1. Kinetics of Rigid Bodies:
From Riley’s Dynamics
Chapter 16
Kinetics of Rigid Bodies:
Newton’s Laws

2. (Q) What are the Euler’s Equations of Motion?
Newton’s Law  applies only to the motion of a single particle
translation R R G G
only translation
translation + rotation
Newton’s 2nd Law
Euler’s Equations of Motion

3. Euler’s Equations of Motion
Rotation of a Rigid Body
moment
∴ Starting Point
Moment of F & f about A
Newton’s 2nd Law
Substitution yields

4. What’s this?
After integration, we can get the general form of the Euler’s equations of motion.
Very general equation about rotation.  Need to unify the coordinate systems to {Axyz}.

5. (Q) Simplified Version  Plane Motion
Mass center G lies in the xy-plane. dm r
Now,
After the similar calculation, we have

6. product of inertia
moment of inertia
Using

7. (Note) The 1st 2 equations are required to maintain the plane motion about z-axis,
especially for non-symmetrical geometry case.
If the body is symmetric
about the plane of motion,

8. If
(symmetry) +
(acceleration of the point A = 0) If
(symmetry) +
(A = G)

9. (Q) More about the Moment of Inertia
For the particle dm
IF widely distributed
THEN larger moment of inertia
For the entire body
It uses the information about its geometry.
∴ THE SAME MASS BUT DIFFERENT GEOMETRY
DIFFERENT MOMENT OF INERTIA

10. There are various ways of choosing this small mass element for integration.
A specific mass element may be easier to use than other elements.

11. You may treat the rigid body as a system of particles.

12. 2nd
moment
of area

13. If the density of the body is uniform,

14. a rigid body  summation of several simple shape rigid bodies
Practical approach
a rigid body  summation of several simple shape rigid bodies
composite body

15. = I : moment of inertia about the axis (the moment of inertia about
gyration [ʤaiəréiʃən] n.
U,C 선회, 회전, 선전(旋轉); 〖동물〗 (고둥 따위의) 나선.
㉺∼al [-ʃənəl] ―a. 선회의, 회전의.
(Q) What is the radius of gyration? = k m m
I : moment of inertia
about the axis
(the moment of inertia about
the axis) = mk2
NO useful physical interpretation!!
Maybe baseball  Home Run !!!!!

16. (Q) What is the Parallel-Axis Theorem for Moments of Inertia?
measurement of the location
of the mass center from the
mass center
= m

18. z’

19. (Q) More about the Product of Inertiadm
In 2-D space y Rz x y x

20. (Q) What is the effect of symmetry on the product of inertia?x y z x y x z z y

21. z y x y z z x y x

22. (Q) What is the Parallel-Axis Theorem for Product of Inertia?
From definition or
But,
mass center from
the mass center
and
Therefore,

23. (Q) What is the Rotation Transformation of Inertia Properties?z’
Consider z y x’ y’ x
We know that
We can represent i’, j’, and k’ w.r.t. i, j, and k.
Substitution yields

24. or
old
new
[R] rotation transformation
matrix from old to new frame
a vector in
the old frame
a vector in
the new frame

25. (Example) y’ y x Θ Θ x’
Rotation about z’-axis
It means that [R] is an orthonormal matrix.

26. z’ z y Θ Θ y’
Rotation about x’-axis x’ x z Θ Θ z’
Rotation about y’-axis

27. Now, the rotational kinetic energy is
This term will be
derived in the next
chapter.
Since energy is invariant
Let : known
old frame
Let : unknown
new frame
old
new
from old to new

28. Claim: [I] = ? (Example) z a = 240 mm b = 120 mm m = 60 kg y c = 90 mm
(Idea) z’ z y G y’ x’ x

29. a c G b

30. z’ a c G b y’ x’
By using the parallel axis theorem,

31. z z’ b a c y Θ y’ Θ x’ x

32. Slender rod

33. Thin rectangular plate

34. Thin circular plate

40. Quiz #1 Y’ X’ Z’
{x’y’z’} 좌표 시스템에 대해 표현된 Inertia matrix를 구하시오.

41. (Q) How to analyze the General Plane Motion of NonSymmetric Bodies?
For Plane Motion

42. For Plane Motion

43. Claim: 5 reactions & T ? 30 mm dia. m = 1.2 kg l = 220 mm
600 rpm ccw
increasing in speed
at the rate of
60 rpm per second
30 mm dia.
m = 1.2 kg
l = 220 mm
= /2-40/2
40 mm dia.
8.5 kg
Claim:
5 reactions
&
T ?
120 mm dia.
m = 7.5 kg
Bearing A resists any motion
in the z-direction.

45. For the entire system The same result for this sphere since
zG and xG are minus sign.
120 mm dia.
m = 7.5 kg
30 mm dia.
m = 1.2 kg
l = 220 mm
= /2-40/2
The same result for this bar since
zG and xG are minus sign.
40 mm dia.
8.5 kg
For the entire system

47. Or
next page

48. x x’ z’ z
Sym.

49. (Q) How to analyze the 3-D Motion of a Rigid Body?x Y O X
All vectors are represented
w.r.t. the body-fixed {xyz}.
Recall
How?

50. Euler’s Equations of Motion
Rotation of a Rigid Body
moment
∴ Starting Point
Moment of F & f about A
Newton’s 2nd Law
Substitution yields

51. What’s this?
After integration, we can get the general form of the Euler’s equations of motion.
Very general equation about rotation.  Need to unify the coordinate systems to {Axyz}.

52. If we use the Cartesian coordinate system,
In vector-matrix form,

53. Or

55. = 75 rad/s
constant
= 25 rad/s
constant

56. = 75 rad/s
constant
= 25 rad/s
constant
or more
mathematically

57. = 75 rad/s
constant
= 25 rad/s
constant
∴ Solvable!

61. Therefore,