- Euler's dynamical equations describe the motion of a rigid body rotating about a fixed point. They relate the angular momentum of the body to the external torque and angular velocity. - Important deductions from Euler's equations include: the angular momentum and kinetic energy of a rigid body are conserved in the absence of external forces/torques, and the magnitude of angular momentum is also conserved. - The general motion of a rigid body involves both translation and rotation. The equations of motion relate the linear/angular acceleration of the body's center of mass to the total external forces/torques acting on the body.

2. ADVANCE
MATH-(IV)
PRESENTED TO:
MA’AM SABA
PRESENTED BY:
AMENAH
SAMEEN
CLASS:
BS.ED VII

3. OBJECTIVES
Today we will discuss:
• EULER’S DYNAMICAL EQUATION
• DEDUCTIONS FROM EULER’S EQUATION
• GENERAL MOTION OF A RIGID BODY

4. MOTION OF A RIGID
BODY IN SPACE

5. RIGID BODY:
“A rigid body is a solid body in which deformation is zero or so small it can
be neglected. The distance between any two given points on a rigid body
remains constant in time regardless of external forces or moments exerted
on it. A rigid body is usually considered as a continuous
distribution of mass.”
Rigid body Non-Rigid body

6. CHARACTERISTICS OF RIGIDBODY MOTION
All lines on a rigid body have the same angular velocity and the same
angular acceleration.
Rigid motion can be decomposed into the translation of an arbitrary
point, followed by a rotation about the point.

7. EULER’S DYNAMICAL
EQUATIONS

8. LEONHARD EULER
• Leonhard Euler15 April 1707 – 18
September 1783) was a Swiss
mathematician, physicist, astronomer,
geographer, logician and engineer who
made important and influential discoveries in
many branches of mathematics.
• Amongst his many discoveries and
developments, Euler is credited for
introducing the Greek letter pi.

9. Let the rigid body be rotating with angular velocity 𝜔 about a point O fixed both in space
and in the body. Let OX, OY, OZ be principal axes at O. If L, 𝐼𝑖𝑗 , 𝜔 be angular
momentum, M.I. matrix and angular velocity at O, then
𝑳 = 𝑰 [𝝎]
𝑳 = 𝑰𝒊𝒋 [𝝎]
𝑳𝒙
𝑳𝒚
𝑳𝒛
=
𝑰𝒙𝒙 𝑰𝒙𝒚 𝑰𝒙𝒛
𝑰𝒙𝒚 𝑰𝒚𝒚 𝑰𝒚𝒛
𝑰𝒙𝒛 𝑰𝒚𝒛 𝑰𝒛𝒛
𝝎𝒙
𝝎𝒚
𝝎𝒛
∵ the axes OX, OY, OZ are principal axes,

10. 𝑰𝒙𝒚= 𝑰𝒙𝒛 = 𝑰𝒚𝒛 = 𝟎
∵so the product of inertia about these axis will be zero
∴ we can write
𝑳𝒙
𝑳𝒚
𝑳𝒛
=
𝑰𝒙𝒙 𝟎 𝟎
𝟎 𝑰𝒚𝒚 𝟎
𝟎 𝟎 𝑰𝒛𝒛
𝝎𝒙
𝝎𝒚
𝝎𝒛
=
𝝎𝒙𝑰𝒙𝒙
𝝎𝒚𝑰𝒚𝒚
𝝎𝒛𝑰𝒛𝒛
Or
𝑳 = 𝝎𝒙𝑰𝒙𝒊 + 𝝎𝒚𝑰𝒚𝒋 + 𝝎𝒛𝑰𝒛𝒌
= 𝝎𝟏𝑰𝟏𝒊 + 𝝎𝟐𝑰𝟐𝒋 + 𝝎𝟑𝑰𝟑𝒌
where 𝑰𝟏, 𝑰𝟐, 𝑰𝟑 are principal moments

11. Now the rate of change of any vector function F in fixed and rotating coordinate systems is
related by
𝒅𝑭
𝒅𝒕 𝒇
=
𝒅𝑭
𝒅𝒕 𝒓
+ 𝝎 × 𝑭
𝒅𝑳
𝒅𝒕 𝒇
=
𝒅𝑳
𝒅𝒕 𝒓
+ 𝝎 × 𝑳
=
𝒅
𝒅𝒕
𝑰𝟏𝝎𝟏𝒊 + 𝑰𝟐𝝎𝟐𝒋 + 𝑰𝟑𝝎𝟑𝒌 + 𝝎 × 𝑳
= 𝑰𝟏𝝎𝟏𝒊 + 𝑰𝟐𝝎𝟐𝒋 + 𝑰𝟑𝝎𝟑𝒌 + 𝝎 × 𝑳
Where the symbols on the R.H.S refer to the rotating coordinate system. But
𝑑𝐿
𝑑𝑡
= 𝐺, the total
external torque (in the fixed or inertial coordinate system).
∵ in rotating, only
derivative 𝜔 , in fixed
derivative L.
∵ Replace F by L

12. Therefore on the substitution,
G= 𝑰𝟏𝝎𝟏𝒊 + 𝑰𝟐𝝎𝟐𝒋 + 𝑰𝟑𝝎𝟑𝒌 + 𝝎 × 𝑳
𝑮𝒙𝒊 + 𝑮𝒚𝒋 + 𝑮𝒛𝒌 = 𝑰𝟏𝝎𝟏𝒊 + 𝑰𝟐𝝎𝟐𝒋 + 𝑰𝟑𝝎𝟑𝒌 +
𝒊 𝒋 𝒌
𝝎𝒙
𝟏 𝝎𝒚
𝟐 𝝎𝒛
𝟑
𝑳𝟏 𝑳𝟐 𝑳𝟑
𝑮𝒙𝒊 + 𝑮𝒚𝒋 + 𝑮𝒛𝒌 = 𝑰𝟏𝝎𝟏𝒊 + 𝑰𝟐𝝎𝟐𝒋 + 𝑰𝟑𝝎𝟑𝒌 + 𝒊 𝝎𝟐𝑳𝟑 − 𝝎𝟑𝑳𝟐 + 𝒋 𝝎𝟑𝑳𝟏 −
---------1

13. Now using the results 𝐿1 = 𝜔1𝐼1, 𝐿2 = 𝜔2𝐼2, 𝐿3 = 𝜔3𝐼3 which are true w.r.t. principal
axes,
we have
𝑮𝒙 = 𝑰𝟏𝝎𝟏 − 𝑰𝟐 − 𝑰𝟑)𝝎𝟐𝝎𝟑
𝑮𝒚 = 𝑰𝟐𝝎𝟐 − 𝑰𝟑 − 𝑰𝟏)𝝎𝟏𝝎𝟑
𝑮𝒛 = 𝑰𝟑𝝎𝟑 − 𝑰𝟏−𝑰𝟐)𝝎𝟐𝝎𝟏
These equations are called Euler’s Dynamical Equations.
They describe the motion of a rigid body fixed at a point.

14. DEDUCTIONS FROM EULER’S
EQUATIONS

15. We will discuss some results which follow directly from the Euler dynamical equations.
In the absence of external forces, G=0, and the Euler equations (1) reduce to
𝒅𝑳
𝒅𝒕
= 𝟎 ⇒ 𝑳 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕
𝑰𝟏𝝎𝟏 − 𝑰𝟐 − 𝑰𝟑 𝝎𝟐𝝎𝟑 = 𝟎 ----- 1
𝑰𝟐𝝎𝟐 − 𝑰𝟑 − 𝑰𝟏 𝝎𝟏𝝎𝟑 = 𝟎 ----- 2
𝑰𝟑𝝎𝟑 − 𝑰𝟏−𝑰𝟐)𝝎𝟐𝝎𝟏 = 𝟎 ----- 3
Multiply 1,2,3 by 𝜔1, 𝜔2, 𝜔3 respectively and adding, we have
𝑰𝟏𝝎𝟏𝝎𝟏 + 𝑰𝟐𝝎𝟐𝝎𝟐 + 𝑰𝟑𝝎𝟑𝝎𝟑 = 𝟎

16. 𝟏
𝟐
𝒅
𝒅𝒕
𝑰𝟏𝝎𝟏
𝟐
+ 𝑰𝟐𝝎𝟐
𝟐
+ 𝑰𝟑𝝎𝟑
𝟐
= 𝟎
Or 𝑰𝟏𝝎𝟏
𝟐
+ 𝑰𝟐𝝎𝟐
𝟐
+ 𝑰𝟑𝝎𝟑
𝟐
= 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕
Now for the kinetic of a rigid body we have the relation 𝑻 = 𝟏 𝟐 𝝎. 𝑳 which when referred
to the principal axes reduces to 𝑻 = 𝟏 𝟐 𝑰𝟏𝝎𝟏
𝟐
+ 𝑰𝟐𝝎𝟐
𝟐
+ 𝑰𝟑𝝎𝟑
𝟐
.
In view of this formula we find that in this case T=constant, i.e., the kinetic energy in the
absence of external forces is a constant of motion.
This is so because no work is being done by the external forces, and therefore the potential
energy is a constant.

17. Another integral of the force-free equations can be obtained by multiplying 1, 2, 3 by
𝐼1𝜔1, 𝐼2𝜔2, 𝐼3𝜔3 respectively and adding. We get
𝑰𝟏
𝟐
𝝎𝟏𝝎𝟏 + 𝑰𝟐
𝟐
𝝎𝟐𝝎𝟐 + 𝑰𝟑
𝟐
𝝎𝟑𝝎𝟑 = 𝟎
Or
𝒅
𝒅𝒕
𝑰𝟏
𝟐
𝝎𝟏
𝟐
+ 𝑰𝟐
𝟐
𝝎𝟐
𝟐
+ 𝑰𝟑
𝟐
𝝎𝟑
𝟐
) = 𝟎
Or 𝑰𝟏
𝟐
𝝎𝟏
𝟐
+ 𝑰𝟐
𝟐
𝝎𝟐
𝟐
+ 𝑰𝟑
𝟐
𝝎𝟑
𝟐
= 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 ------4
Now the angular momentum L w.r.t. the triad of principal axes is given by
𝑳 = 𝑰𝟏𝝎𝟏𝒊 + 𝑰𝟐𝝎𝟐𝒋 + 𝑰𝟑𝝎𝟑𝒌
And therefore 𝑳𝟐 = 𝑳 𝟐 = 𝑰𝟏
𝟐
𝝎𝟏
𝟐
+ 𝑰𝟐
𝟐
𝝎𝟐
𝟐
+ 𝑰𝟑
𝟐
𝝎𝟑
𝟐
Hence eq.4 expresses the fact that the magnitude of L is a constant of motion.
This can be related to the absence of the external torque about the point O.

18. GENERAL MOTION
OF A
RIGID BODY

19. General Motion = Translational Motion + Rotational Motion
TRANSLATIONAL MOTION
“The motion by which a body shifts from one point in
space to another.” Or
“A body moves along a line without any rotation. The
line may be straight or curved.”
Example: A bullet fired from a gun, Ferris wheel etc.
ROTATIONAL MOTION
“ A kind of motion in which the body follows a
curved path.”
“The spinning motion of a body about its axis”
Example, the wheel of the car , Earth etc.

20. In the general motion of a body (i.e., no point of the body is fixed in space).
Let 𝑭𝒄
𝒆𝒙𝒕 be the total external force on the rigid body and 𝑮𝒄
𝒆𝒙𝒕 the total external torque about
is mass centre (i.e., centroid).
Then the equations of motion are
𝑴𝒂𝒄 = 𝑭𝒄
𝒆𝒙𝒕
𝑳𝒄 = 𝑮𝒄
Where 𝑎𝑐 is the acceleration of the centre mass and 𝐿𝑐 is the total angular momentum about it.
Now we resolve the vectors 𝑎𝑐, F, 𝐺𝑐 and L along the unit vector 𝑖, 𝑗, 𝑘 taken along the principal
axes at the mass centre. The triad of vectors 𝑖, 𝑗, 𝑘 may be referred to as a principal triad. It will
be assumed to be permanently a principal triad. Let 𝛺 be the angular velocity. If the triad is
fixed in the body then Ω = ω, the angular velocity of the body. Now using the relation
𝒅𝑭
𝒅𝒕 𝒇
=
𝒅𝑭
𝒅𝒕 𝒓
+ 𝝎 × 𝑭
------------ (1)
------------ (2)
𝑻𝒐𝒓𝒒𝒖𝒆 = 𝑮 = 𝑳
𝑭 = 𝒎𝒂

21. Which relates the rate of change of a vector in a fixed (i.e., inertial) frame and a rotating
frame, we have (on dropping the suffix r)
𝒂𝒇 =
𝒅𝒗
𝒅𝒕 𝒇
=
𝒅𝒗
𝒅𝒕
+ 𝜴 × 𝝂
𝒂𝒇 =
𝒅𝒗
𝒅𝒕
+ 𝜴 × 𝝂
Where 𝑣 = 𝑣1𝑖 + 𝑣2𝑗 + 𝑣3𝑘 is the velocity of the mass centre (in the rotating coordinate
system).
𝝂𝒇 = 𝝂𝒓 = 𝝂
------------ (3)
F = v

22. Substituting for 𝑎𝑓 = 𝑎𝑐 form (3) into (1), we obtain
𝑴𝒂𝒄 = 𝑭𝒄
𝒆𝒙𝒕
𝑴
𝒅𝒗
𝒅𝒕
+ 𝜴 × 𝝂 = 𝑭
𝑴 𝝂𝟏𝒊 + 𝝂𝟐𝒋 + 𝝂𝟑𝒌 +
𝒊
𝜴𝟏
𝝂𝟏
𝒋
𝜴𝟐
𝝂𝟐
𝒌
𝜴𝟑
𝝂𝟑
= 𝑭𝟏𝒊 + 𝑭𝟐𝒋 + 𝑭𝟑𝒌
𝑴 𝝂𝟏𝒊 + 𝝂𝟐𝒋 + 𝝂𝟑𝒌 + 𝒊 𝜴𝟐 𝝂𝟑 − 𝜴𝟑 𝝂𝟐 − 𝒋 𝜴𝟏 𝝂𝟑 − 𝜴𝟑 𝝂𝟏 + 𝒌 𝜴𝟏 𝝂𝟐 − 𝜴𝟐 𝝂𝟏
= 𝑭𝟏𝒊 + 𝑭𝟐𝒋 + 𝑭𝟑𝒌
𝑴 𝝂𝟏𝒊 + 𝒊 𝜴𝟐 𝝂𝟑 − 𝜴𝟑 𝝂𝟐 + 𝝂𝟐𝒋 − 𝒋 𝜴𝟏 𝝂𝟑 − 𝜴𝟑 𝝂𝟏

23. Which is equivalent to
𝑴 𝝂𝟏 + 𝜴𝟐 𝝂𝟑 − 𝜴𝟑 𝝂𝟐 = 𝑭𝟏
𝑴 𝝂𝟐 + 𝜴𝟑 𝝂𝟏 − 𝜴𝟏 𝝂𝟑 = 𝑭𝟐
𝑴 𝝂𝟑 + 𝜴𝟏 𝝂𝟐 − 𝜴𝟐 𝝂𝟏 = 𝑭𝟑
From 2, on using
𝑑𝐿
𝑑𝑡 𝑓
=
𝑑𝐿
𝑑𝑡
+ Ω × 𝐿
𝑑𝐿
𝑑𝑡 𝑓
= 𝐼1𝜔1 𝑖 + 𝐼2𝜔2 𝑗 + 𝐼3𝜔3 𝑘 +
𝑖
Ω1
𝐿1
𝑗
Ω2
𝐿2
𝑘
Ω3
𝐿3
And the relation 𝑳 = 𝑰𝟏𝝎𝟏 𝒊 + 𝑰𝟐𝝎𝟐 𝒋 + 𝑰𝟑𝝎𝟑 𝒌, we obtain the equation
∴ 𝑳𝒄 = 𝑮𝒄
------------ (4)

24. 𝒅𝑳
𝒅𝒕 𝒇
= 𝑰𝟏𝝎𝟏 𝒊 + 𝑰𝟐𝝎𝟐 𝒋 + 𝑰𝟑𝝎𝟑 𝒌 + 𝜴𝟐 𝑳𝟑 − 𝜴𝟑 𝑳𝟐 𝒊 − 𝜴𝟏 𝑳𝟑 − 𝜴𝟑 𝑳𝟏 𝒋 + 𝜴𝟏 𝑳𝟐 − 𝜴𝟐 𝑳𝟏 𝒌 = 𝑮
𝑰𝟏𝝎𝟏 𝒊 + 𝑰𝟐𝝎𝟐 𝒋 + 𝑰𝟑𝝎𝟑 𝒌 + 𝜴𝟐 𝑳𝟑 − 𝜴𝟑 𝑳𝟐 𝒊 − 𝜴𝟏 𝑳𝟑 − 𝜴𝟑 𝑳𝟏 𝒋 + 𝜴𝟏 𝑳𝟐 − 𝜴𝟐 𝑳𝟏 𝒌
= 𝑮𝟏𝒊 + 𝑮𝟐𝒋 + 𝑮𝟑𝒌
From this vector equation we obtain the following three scalar equations
𝑰𝟏𝝎𝟏 + 𝜴𝟐 𝑳𝟑 − 𝜴𝟑 𝑳𝟐 = 𝑮𝟏
𝑰𝟐𝝎𝟐 + 𝜴𝟑 𝑳𝟏 − 𝜴𝟏 𝑳𝟑 = 𝑮𝟐
𝑰𝟑𝝎𝟑 + 𝜴𝟏 𝑳𝟐 − 𝜴𝟐 𝑳𝟏 = 𝑮𝟑
Where we have used the results
𝑳𝟏 = 𝑰𝟏𝝎𝟏, 𝑳𝟐 = 𝑰𝟐𝝎𝟐, 𝑳𝟑 = 𝑰𝟑𝝎𝟑

25. We have
𝑰𝟏𝝎𝟏 + 𝜴𝟐 𝑰𝟑𝝎𝟑 − 𝜴𝟑 𝑰𝟐𝝎𝟐 = 𝑮𝟏
𝑰𝟐𝝎𝟐 + 𝜴𝟑 𝑰𝟏𝝎𝟏 − 𝜴𝟏 𝑰𝟑𝝎𝟑 = 𝑮𝟐
𝑰𝟑𝝎𝟑 + 𝜴𝟏 𝑰𝟐𝝎𝟐 − 𝜴𝟐 𝑰𝟏𝝎𝟏 = 𝑮𝟑
Which are the same as for a rigid body with a fixed point.
In these equations 𝐼1, 𝐼2, 𝐼3 denote principal moments of inertia at the centroid of the body.
The set of equations 4 and 5 constitute six equations for the components of velocity of the mass
centre and the components of angular velocity of the body.
------------ (5)

26. For any these six equations, we can substitute the law of conservation of energy, viz.
𝑻 + 𝑽 = 𝑬
provided the external forces are conservative.
The last equation is equivalent to
𝟏
𝟐
𝑴 𝒗𝟏
𝟐 + 𝒗𝟐
𝟐 + 𝒗𝟑
𝟐 +
𝟏
𝟐
𝑰𝟏𝝎𝟏
𝟐 + 𝑰𝟐𝝎𝟐
𝟐 + 𝑰𝟑𝝎𝟑
𝟐 + 𝑽 = 𝑬
K.E + P.E = Total Energy
𝑪𝒐𝒏𝒔𝒆𝒓𝒗𝒂𝒕𝒊𝒗𝒆 ⇒ 𝑭 = 𝟎
𝑻 = 𝑻𝒕𝒓 + 𝑻𝒓𝒐𝒕
𝑻 =
𝟏
𝟐
𝑴𝒗𝟐 +
𝟏
𝟐
𝝎. 𝑳

27. ANY
QUESTION