5 5 2h 5 5
3Ma 2 10 (ii). Note that for disk I X ' IY '
....(1)
Now by axis theorem I X ' IY ' I Z '
I X ' IY ' & I Z ' I Z
2 I X ' IZ
IX'
1 I X ' r 2 dm 4
1 1 1 I Z r 2 dm 2 2 2
Now by // axis theorem M.I of disk about x-axis (as shown in fig)
I X ' z 2 dm
1 2 r dm z 2 dm 4
1 2 r ( r 2 dz ) z 2 ( r 2 dz ) 4
1 r 4 dz z 2 r 2 dz 4
a 4 (h z ) 4 a 2 (h z ) 2 1 2 dz z dz 4 2 4 h h
dm r 2 dz
55 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
a 4 4h
4
a 4 4h
4
a 2
(h z )4 dz
4h
2
.z 2 (h 2 z 2 2hz )dz
a 2
(( z h)) 4 dz
4h
2
(h 2 z 2 z 4 2hz 3 )dz
Now M.I of cone about x-axis (as shown in fig) =
h 1 a 4 a 2 2 2 4 4 3 ( z h ) dz ( h z z 2 hz ) dz 4 2 2 0 4h h
a 4 4h
4
h
( z h)
4
dz
a 2 h
0
a 4 ( z h)5 4h 4
5
h
2
h2
2
z 4 2hz 3 )dz h
z3 z5 z4 h 2 h 3 5 4 0
a 4 4h 4
2
(h z 0
a 2
0
h
h5 a 2 0 2 5 h
2
2 h3 h5 h4 h 3 5 2h 4
a 4 3M h 5 3M a 2 h 5 h 5 h 5 4 0 4h a 2 h 5 a 2 h h 2 3 5 2
a 4 3M h 5 3M a 2 h 5 h 5 h 5 4h4 a 2 h 5 a 2 h h 2 3 5 2
3Ma 2 3M 20h5 12h5 30h5 3 20 h 60
3Ma 2 3M 2h5 3 20 h 60
3Ma 2 3M 3 2h 2 20 h
M 3a 2 2h 2 20
56 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
[email protected] 0316-7017457
By Muzammil Tanveer
Lecture # 1
Mechanics: Mechanics is the branch of science which studies the state of rest and motion of objects and laws governing rest, equilibrium and motion. Since material objects exist in the form of liquids gases and solids there are corresponding types of mechanics to deal with them. (i) (ii) (iii)
Kinematics Dynamics Statics
Kinematics: Kinematics is the branch of mechanics which describe the motion of objects without consideration of their masses and force acting on them.
Dynamics: Dynamics is the branch of mechanics concerned with the motion of objects under the action of force.
Statics: Statics is the branch of mechanics concerned with objects at rest or in equilibrium under the action of forces.
Ch# Rigid Body Motion: Particle: An extremely small part of matter or an infinitesimal part of matter having negligible dimension is called a particle.
Rigid Body: A collection of particles such that distance between any two particles remains same, irrespective to the motion of the body or force acting upon. e.g.
metallic balls, wood, stones etc.
Note:
The concept of rigid body is an idealization.
Displacement: Directed change in position of particles of a rigid body is called displacement.
1 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Translational Motion: The motion in which all particles of a rigid body are displaced such that the line joining initial and final position of particles are parallel to each other.
Rotational Motion: If during the motion particles of the rigid body move in curved path or circular path or are displaced through some angle about (an imaginary) line called axis of rotation. Then that motion is called Rotational motion.
Chasles’ Theorem: The most general displacement of a rigid body is composed of pure translation followed by a rotation about some point (base point). ⃗ = (displacement)
+ (displacement)
⃗ = dt ⃗ + ( ω⃗ × r⃗) ⃗
= ⃗ + ( ω⃗ × r⃗)
where ω⃗ is angular velocity of body and r⃗ is position vector of P with respect to A (base point). ⃗ =
⃗ + ( ω⃗ × r⃗)
since
⃗
= ⃗
2 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Lecture # 2
Angular Equation motion: We know that = ω = dt By integrating ω = t+c (i) ; where c is constant of integration and can be found by using initial condition t = 0 then ω = ω ω = (0)+c
ω = t+ω
Put in (i)
c=ω (ii)
called first angular equation of motion. Now from (ii)
= t+ω
since ω =
d = (t+ω )dt = + ω
+c
(iii)
When t = 0 , = 0 put in (iii) 0 = a (0) + ω (0) + c c =0
Put in (iii)
= + ω
+0
= + ω
(iv)
Called Second Angular equation of motion. Now from (ii) t = ω − ω (
Put in (iv) =( =(
a
a
a
= ) )
a .
a
)
+ ω (
a
+ ω =
a
) (
a
t=
a
)
+ ω
or 2 = ω − ω
Called the third Angular equation of motion. 3 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Note: These equations are only useful when a rigid body is rotating with angular velocity about a point ‘o’.
Question: A rigid body is rotating about at a point say ‘0’. Find ⃗ (i) (ii)
ω⃗ = 2 ̂+3 ̂ − ω⃗ = ̂ −
r⃗ = ̂+ ̂ + r⃗ = 2 ̂ − ̂ − 2
, ,
Solution: (i)
ω⃗ = 2 ̂+3 ̂ −
,
r⃗ = ̂+ ̂ +
⃗ = ω⃗ × r⃗ ̂ ̂ ⃗ = 2 3 −1 1 1 1 ⃗ =4 ̂−3 ̂− (ii)
ω⃗ = ̂ − ̂ ̂ ⃗ = 1 0 −1 2 −1 −2 ⃗ =− ̂−0 ̂−
̂(3 + 1) − ̂(2+1)+ (2 − 3)
r⃗ = 2 ̂ − ̂ − 2
,
̂(0 − 1) − ̂(-2+2)+ (−1 − 0)
Screw Motion: The motion which consist of translation and rotation about a line along the translation is called screw motion. In this motion linear velocity of each particle on the axis of rotation is parallel ( or anti parallel) to the angular velocity.
Theorem: Find the equation of axis of rotation in vector form in case of screw motion Or Show that equation of axis of rotation in case of screw motion is r⃗ = a⃗ + ω⃗ where r⃗ is position vector of any point on the axis of rotation and a⃗ is any vector, is scalar and ω⃗ is the angular velocity Or Show that instantaneous general motion is screw motion. Proof: Consider a rigid body in general motion. Let A be the base point of the body with linear velocity ⃗. Let B be any other particle then Linear velocity of B is 4 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
⃗ =
⃗ + ( ω⃗ × r⃗)
(i) ; where r⃗ is position vector of B from base point .
In general, ⃗ and ω⃗ are not parallel (or antiparallel) but we can choose B such that Linear velocity ⃗ of B is parallel to angular velocity ω⃗ of the rigid body. Taking cross product of (i) with ω⃗ ω⃗ × ⃗ = ( ω⃗ × ⃗) + ω⃗ × ( ω⃗ × r⃗ ) Since ω⃗ ‖ ⃗ so ω⃗ × ⃗ = 0 0 = ( ω⃗ × ⃗) + ( ω⃗ . r⃗ ) ω⃗ −( ω⃗ . ω⃗ )r⃗ = ( ω⃗ × ⃗) + ( ω⃗ . r⃗ ) ω⃗ − ω r⃗ Since ω⃗ . ω⃗ = | ω⃗|| ω⃗|
0=ω
ω r⃗ = (ω⃗ × ⃗) + ( ω⃗ . r⃗ ) ω⃗ ⃗ × ⃗
r⃗ =
+
⃗ .⃗
ω⃗
r⃗ = a⃗ + ω⃗ ⃗ × ⃗
where a⃗ =
and =
⃗ .⃗
Question: Show that ω⃗ = curl ⃗ (for rotation). Solution: ∇=
̂+
̂+
⃗=
̂+
̂+
̂
̂
∇× ⃗=
̂+ (ω
−
̂+
̂+
̂ = ω
−ω
) ̂+ (ω
−ω ) + ̂ (ω
−ω
=ω
−ω
,
=ω
,
=ω ,
−
̂+
−
… (i)
̂ ω
= −ω ,
ω
−ω
= −ω ,
) =ω
=ω ,
−ω
= −ω ,
=ω
5 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Putting above values in eq. (i) ∇ × ⃗ = (ω − (−ω )) ̂ +(ω − (−ω )) ̂+(ω − (−ω )) = (ω + ω ) ̂ +(ω + ω ) + ̂ (ω +ω ) = 2ω ̂ +2ω + ̂ 2ω
or 2[ω ̂ +ω ̂+ω
]
(∇ × ⃗ ) = ω⃗ Or ω⃗ = curl ⃗ proved.
Question: A rigid body is rotating about a fix point ‘o’. The points A(0,-1,2) and B(2,0,0) are moving with velocities ⃗ = [7,-2,-1] and ⃗ = [ 0,6,-4] respectively. Find the angular velocity of the body. Solution: Position vector of ‘A’ ⃗ = [0,-1,2] ⃗ = ω⃗ × ⃗ ̂ [7,-2,-1] = ω 0
̂
̂ (−ω − 0) ω ω = (2ω + ω ) ̂ − (2ω − 0) + −1 2
[7,-2,-1] = (2ω + ω ) ̂ − 2ω ̂−ω 7 = 2ω + ω
-2 = −2ω
,
ω = 1
Position vector of ‘B’ ⃗ = [2,0,0] ⃗ = ω⃗ × ⃗ ̂ [0,6,-4] = ω 2
̂ ω 0
̂ (0 − 2ω ) ω = (0 − 0) ̂ + (2ω − 0) + 0
[0,6,-4] = 0 ̂+2ω ̂-2ω 0=0
,
6 = 2ω
−2ω = -4
ω =2
Angular Velocity ω⃗ = ω ̂ +ω ̂+ω
ω =3
ω⃗ = ̂ +2 ̂+ 3
6 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Lecture # 3
Question: The instantaneous linear velocity of three particles A(a,0,0) , B(0,
√
, 0) and
⃗ = (u,0,0), ⃗( , 0, ) and ⃗(u+v,−√3 , ) respectively w.r.t the cartesian coordinate system. Find |ω⃗| of rigid body. C(0,0,2a) are Solution: Let A(a,0,0) be a base point (fix point). We know that linear velocity of any particle ‘P’ of a rigid body in case of general motion ⃗ = ⃗ + ( ω ⃗ × ⃗ )
⃗ is position vector of P w.r.t A
Now ⃗ = (− , ,0) and ⃗ = (− , 0,2 ) √
,ω ,ω )
⃗ − ⃗ = ( ω ⃗ × ⃗ )
ω ⃗ = ( ω
⃗ = ⃗ + ( ω ⃗ × ⃗ )
Then
̂ ω −
( 0,0,V) = = ̂ (0− ( 0,0,V) =
̂ (−
̂ ω
ω 0
√
) + ̂(− ω − 0) + (
√
√
) + ̂(− ω ) + (
√
√
+ ω ) + ω )
On comparig −
√
=0 , − ω =0 ,
√
+ ω =V
ω =0 Now
⃗ = ⃗ + ( ω ⃗ × ⃗ ) ⃗ − ⃗ = ( ω ⃗ × ⃗ )
̂ (V , −√3V , ) = ω −
̂ ω 0
ω 2
= ̂ (2aω − 0)+ ̂(− ω − 2 ω ) + (0+aω ) (V , −√3V , )
= ̂ (2aω )+ ̂(− ω − 2 ω ) + (aω )
On comparing 7 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
2aω = V , ω =
− ω − 2 ω = −√3V
,
− (0) − 2 ω = −√3V
,
aω = ω =0
−2 ω = −√3V ω = √
ω ⃗ = (
,
, 0) √
Now |ω ⃗| =
√
+ (0)
+
|ω ⃗| =
=
+
|ω ⃗| =
Ans
Question: The instantaneous linear velocity of three particles A(a, 2a, − ) , B(− , − , ) and C(a,a,a) are
⃗ = (√ , 0 , √ ), ⃗ −
√
, 0, −
√
and ⃗(0, −
, )
√ √
respectively. Find direction of cosine of axis of rotation. Solution: Let A(a, 2a, − ) be a base point. We know that Linear velocity of any particle ‘P’ of a rigid body in general motion is ⃗ = ⃗ + ( ω ⃗ × ⃗ ) Now ⃗ = (−2 ,−3 ,2 ) and ⃗ = (0, − , 2 ) ,ω ,ω )
⃗ − ⃗ = ( ω ⃗ × ⃗ )
−
ω ⃗ = ( ω
⃗ = ⃗ + ( ω ⃗ × ⃗ )
Then
√
, 0, −
√
̂ = ω −2
̂ ω −3
ω 2
= ̂ (2aω + 3aω ) + ̂(−2 ω −2aω ) + (−3 ω −2aω ) On Comparing 2aω + 3aω = −
√
,
−2 ω −2aω = 0 , −3 ω −2aω = −
√
−ω = ω Now
⃗ = ⃗ + ( ω ⃗ × ⃗ ) 8
Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
⃗ − ⃗ = ( ω ⃗ × ⃗ ) (
√
,
√
,
√
̂ )= ω 0
̂ ω −
ω 2
= ̂ (2aω + aω ) + (̂ 0 −2aω ) + (− ω −0) = ̂ (2aω + aω ) + (̂ −2aω ) + (−aω ) On Comparing √
2aω + aω =
, −2aω = ω =
ω =
And 2aω + a( ω = ω ⃗ = (
√ √
√
,
√
√
√
)=
√
, −aω =
√
√
Now |ω ⃗| =
,
√
)
+
√
|ω ⃗| =
√
+
=
+
√
|ω ⃗| =
+ is the
magnitude of Angular velocity. ⃗ | ⃗|
=(
√
,
√
,
√
)
(
√
,
√
,
√
)
We know that direction of cosine of the axis of rotation are the component of unit vector along the angular velocity and are (
√
,
√
,
√
).
Inertia: Tendency of a rigid body to remain at rest or of a body in motion to stay in motion unless an external force acted upon. Or Measure of resistance in Linear acceleration.
Moment of Inertia (Rotational Inertia) Measure of resistance of a rigid body in angular acceleration. 9 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Mathematically I = m Where r is perpendicular distance of rigid body from a line about which it is rotating and r =
is radius of Gyration (Latin word means rotation or rapid
motion in circular path). In case of system of particles +
I=
+……..+
I=∑ Radius of Gyration =
∑
∑
=
Where M = ∑
For Continuous or Uniform mass distribution: I=∫
dm
Z
= m ( distance from y-axis) =m(
mass(m)
)
+
P(x,y,z)
Momentum: P⃗ = mV⃗
z X
Angular Momentum: L⃗ = r⃗ ´ P⃗
x
For system of particles L⃗ = ∑
⃗ × ⃗
=∑
Y
⃗)
⃗ × (
⃗ × ⃗
=∑ =∑
( ⃗ × ( ω⃗ × ⃗ ))
=∑
[( ⃗ . ⃗ )ω⃗ − ( ⃗ . ω⃗) r⃗)] If ⃗ = ( ,
=∑
∑
, ) and ω⃗ = (ω , ω , ω )
[( ⃗) ω , ω , ω [(
+
V⃗ = ω⃗ × r⃗
+
− {( ,
) ω ,ω ,ω
, ). ω , ω , ω }( ,
, )]
−( ω + ω + ω ) ( ,
, )]
10 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
[(
=∑
+
ω +
(
ω +
=∑
[(
[
=∑
ω ), ( ω +
= [{∑
ω +
ω +
ω ),
−( ω +
ω +
ω ),
ω
ω +
ω +
ω )−(
ω +
ω +
ω +
ω −
ω −
ω −
ω ,
ω +
ω +
ω −
ω −
ω −
ω ,
ω +
ω +
ω − −ω
) ω − ω
−ω
) ω − ω
−ω
) ω + (− ∑
+
(
ω −
ω −
, ] )ω + (− ∑
)ω }, , )ω } ,
(
+
) ω + (− ∑
)ω ) + (− ∑
{∑
(
+
) ω + (− ∑
)ω + (− ∑
ω +
ω +
ω ,
=
ω +
ω +
ω
=
ω +
ω +
ω
=
ω +
ω +
ω
Or
ω +
ω ]
,
{∑
=[
ω )]
ω )]
ω +
) ω − ω
)ω −
+ ω +
ω )−(
ω ), (
+ ω +
ω +
+
(
ω +
ω +
+
(
)ω , (
+
ω +
+
[(
+
ω +
( =∑
)ω ,(
+
ω +
ω ,
)ω }] ω +
ω +
ω ]
ω ω ω
=
Diagonal element (
,
,
) called Moment of Inertia about coordinate axis.
Upper and lower triangular matrix element called product of inertia about coordinate axis. L = I ω⃗ 11 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Where
=∑
+
(
)
=∑
(
+
)
=∑
(
+
)
= (− ∑
)
= (− ∑
)
= (− ∑
)
For continous (uniform mass distribution = ∫(
+
= −∫
)
, ,
= ∫( = −∫
)
+
,
,
= ∫(
= −∫
+
)
12 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Lecture # 4
Question: Find moment of inertia of a uniform rod about an axis passing through one of its extremities or passing through its end points and perpendicular to the length rod. Solution: Let the mass of the rod is ‘m’ and length ‘a’. We take small particle of mass dm Y and length dx and coordinate (x,0,0). mass ‘m’ dx dm Moment of Inertia of rod about Y-axis = ∫(
+
=∫
dm
) dm
O
z=0
(x,0,0)
As mass dm dm dx length
dx
a
a
0
0
I yy x 2 dx x 2 dx 3 a
a3 x 0 3 0 3
m a
m a 3 ma 2 a 3 3 mass lengthof rod I yy 3
2
If length of the rod is 2a mass (length of rod ) 2 m(2a ) 2 I yy 3 3 4ma 2 3 4ma 2 I yye ...........( A) 3
I yye , e end po int
13 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
a
X
Question:Find M.I of uniform rod about an axis passing through centre of rod and to the length of rod. Solution: Let the mass of rod in ‘m’ and length ‘2a’. Take a small particle of mass ‘dm’ having length dx and coordiantte (x,0,0) Y M.I about Y-axis = ∫(
+
=∫
dm
) dm
(x,0,0)
z = 0 -a o
a
As mass dm dm dx length
X
dm
dx
dx
a
a
a
a
I yy x 2 dx x 2 dx 3 a
a3 a3 3 3 a
x 3
m a
m 2a 3 ma 2 2a 3 3 ma 2 ........( B ) 3 From Aand B
I yyc , c centre
I yyc
I yyc 4 I yye
Question: Find M.I of rectangular lemina about an axis in the plane of lemina and passing through any of its edges. Soluiton: Let ‘a’ be the length and ‘b’ be the width of rectangular lemina. Take a point having mass dm and area dxdy and coordinates (x,y,0). Moment of inertia about width I yy ( x 2 z 2 ) dm
Y z 0
x 2 dm
b
0
a
0
b
dm (x,y,0)
O
a
dm dxdy
2
x dxdy
Area = dxdy
14 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
X
a
b
0
x3 dy 3 0
3 a a I yy 0 dy 0 3 3 3 a b a y 0 b 0 3 3 m a 3b ma 2 ab 3 3
m ab
ma 2 mass(lengthof Re c tan gular le min a) 2 3 3 Moment of inertia about length
I yyc
I xx ( y 2 z 2 ) dm z 0
y 2 dm
b
0
a
0
dm dxdy
y 2 dxdy
b
a
y 2 x 0 dy 0
b
y 2 (a 0)dy 0
b
y 2 dy 0
3
a
y 3
b
0
3
b a 0 3 3 m ab . ab 3 mb 2 mass( widthof rec tan gular le min a) I xx 3 3
m ab
15 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Question:Find M.I of square lemina about an axis in the plane of lemina and passing through any of its edges. Solution: Let the length and the width of square lemina is ‘a’. Take a small particle of mass dm and area dxdy and coordinates (x,y,0). Y Moment of inertia about width I yy ( x 2 z 2 ) dm
a z 0
x 2 dm
a
0
a
0
a
0
O
a
dm dxdy
2
x dxdy
x3 3
dm (dxdy) (x,y,0)
a
0
a3 I yy 0 dy 0 3 a3 a a3 y 0 a 0 3 3 4 m a ma 2 2 a 3 3 a
m m a.a a 2
Similarly, ma 2 I xx 3
Question: Find M.I of a rectagular lemina about an axis in the plane of lemina parallel to its edges and passing through its centre. Solution: Let the length of rectangular lemina is ‘a’ and width ‘b’ . Take a small particle of mass dm and area dx,dy and coordiante (x,y,0) M.I about width I yy ( x 2 z 2 ) dm z 0
x 2 dm 16 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
X
b
a 2 2 b a 2 2
3
b
b2
2
x 3
a
dm dxdy
x2 dxdy
Length = a b/2
2
a
Width = ab
dm (x,y,0)
a/2
-a/2
2
-b/2
a 3 a 3 b 2 2 I yy b2 dy 3 2 3 3 3 b 2a 2a b b y b2 2 24 24 2 2
m ab
m 2a 3 ma 2 .b ab 24 12 ma 2 mass (length of rec tan gular le min a ) I yy 12 12
M.I about length I xx ( y 2 z 2 ) dm z 0
y 2 dm b
b2
a
2 a 2 2
b a
x a2
2
2
b
y 2dy
2
3 a a y 2 2 3 3
dm dxdy
y 2 dxdy
b
2
b
2
3
b b a 24 24 am 2b3 mb 2 ab 24 12
m ab
mb 2 mass(lengthof rec tan gular le min a) I xx 12 12 17 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
For MCQ I xxe
mb 2 3
e edge , end poin t
I xxc
mb 2 12
c centre
mb2 2b3 I 4 4I xxc 3 12 e xx
I xxe 4I xxc
Question: Find M.I of square lemina about an axis in the plane of lemina parallel to its edges and passing through its center. Solution: Let the length and width of square lemina is ‘a’.Take a small particle of mass dm area dxdy and coordinates (x,y,0). M.I about width
a/2 dm
I yy ( x 2 z 2 ) dm z 0
x dm a
a 2 2 a a 2 2
a
a2
2
x3 3
(x,y,0) a/2
-a/2
2
a
dm dxdy
x2 dxdy
-a/2 2
a
2
a 3 a 3 a 2 2 2 I yy a dy 3 2 3 2a 3 b 2 a3 a a y b 2 24 12 2 2
m a2
ma 3 a ma 2 2 a 12 12 I yy
ma 2 ma 2 Similarly, I xx 12 12 18
Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Lecture # 5
Wallis Formula’s: (i)
If ‘m’ is even then =
(ii)
=
(
− 1)( − 3) … … … 1 . ( − 2)( − 4) … . .2 2
If ‘n’ is odd then =
(iii)
=
( − 1)( − 3) … … … 1 ( − 2)( − 4) … . .2
If both ‘m’ and ‘n’ are even then =
(iv)
(
− 1)( − 3) … … … 1. ( − 1)( − 3) … … 1 . ( + )( + − 2)( + − 4) … . .2 2
If both ‘m’ is even and ‘n’ is odd then =
− 1)( − 3) … … … 1. ( − 1)( − 3) … … 2 ( + )( + − 2)( + − 4) … . .2 If both ‘m’ and ‘n’ are odd then =
(v)
=
(
(
− 1)( − 3) … … … 2. ( − 1)( − 3) … … 2 ( + )( + − 2)( + − 4) … . .2
Theorem: (Only statement used) If f(x) is even function then ∫
l l
( )
= 2 ∫
l
( )
If f(x) is odd function then ∫
l l
( )
=0
Question: Find M.I of a circular lemina about an axis in the place of lemina and passing through its centre. Solution: Let ‘m’ be the mass of circular lemina and radius ‘a’. Take a small area element having dm and coordinates (x,y). 19 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Y a
= ∫(
+
) dm
= ∫
dm
(x,y) -a
a
2 2 a y a x
=
X
-a
y 2 dydx
Area dm
a y a 2 x 2
3
y = 3 a
y a2 x2
dm = dxdy dx
x changes from −a a
a
=
=
a 3 a
2
a x
=
2
3
2
a x
2
3
dx
a
3 3 2 2 2 2 2 2 a x a x dx 3 a a
=
+
y a 2 x 2
2 2a x 3 a
3 2 2
= −
= y=√
−
y changes from −√
−
√
−
dx
a 3 2 2 2 2 a x dx 3 a
Put x = asin , dx = acosd x = −a
=
(−1) = −
x=a
=
(1) =
2 = 3
3 2 2 2 2 .a cos d a a sin 2
2
=
2 3
3 2 2 a cos 2 .a cos d 2
2
20 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
2 = 3
2
a
4
cos4 d
2
2 a 3
=
4 2 4
cos d
2
By theorem f(x) is even
2 a4 2 = .2 cos 4 d 3 0
∫
l l
( )
=
= 2 ∫
l
( )
=
By Wallis cosine formula 4 a 4 3.1 . . 3 4.2 2
=
a 4
= =
4 m a 4 . a2 4
=
ma2 4
Similarly
=
ma2 4
Question: Find M.T of a elliptic lemina about an axis in the plane of lemina through its major and minor axis. Y Solution: dm √ − = ∫(
+
-a
) dm
dm
= ∫ a
=
y
a z=0
b 2 2 a x a
(x,y)
√
−
2
y dydx
a y b a2 x2 a
21 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
X
x changes from −a a
a
=
a
y
3
y 3
+
b 2 2 a x a
y= √
dx y
b 2 2 a x a
3
b a x a2 x2 = 3 a a a =
2
−
y changes from 3
a b
=1
2
− √ dx
−
√
a b3
3 3 b3 2 2 2 2 2 2 a x a x a3 dx 3 a a 3
a 2 b3 2 2 32 . = a x dx 3 a3 a
Put x = asin , dx = acosd x = −a
=
(−1) = −
x=a
=
(1) =
3 2 2 2 a a sin 2 .a cos d
3 2
=
2 b . 3 a3
2
3 2 2 a cos 2 .a cos d
3 2
2 b . = 3 a3
2
3 2
=
2 b 3 a3
a
4
cos4 d
2
3
2 b a = 3
2 4
cos d
2
By theorem f(x) is even
=
2 b 3a 2 .2 cos 4 d 3 0
∫
l l
( )
=
= 2 ∫
l
( )
=
By Wallis cosine formula 22 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
−
4 b 3a 3.1 = . . 3 4.2 2
ab3
= =
4 m ab 3 . ab 4
mb2 4
=
Similarly
=
ma2 4
Question: Find M.I of rectangular parallelepiped with respect to its edges with one corner at the origin. Z (0,0,c) Solution: = ∫(
+
) dm (0,0,0)
= dm = dxdydz
Y
c b a
(0,b,0)
(a,0,0) X
= ( y 2 z 2 )dxdydz 0 0 0
c b
= x( y
2
a
2
z )
dydz 0
0 0
c b
= (a 0)( y
2
2
z )
a
dydz 0
0 0
c b
= a ( y 2 z 2 )dydz 0 0
c
3
b
y yz 2 dz 3
= a 0
0
23 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
c b3 = a bz 2 0 dz 3 0
c
= a ( 0
b3 bz 2 )dz 3 3
= a
3
b z z b 3 3
c
0
cb3 bc3 = a 3 3 =
m abc 2 2 . b c abc 3
=
m 2 2 b c 3
=
=
Similarly, m 2 a c2 3
= = Find
,
m 2 a b2 3
, = xydm c b a
= xy dxdydz 0 0 0
x2 = y 2
a
c b
0 0
a2 = 2
dydz 0
c b
y dydz 0 0
24 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
2 c
=
a 2
0
2
y 2
b
dz 0
c
a 2 b2 = . dz 2 2 0 a2 b2 = . 2 2
z
c 0
m a 2b 2 c = . abc 4
=
=
=
mab 4
Similarly, =
mac 4
=
mbc 4
Find matrix
I xx I yx I zx
I xy I yy I yz
⎡ ⎢ I xz ⎢ I yz ⎢ ⎢ I zz ⎢ ⎢ ⎣
m 2 2 b c 3
mab 4 mac 4
mab 4
mac 4 mbc 4
⎤ ⎥ ⎥ ⎥ ⎥ ⎥ m 2 2 ⎥ a b ⎦ 3
m 2 a c2 3
mbc 4
Question: Find M.I of a cube w.r.t its edges with one corner at the origin. Solution: = ∫(
Z +
(0,0,a)
) dm
(0,0,0)
= dm = dxdydz
Y
(0,a,0)
(a,0,0) X
25 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
=
a
a
a
0
0
y
0
a
a
= 0 0 =
a
0
z 2 dxdydz
2
x( y z )
a
0
dydz
(a 0)( y 2 z 2 )dydz a
3
= a
a
2
0
a
0
2
y yz 2 3
dz 0
3 a a = a az 2 (0) dz 0 3
3
= a
a z az 3 3
3
a
0
a 3 .a a.a 3 = a 3 3 a4 a4 = a 3 3 2a 4 = a 3 ma 2a4 = 3 a 3 =
2ma 2 3
Similarly, =
2ma 2 , 3
=
2ma 2 3
= xy dm
a
0
a
a
0
0
dxdydz
26 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
=
x2 a y dydz 2 0
a a
0 0
a
a2 = 2
a
a
0
0
ydydz
=
a4 a dz 4 0
=
a4 a z 4 0
=
a2 a y 2 dz = 2 0 2 0
m a4 (a ) a3 4
ma 2 = 4
Similrarly, =
ma 2 4
,
=
ma 2 4
Matrix form 2
I xx I yx I zx
I xy I yy I yz
⎡ 2ma ⎢ 3 I xz ⎢ ma 2 I yz ⎢ ⎢ 4 I zz ⎢ 2 ⎢ ma ⎣ 4
ma 2 4 2ma 2 3
ma 2 4
ma 2 ⎤ 4 ⎥ ⎥ ma 2 ⎥ 4 ⎥ ⎥ 2ma 2 ⎥ 3 ⎦
27 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Lecture # 6
Question: Find M.I of a circular ring about a line passing through its Centre and perpendicular to its circumference. Solution: Consider the circular ring of radius ‘a’ in xy-plane. Consider a mass particle dm with arc length dl with central angle d Y =
dl
d =
=
dm
O a
X
dm = a d a M.I about z-axis =
=∫ ∫
=
0
d
O
=
Since l = r
2
l = a
=
(2 − 0)
=
2
=
dm = ∫
dl = ad
.2
=
= m
Perpendicular axis theorem: Statement: The M.I of Lemina (or a plane rigid body) about a normal axis (about an axis to its plane) is equal to the sum of M.Is about two mutually axis in the plane of lemina and passing through the intersection of lemina with the normal axis. Proof: Consider place Lemina in XY-plane. X-axis and Y-axis are axis. Z-axis is to X-axis and Y-axis. Let
be the ith particle of mass
with
28 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
dl
Position vector ⃗ M.I about z-axis n
I zz ( xi2 yi2 )dm i 1 n
n
i 1
i 1
I zz xi2 dm yi2 dm I zz I yy I xx I zz I xx I yy Pr oved 2
Or I zz ( x 2 y 2 ) dm x 2 dm y dm I yy I xx
K.E in general motion: Consider a rigid body in general motion with base point at the center of mass of the rigid body. Then linear velocity of the ith particle of the rigid body is ⃗
= ⃗ + ( ⃗ × ⃗) And its K.E is
⃗
= ∑
1 n m i V ( ´ ri ) . V ´ ri 2 i 1 1 n i 1 mi V 2 V .( ´ ri ) V . ´ ri 2
´ ri
2
2 1 n 2 i 1 mi V 2V .( ´ ri ) ´ ri 2
T
2
n 1 n 1 n mi V 2 mi ´ ri V . ´ mi ri 2 i 1 2 i 1 i 1
(1) n
If center of mass (base point) is taken at origin then
mi ri 0 i 1
Then from (1)
T=
1 n 1 n mi V 2 mi ´ ri 2 i 1 2 i 1
2
29 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
1 1 n = M V 2 mi ´ ri 2 2 i 1
2
n
where M mi i 1
T = Ttranslational K .E Trotaional K .E
Now Let ´ ri a 2
´ ri ´ ri . ´ ri ´ ri . a
.( ri ´ a )
.[ ri ´ ( ´ ri )]
Now
=
=
1 n m i ´ ri 2 i 1
2
1 n 1 n m .[ r ´ ( ´ r )] m [ = i i ri ´( ´ ri )] i i 2 i 1 2 i 1
1 .L = 2
where L
Now
=
mi [ ri ´( ´ ri )] i 1
1 MV 2 = 2
Hence
n
And
1 .L 2
=
1 [x y z ].[Lx Ly Lz ] 2
1 = [x Lx yLy z Lz ] 2
1 [x (xIxx y Ixy z Ixz ) y (xI yx y I yy z I yz ) z (xIzx y Izy z Izz ] 2 1 [(x2 )Ixx xy Ixy xz Ixz yxI yx (y2 )I yy yz I yz zxIzx zyIzy (z2 )Izz ] 2
Ixy Iyx , Ixz Izx , Iyz Izy 1 [(x )2 Ixx (y )2 I yy (z2 )Izz 2xy Ixy 2xz Ixz 2yz I yz ] 2 30 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Lecture # 7
Question: Find M.I of a rigid body about line having direction cosine , and v Solution: Consider a line ‘l’ through origin having direction cosine ,
and v. Let
̂ be a unit vector along line ‘l’. Then we can write ̂ =̂+ Let
̂ +
( , , ) be the particle of the rigid body and di be the distance of
From the line ‘l’ then from fig. di = | ⃗| sin where ⃗ =
̂+
̂+
Z
= | ̂ ´ ⃗| l
Now M.I of the rigid body about line ‘l’
Y
n
I
mi di
2
i 1
X
2 mi e ´ ri n
......(1)
i 1
Now i e ´ ri l xi
j
k
yi
zi
e ´ ri zi vyi i vxi l zi j l y i xi k
Put in (1) n
2 2 2 I mi zi vyi vxi l zi l yi xi i 1 n
I mi 2 zi2 v2 yi2 2vzi yi v2 xi2 l 2 zi2 2vl xi zi l 2 yi2 2 xi2 2l xi yi i 1
31 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
n
I mi ( yi2 zi2 )l 2 ( xi2 zi2 ) 2 ( xi2 yi2 )v2 2l xi yi 2vyi zi 2vl xi zi i 1
n n n n n n I l2 mi yi2 zi2 2 xi2 zi2 v2 xi2 yi2 2lmx y 2 v my z 2 v l mx iii 1i i i i zi i1 i1 i1 i1 i1 i1
I l 2 I xx 2 I yy v 2 I zz 2l I xy 2 vI yz 2vl I xz
Which is M.I of rigid body about a line having direction cosine ,
and v
Momental Ellipsoid: We know that M.I of a rigid body about a line having direction cosine , v is given as
and
I l 2 I xx 2 I yy v 2 I zz 2 l I xy 2 vI yz 2 vl I xz
Let
̂ √
⃗= Also
be vector along ‘l’ and Q(x,y,z) be the point on line ‘l’ such that ̂ √
and ⃗=
⃗ = ̂+
̂+
Now direction cosine of ⃗
i.e.
,
⃗
,I0
√
,
⃗ are
⃗
√ , √ , √
Since line ‘l’ and
⃗ have same direction cosines therefore
= √ , = √ and v = √ Put in eq (1) I x 2 I I xx y 2 I I yy z 2 I I zz 2 xyI I xy 2 yzI I yz 2 zxI I xz
Divide both side by I x 2 I xx y 2 I yy z 2 I zz 2 xyI xy 2 yzI yz 2 zx I xz 1
Which represents equation of ellipsoid thus ellipsoid is called Momental ellipsoid or ellipsoid of inertia (Ass , , are +ve) 32 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Question Show that Momental ellipsoid at the center of elliptic disc is
x2 y2 1 1 2 z cons tan t a2 b2 a2 b2
√
−
-a
a
Solution: We know that =
mb 4
√
2
, = = =
ma 4
= +
2
By axis theorem
mb2 ma2 + 4 4
m(a 2 b2 ) 4 y
x a
= xy dm
xa
y2 x 2 x a
b 2 2 a x a
x a
y
xy dydx
b 2 2 a x a
b 2 2 a x a
dx b 2 2 a x a
xa
b2 2 b2 2 2 x 2 a x 2 a x 2 dx a a x a xa
x (0) dx 0
x a
Similarly,
=
=0
Now by Momental ellipsoid formula x 2 I xx y 2 I yy z 2 I zz 2 xyI xy 2 yzI yz 2 zx I xz 1
By putting the above value
33 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
−
m a2 b2 2 2 mb ma 2 2 x2 y z 4 4 4
2xy(0) 2yz(0) 2zx(0) 1
2 ma2 2 ma2 mb2 x y z 1 4 4 4 4 2 mb
2
Divide by
x2
ma 2 b 2 4
y2
1 4 2 1 z a2 b2 a2 b2 ma2b2 x2 y2 1 1 2 z cons tan t a2 b2 a2 b2
Question: Write inertia matrix if equation of Momental ellipsoid is 3x2 3 y2 5z2 xy 2 yz 5xz 3
Solution: Given 2x2 3y2 5z2 xy 2 yz 5xz 3
Divide by 3 2 2 2 5 2 1 2 5 x y z xy yz xz 1 3 3 3 3 3
..................(1)
As we know x 2 I xx y 2 I yy z 2 I zz 2 xyI xy 2 yzI yz 2 zx I xz 1
……….(2)
Comparing (1) and (2) =
,
=1,
= ,
I xx Now Inertia Matrix is I yx I zx
= − , I xy I yy I yz
=
I xz I yz I zz
2 3 1 6 5 6
,
= 1 6 1 1 3
5 6 1 3 5 3
34 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Lecture # 8
Question: Find equation of momental ellipsoid of a uniform rectangular parallelepiped. Solution: x 2 I xx y 2 I yy z 2 I zz 2 xyI xy 2 yzI yz 2 zx I xz 1
As
…..(1)
We know that for a parallelopiped (
=
)
,
(
=
)
,
(
=
)
,
=−
,
=
,
=− Put these in (1) (
)
+
(
)
(
+
)
)xy+2(
+2(
)yz = 1
)xz +2(
Multiplying by 6/m both side 2(
+
)
+ 2(
+
)
2(
+
)
+ 2(
+
)
+( +(
+
)
+
)
-3abxy-3acxz-3bcyz = 6/m -3abxy-3acxz-3bcyz = c
c = 6/m Which is required equation of momental ellipsoid of a uniform parallelopiped.
Question: Find M.I of solid cylinder about (i) (ii)
its axis of symmetry or an axis passing through the Centre of cylinder and parallel to its length. its central diameter or an axis passing through its Centre and perpendicular to the axis of symmetry.
Solution: Let h be the height of cylinder ‘O’ be the Centre and z-axis be its axis of symmetry and x-axis & y-axis are the lines perpendicular to its axis of symmetry through its centre in cylindrical coordinate (r,,z) x = r cos
;
0ra
V=
y = rsin
;
0 2
=
z=z
;
-h/2 2 h/2
h
Z h/2
O
X
-h/2
35 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
h
Y
M.I about the axis of symmetry = ∫(
+
=
) dm
x
2
y2 dV
h /2 2 a
=
(r
2
cos2 r 2 sin 2 ) r drd dz
h /2 0 0
h /2 2 a
h/2 2 a 2
=
r .r drd dz
h /2 0 0
=
4
r 4
h / 2 2
r
3
drd dz
h/2 0 0
a
d dz
h /2 0
0
h /2 2
a4 = 0 d dz 4 h /2 0 a4 h/2 2 = d dz 4 h/2 0 a 4 h /2 = 4 h /2
2
dz 0
a 4 h/2 = (2 0)dz 4 h/2 a4 h /2 = .2 z h/2 4 a4 h h .2 ( ) 2 2 4
= =
m a4 a 2 h 4 =
.2 h
m a2 h
ma2 2
36 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
(ii). M.I about central diameter = = ∫( =
+
) dm
2 2 x z dV =
h /2 2 a
(r
2
cos 2 z 2 ) r drd dz
h /2 0 0
h /2 2 a
=
(r
3
cos 2 rz 2 )drd dz
h /2 0 0 h /2 2
=
a4 a2 2 2 cos z d dz 4 2 h /2 0 h /2 2
=
a 4 1 cos 2 a 2 2 z d dz 4 2 2 h /2 0
sin 2 h /2 a4 2 = 4 2 h /2
2
a2 z 2 d dz 2 0
sin 4 a 4 2 2 = 4 2 h /2 h /2
a2 z 2 .2 0 dz 2
h /2
=
a 4 2 0 ) a 2 z 2 dz ( 4 2 h /2 h /2
a 4 a 2 z 2 dz = 4 h /2 a 4 = 4
h /2
h /2 2
dz a
h /2
h /2
a4 h/2 z3 2 z a = 4 h/2 3 =
z 2 dz h/2
h/2
3 a 4 h h h3 2 h a 4 2 2 24 24
37 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
=
3 a 4 2 h h a 4 12
=
m 2 a 2 h2 .a h a2 h 4 12
=
ma 2 mh2 4 12 =
m a2 h
ma 2 mh 2 4 12
Question: Find M.I of a sphere about its diameter or about axis passing through the centre of sphere. Y Solution: = ∫( =
+
) dm
x
2
a
y dV ….(1) 2
x = rsincos
Z
y = r sinsin z = rcos Put in Eq (1) 2 a
I zz r 2 sin 2 cos 2 r 2 sin 2 sin 2 r 2 sin drd d 0 0 0
2 a
I zz r 2 sin 2 cos2 sin 2 r 2 sin drd d 0 0 0
2 a
I zz r 4 sin 3 drd d 0 0 0
2
a
r5 I zz sin3 d d 50 0 0
38 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
X
2
a5 0 sin 3 d d 5 0 0 a5 5
2
a5 5
2
a5 5
2
3
sin d d 0 0
2
sin sin d d 0 0
2
sin (1 cos )d d 0 0
Put
cos = u -sin d = du
Where = 0 u = 1 = u = -1
a5 5
2 1
a5 5
2
a5 5
2
2
(1 u )(du)d
a5 5
0 1
u3 u 0 3
2 1
(1 u
2
)dud
0 1
1
d 1 2
1 1 1 1 0 3 3 d
m a5 4 2 4 3 5 3 0 a 3
a5 4 d 5 3 0
m a5 (2 0) a3 5
2ma 2 I zz 5
39 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Lecture # 9
Theorem: Show that in matrix notation L I ´ L where '' mean derivative w.r.t ' t ' Proof: We know that for a rotating rigid body L⃗ = ∑
⃗ × ⃗
=∑
⃗)
⃗ × (
⃗ × ⃗
=∑ =∑
………(1)
V⃗ = ω⃗ × r⃗
( ⃗ × ( ω⃗ × ⃗ )) ……..(2)
In components form ω +
ω +
ω
=
=
ω +
ω +
ω
=
ω +
ω +
ω
Or
ω ω ω
=
[L] = [I] [ω]
Now from (1) Diff. w.r.t ‘t’ n L mi [ ri ´Vi ri ´Vi ] i 1
ri Vi and Vi ´ ri
mi [Vi ´Vi ri ´ i ´ ri ] n
i 1
mi [ ri ´ ´ ri ´ ri ] n
i 1
mi [ ri ´ ´ ri ] n
i 1
ri Vi
m [ r i i ´ ´ ri ] n
i 1
40 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
m [ r i i ´ ´ ri ] n
i 1
m [ r i i ´ ´ ( ´ vi ) ] n
i 1
From (2) and (3) we note that = ∑ matrix notation as [I] [ω] .In
m [ r i i ´ ´ ri ] n
i 1
...( A)
( ⃗ × ( ω⃗ × ⃗ )) can be written in the same way we can write
as
[I] [ ⃗∙ ] .
Now ri ´ ´ ( ´ ri )
= ri ´ (.ri ) ( ´ ) ri
r ´ ( . r ) () r
2 = ri ´ ( . ri ) () ri
=
2
i
i
i
= ( . ri )(ri ´ ) ()2 (ri ´ ri )
( . ri )( ´ vi ) =
...(4)
ri ´ ri 0 and ri ´ ( ´vi )
Similarly (or replace ⃗ × ω⃗ )
´ r ´ ( ´ r ) ´ ri ´ (ri ´ ) i
i
= ( . ri ) (ri ´ ) = ( . ri ) ( ´ ri )
Comparing (4) and (5)
ri ´ ´ ( ´ ri )
Now Apply ∑
ri ´ ´ ri
both side
m ( r i i ´ ´ ( ´ ri ) ) n
= ´ ri ´ ( ´ ri )
..(5)
i 1
m ( = i ´ ri ´ ( ´ ri ) ) n
i 1
n
= ´( mi ri ´ ( ´ ri ) ) i 1
n L mi (ri ´ ´ ri
= ´L
i 1
41 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
m ( r i i ´ ´ ( ´ ri ) ) n
So
i 1
can be written in matrix form as [ ⃗´ ⃗]
Hence from (A) L [ I ][ ] ´ L proved
Question:
X
Find M.I of a tringular lemina about one of its edges. A
Solution: p
M.I about y-axis )
(
=∫
x P R p-x
Q
B
∫( − ) dA
A
dm dA
…..(1)
PQRS become a rectangle
x
p
Area = length ´ width dA = |
Y
C
a
= ∫( − ) dm =
S
Q
P
|dx ⃗
⃗
=
⃗ ⃗
⃗
Area = ap
=
⃗ =
⃗ dx =
dA =
=
I yy
Put in (1)
( p x)2
x 0
a p
a p
2
x( p x 2 px)dx
x0
2
3
xp x 2 px
a p2 p 2
. p2
=
ax dx p
x p 2
=
dx x p
a
B
Since the APQ and ABC are similar so
2 p 0
a p
a p2
x p
( xp 2 x3 2 px 2 )dx
x0
p4 2 p4 .p (0) p 2 4 3
2
p4 2 p4 a 6 p4 3 p4 8 p4 4 3 p 12
I yy
mp 2 6
42 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
C
Lecture # 10
Equimomental System: Two systems are said to be Equimomental if they have same M.I about any line space.
Principle axis: Three mutually perpendicular axis are called Principle axis if product of inertia realtive to these axis if product of inertia relative to these axis is zero and corresponding M.I’s are called Principle moment of inertias i.e. A set of three mutaully axis relative to which Inertia matrix is diagonal.
Theorem: Two systems are said to be equimomental iff (i) (ii)
They have same masses They have same centroid (center of mass of an object with uniform density) They have same Principle axis and same Principle M.I at cenroid
(iii) Proof:
Suppose that condition (i) , (ii) and (iii) holds. Let and be two systems each having same masses ‘m’ same centroid ‘C’ and same Principal axis at centroid. Consider a line ‘l’ having direction cosines , and v passing throug centroid C. The M.I of system
about line ‘l’ is
I l l 2 I xx 2 I yy v 2 I zz 2l I xy 2 vI yz 2l vI xz 1
Since given axis are Principle axis
=
=
.....(1)
=0)
Therefore from (1) we have I l1 l 2 I xx 2 I yy v 2 I zz
Similarly M.I of system
....(2)
about line ‘l’ is
I l l 2 I xx 2 I yy v 2 I zz 2
....(3)
From (2) and (3) I l1 I l 2 I 43 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Let
be the line parallel to ‘l’ at a distance ‘d’. then M.I of
Il(1) I md 2 ' Similary M.I of
about line ‘ ’ is
..(4) ( By Parallel Axistheorem)
about line ‘ ’ is
Il(2) I md 2 '
..(5) ( By Parallel Axistheorem)
From (4) and (5) Hence both system
and
are equimomental
Conversely: (i)
Let
and
are equimomental
Consider a line ‘ ’ passing through centroids and of system respectively. Since system and are equimomental therefore
and
I l1 I l 2 I
Let ‘ ’ be the line parallel to ‘ ’ at a distance ‘ ’. Then M.I of ‘ ’ is
I l(1) I m1d12 2 Similarly M.I of
..(6) ( By Parallel Axistheorem)
about line ‘ ’ is
Il(2) I m2 d12 2 As
and
about line
..(7) ( By Parallel Axistheorem)
are equimomental so
I l(1) 2
=
I m1d12
I l(2) 2 = I m2 d12
m1 = m2
They have same masses. (ii)
Now consider a line ‘ ’ passing through centroid and to ‘ ’. Also consider a line ‘ ’ passing through and to ‘ ’. This means ‘ ’ and ‘ ’ are parllel
Now M.I of
about line ‘ ’ is
Il(1) Il(1) md 2 2 4 3 Similarly M.I of
Il(2) I l(2) md2 2 3 4
( By Parallel Axistheorem) about line ‘ ’ is
( By Parallel Axistheorem) 44
Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
As
and
are equimomental. So = I l(2) I l(1) 4 4
I l(1) md 2 2 3
2md 2 2 0
d2 0
and
I l(2) md 2 2 4
coincides.
=
same centroid (iii)
Now the two systems have same centroid and same masses if we take any line through the common centroid say G i.e. G =
=
The two systems have the same M.I about that line. It means that we have same Principle axis and same M.I at point G. Hence all the conditions are satisfied which complete the proof.
Question: Show that a circular plate of mass ‘m’ and radius ‘a’ is equimomental with a loop of same mass ‘m’ and radius √
Solution: We know M.I of circular plate about an axis passing through its Circular plate
centroid and to its plane =
loop
M.I of loop about same axis =
√
=
Hence circular plate and loop are equimomental.
√2
a
45 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Lecture # 11
Theorem (Existence of Principle axis theorem): Statement: For a rigid body a set of three mutually perpendicular axis relative to which product of inertia are zero and angular velocity ⃗ and a linear momentum ⃗ are in same direction. Proof: Consider an axis through a point ’0’ such that ⃗ and ⃗ are parallel to it. Then we can write ⃗= ⃗
; where is contant of proportionality
= = =
………(1)
Next we know that (In general )
L1 I111 I122 I133 L2 I 211 I 222 I 233 ....(2) L3 I 311 I322 I 333 From (1) and (2) we can write
( I11 l )1 I122 I133 0 I 211 ( I 22 l )2 I 233 0 ....(3) I311 I 322 ( I 33 l )3 0 This is homogenous system of equation which unknown , , . It has nontrivial soltution when determinant of the matrix made by the coefficient in set of equation (3) is given i.e.
=0 I11 l I 21
I12 I 22 l
I13 I 23
I 31
I 32
I 33 l
0
Which is cubic equation in which has three relation = , , . These relation are actually eigen value of the inertia matrix and these values are
46 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
principal moment of inertia of the rigid body. The corresponding eigen vectors give the direction of principal axis.
Working rule of finding Principle M.I and Principle Axis: (i) (ii)
(iii)
Find inertia matrix at given point (say origin) If inertia matrix is diagonal then corresponding axis are Principle. If inertia matrix is not diagonal the eigen values matrix is not diagonal the eigenvalues of this matrix give values of Principle moment of Inertia. Eigenvectors corresponding to each eigenvalue gives the direction of Principle axis.
Question: Consider a system of particle having masses m,2m,3m,4m located at points (a,a,a) . (a,-a,-a),(-a,a,-a),(-a,-a,a) respectively. Find Principle moment of inertia of the system about origin. Solution: n
I xx mi ( yi2 zi2 ) i 1
m (a 2 a 2 ) 2m (a 2 a 2 ) 3m (a 2 a 2 ) 4m (a 2 a 2 ) 2ma 2 4ma 2 6ma 2 8ma 2 20ma 2 n
I yy mi ( xi2 zi2 ) i 1
m (a 2 a 2 ) 2m (a 2 a 2 ) 3m (a 2 a 2 ) 4m (a 2 a 2 ) 2ma 2 4ma 2 6ma 2 8ma 2 20ma 2 n
I zz mi ( xi2 yi2 ) i 1
m (a 2 a 2 ) 2m (a 2 a 2 ) 3m (a 2 a 2 ) 4m (a 2 a 2 ) 2ma 2 4ma 2 6ma 2 8ma 2 20ma 2
47 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
n
I xy mi xi yi i 1
[m (a 2 ) 2m ( a 2 ) 3m ( a 2 ) 4m (a 2 )] [ m a 2 2m a 2 3m a 2 4m a 2 ] I xy 0 I yx n
I yz mi yi zi i 1
[m (a 2 ) 2m (a 2 ) 3m (a 2 ) 4m ( a 2 )] [ m a 2 2m a 2 3m a 2 4m a 2 ] I yz [4ma 2 ] 4ma 2 I zy n
I xz mi xi zi i 1
[m (a 2 ) 2m ( a 2 ) 3m (a 2 ) 4m ( a 2 )] [ m a 2 2m a 2 3m a 2 4m a 2 ] I xz [2ma 2 ] 2ma 2 I zx I xx I yx I zx
I xz 20ma 2 I yz 0 I zz 2ma 2
I xy I yy I zy
0 20ma 2 4ma 2
2ma 2 4ma 2 20ma 2
For Eigenvalue I xx l
I xy
I xz
I yx
I yy l
I yz
I zx
I zy
I zz l
0
20ma 2 l
0
2ma 2
0
20ma 2 l
4ma 2
2ma
2
4ma
2
0
2
20ma l
By expanding we get
20ma
2
l 384m 2 a 4 l 2 40ma 2 l 4m 2 a 4 0
48 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
20ma
2
l 0 and 380m 2 a 4 l 2 40ma 2 l 0
l 20ma 2
,
l
l 2 40ma 2 l 380m2 a 4 0
( 40ma 2 ) 1600m 2 a 4 1520m 2 a 4 2(1)
40ma 2 80m2 a 4 l 2(1)
l
40ma 2 4 5ma 2 2
l 20ma 2 2 5ma 2 l 20ma 2 2 5ma 2
,
l 20ma 2 2 5ma 2
l 2(10 5)ma 2
,
l 2(10 5)ma 2
l 20ma 2
,
l 20ma 2 2 5ma 2
are the Principle M.I about origin.
49 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Lecture # 12
Spherical Top: A rigid body is said to be spherical top if its all Principle moment of inertias are equal. i.e. = = . A sphere is an example of spherical top with axis passing through the centre of sphere.
Symmetrical Top: A rigid body is said to be symmetircal top if
= . A cylinder is an example of symmetrical top (taking third axis along the axis of cylinder).
Asymmetrical top: A rigid body is called Asymmetrical top if
. A rigid body in general is example of Asymmetircal top.
Rotor: A rigid body is called rotor if its two Principle M.I’s are equal and third M.I is zero i.e. = and = 0 or molecule is an example of rotor.
= 0 and
=
. A diatomic l
Question:Find equimomental system of particle for a rod of mass M.
m A
-a
M-2m m aB 0
Solution: Consider a uniform rod of length 2a. If ‘0’ be the centre of mass. Let the masses m, M-2m , m are located at point A,O,B respectively. The system of particle will be equimomental with the rod. If its M.I about any line is equal to M.I of the rod about the same line. Let ‘l’ be the line through ‘0’ perpendicular to this rod. Then M.I of rod about y-axis (axis passing through cenroid of rod) = = (Mass) (ℎ
ℎ) = M
M.I of system of particle about y-axis = m(− ) +(M-2m) (0) +m( ) = 2m If both system are equimomental then
=
Hence if we take two particles each having mass
M
= 2m
m=
at end points of rod and a
particle mass M at the centre of rod then this system of three particles is equimomental with the given rod of mass M 50 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Lecture # 13
Euler’s Dynamical equations: Consider a rigid body is rotating in OXYZ (Cartesian) coordiante system such that OX, OY & OZ are Principle axes. Then we know that Or in matrix form LI L1 I11 L I 2 21 L3 I 31 L1 I11 L 0 2 L3 0
I12 I 22 I 32 0 I 22 0
OX, OY and OZ are Principle axes. So
L1 I11 L 0 2 L3 0
0 I 22 0
I13 I 23 I 33
1 2 3
0 0 I 33
1 2 3
=
=
0 0 I 33
=0
I111 I 22 2 I 333
In vector form angular momentum is given below L I111 i I 222 i I 333 k We know that rate of change of vector function in fix and rotating coordiante system is related as follow dV dV ´V dt f dt r
Putting ⃗ = ⃗ Where V is any vector function dL dL ´ L dt f dt r dL I111 i I 222 i I 333 k ´ L dt f
51 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
dL Here is external torque acting on the system dt f d L d r ´ P d P V ´P r dt dt dt = V ´ P r ´ F = V ´ mV r ´ F = 0 + r´ F = So I111 i I 222 i I 333 k ´ L ….(A)
i
j
´ L 1 2 I111 I 222
k
3 I 333
´ L I3323 I 2223 i I 3313 I1113 j I 2212 I1112 k
´ L I 22 I 33 23 i I 33 I11 13 j I11 I 22 12 k
Put in (A)
I111 i I 222 i I 333 k I22 I33 23 i I33 I11 13 j I11 I22 12 k
1 I111 ( I 22 I 33 )23 2 I 222 ( I33 I11 )13 3 I 333 ( I11 I 33 )12 Are Eulers Dynamic equations Results: If there is no external torque then I111 ( I 22 I 33 )23 0 I 222 ( I 33 I11 )13 0
(ii)
I 333 ( I11 I 33 )12 0
Multiplying (i) by
, (ii) by
(i)
and (iii) by
(iii) and adding
I1111 I 22 I33 123 I 2222 I33 I11 123 I 3333 I11 I 22 123 0
52 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
I11 1 1 I22 1 23 I33 1 23 I2222 I33 1 23 I11 1 23 I33 3 3 I11 1 23 I22 1 23 0
I 1111 I 22 2 2 I 33 3 3 0
1 I11 .211 I 22 .2 2 2 I 33 .2 3 3 0 2
1 d I1112 I 22 22 I 33 32 0 I1112 I 2222 I 3332 = constant 2 dt
Now we know that Rotational K.E is T =
1 L 2
=
1 1 L1 2 L2 3 L3 2
=
1 1 I111 2 I 222 3 I333 2
=
1 I1112 I 2222 I 3332 = constant 2
In the absence of external torque the rotational K.E of system is constant. Now multiplying (i) by
, (ii) by
and (iii) by
and adding
2 2 I112 1 1 I11 I22 I33 123 I2222 I22 I33 I11 1 23 I3333 I33 I11 I22 1 23 0
2 I112 11 I11I22123 I11I33123 I222 22 I22 I33 1 23 I11 I22 1 23 I3333 I11 I33123 I22 I33 1 23 0
2 I 112 11 I 22 2 2 I 332 3 3 0
1 2 2 I11 .211 I 22 .2 2 2 I 332 .2 3 3 0 2
1 d 2 2 2 I111 I 22 22 I 332 32 0 2 dt
I
12 I 222 22 I 332 32 = constant
2 11
Now we know L I111 i I 222 i I 333 k L I112 12 I 222 22 I 332 32 = constant
In the absence of external torque magnitude of angular momentum is constant. 53 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
Lecture # 14
Question: Find M.I of a solid right circular cone. (i) (ii)
About its axis of symmetry About an axis passing through its base and also through its center OR About the diameter of its base.
Solution: (i) Let M be the mass ‘a’ the radius and ‘h’ be the height of right circular cone. We choose the z-axis along the axis of symmetry and consider a typical disc of radius r and width dz at a distance z from the base. Z Since ∆OAC and ∆BCD are similar triangles. C
So r=
= (
)
h
B
=
=
r
D
dz
Now mass of disk = dm = dv = dz Also
Vertex
=
O a
z A
We know that Y
M.I of disk about z-axis (as shown in fig) =
dm
Now M.I of cone about z-axis (as shown in fig) h
1 1 r 2 dm r 2 r 2 dz 2 20 h
1 r 4 dz 2 0 h
h
1 a 4 (h z ) 4 1 a 4 dz ( ( z h)) 4 dz 4 4 2 h 2 h 0 0
1 a 4 ( z h)5 2 h4 5
h
0
54 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
X
1 3M a 4 ( h h)5 (0 h)5 . 2 a 2 h h 4 5 5
3Ma 2 2h 5
(h)5 3Ma 2 (h)5 5 2h 5 5
3Ma 2 10 (ii). Note that for disk I X ' IY '
....(1)
Now by axis theorem I X ' IY ' I Z '
I X ' IY ' & I Z ' I Z
2 I X ' IZ
IX'
1 I X ' r 2 dm 4
1 1 1 I Z r 2 dm 2 2 2
Now by // axis theorem M.I of disk about x-axis (as shown in fig)
I X ' z 2 dm
1 2 r dm z 2 dm 4
1 2 r ( r 2 dz ) z 2 ( r 2 dz ) 4
1 r 4 dz z 2 r 2 dz 4
a 4 (h z ) 4 a 2 (h z ) 2 1 2 dz z dz 4 2 4 h h
dm r 2 dz
55 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
a 4 4h
4
a 4 4h
4
a 2
(h z )4 dz
4h
2
.z 2 (h 2 z 2 2hz )dz
a 2
(( z h)) 4 dz
4h
2
(h 2 z 2 z 4 2hz 3 )dz
Now M.I of cone about x-axis (as shown in fig) =
h 1 a 4 a 2 2 2 4 4 3 ( z h ) dz ( h z z 2 hz ) dz 4 2 2 0 4h h
a 4 4h
4
h
( z h)
4
dz
a 2 h
0
a 4 ( z h)5 4h 4
5
h
2
h2
2
z 4 2hz 3 )dz h
z3 z5 z4 h 2 h 3 5 4 0
a 4 4h 4
2
(h z 0
a 2
0
h
h5 a 2 0 2 5 h
2
2 h3 h5 h4 h 3 5 2h 4
a 4 3M h 5 3M a 2 h 5 h 5 h 5 4 0 4h a 2 h 5 a 2 h h 2 3 5 2
a 4 3M h 5 3M a 2 h 5 h 5 h 5 4h4 a 2 h 5 a 2 h h 2 3 5 2
3Ma 2 3M 20h5 12h5 30h5 3 20 h 60
3Ma 2 3M 2h5 3 20 h 60
3Ma 2 3M 3 2h 2 20 h
M 3a 2 2h 2 20
56 Collected By : Muhammad Saleem
Composed By: Muzammil Tanveer
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