Download Plane Kinetics of Rigid Bodies

Plane Kinetics of Rigid Bodies
:: Relates external forces acting on a body with the translational and
rotational motions of the body
:: Discussion restricted to motion in a single plane (for this course)
 Body treated as a thin slab whose motion is confined to the
plane of slab
 Plane containing mass center is generally considered as plane of
motion
 All forces that act on the body get projected on to the plane of
motion
 All parts of the body move in parallel planes
 A body with significant dimensions normal to the plane of
motion may be treated as having plane motion if the body is
symmetrical about the plane of motion through the mass center
 Idealizations suitable for a very large category of rigid body motions
ME101 - Division III
Kaustubh Dasgupta
1
Plane Kinetics of Rigid Bodies
:: Earlier discussion on rectilinear/curvilinear motion
- 2 equations of motion
 Fx  ma x
 Fy  ma y
:: Plane kinetics of rigid bodies
- Additional equation of motion
- Account for the rigid body rotation
ME101 - Division III
Kaustubh Dasgupta
2
Plane Kinetics of Rigid Bodies
General Equations of Motion
G is the mass center of the body
Dynamic
Response
Action
ME101 - Division III
Kaustubh Dasgupta
3
Plane Kinetics of Rigid Bodies
• Force/mass/acceleration
– Free Body Diagram
• Work-energy principles
– Active force diagram
• Showing only the (active) forces which contribute to
the work done
• Impulse-momentum method
– Impulse-momentum diagram
ME101 - Division III
Kaustubh Dasgupta
4
Rigid Body Kinetics :: Force/Mass/Acc
Plane Motion Equations
ω = ωk ; α = αk ; α = ω̇
Angular momentum @ G
Vel of mi relative to G
is a vector normal to the x-y plane along ω
(magnitude = ρi2ω)
Magnitude of HG:
The summation:
Mass Moment of Inertia (Ῑ ) of the body about z-axis through G
Measure of the rotational inertia, which is the resistance to change in
rotational velocity due to the radial distribution of mass around the z-axis
through G

 Generalized laws of motion:
ME101 - Division III
5
Rigid Body Kinetics :: Force/Mass/Acc
Alternative Derivation
Using mass center G as the reference point:
Accln of mi is vector sum of three terms:
a̅, and relative accln terms ρi ω2 and ρi α
Sum of moments of these force comp @ G:


Since origin is taken as mass center:

Same equation  moment of only the external forces
The force comp mi ρi ω2 passes through G
 ω has no influence on the moment eqn @ G
ma̅ = translational dynamic response
Ῑ α = rotational dynamic response
ME101 - Division III
Kaustubh Dasgupta
Rigid Body Kinetics :: Force/Mass/Acc
Alternative Moment Equations
Moment @ any arbitrary point P:
ρ̅ is the vector from P to mass center G,
and a̅ is the mass center accln.
ρ̅ x ma̅ = moment of magnitude of ma̅ @ P
 ma̅ d

Another eqn was developed for system of particles:
For rigid body plane motion, if P is fixed to the body:
Magnitude of
= IP α (IP is mass moment of inertia @ P)

For rigid body rotating @ a nonaccelerating point O fixed to the body:
(Point P becomes O and aP = 0)
ME101 - Division III
Kaustubh Dasgupta
7
Rigid Body Kinetics :: Force/Mass/Acc
Constrained and Unconstrained Motion
:: Motion of a rigid body may be constrained or unconstrained
a̅ x, a̅y, and α can be
determined independently
using plane motion eqns
ME101 - Division III
Kinematic relationship betn the accln comp
of mass center (linear accln) and the
angular accln of the bar to be determined
first and then apply the plane motion eqns.
Kaustubh Dasgupta
8
Rigid Body Kinetics :: Force/Mass/Acc
Systems of Interconnected Bodies
:: If motion of connected rigid bodies are related kinematically
analyze the bodies as an entire system
Two rigid bodies hinged at A
Forces in the connection A are
internal to the system
:: No. of remaining unknowns in the system > 3 (3 eqns of plane motion
insufficient)
 Use Virtual Work method (discussed later)
Or dismember the system and analyze each part separately
ME101 - Division III
Kaustubh Dasgupta
9
Rigid Body Kinetics :: Force/Mass/Acc
Application to three cases of rigid body motion:
Translation
No angular motion of body (ω and α will be zero)
Mass moment of inertia will not be effective
ME101 - Division III
Kaustubh Dasgupta
10
Example on Translation
Solution:
Motion of bar is curvilinear translation
since the bar does not rotate.
Motion of G is circular  choose n-t coordinates
Choosing the reference axes coincident with
the directions in which the comp of accln of
mass center are expressed
 Better choice
ME101 - Division III
Kaustubh Dasgupta
11
Example on Translation
Draw the FBD and the Kinetic Diagram
From FBD of AC (negligible mass eqn of equilibrium)
 At = M/1.5 = 5/1.5 = 3.33 kN
Member BD also has a negligible mass
Two force member in equilibrium
The force at B will be along the link
θ = 30o
ME101 - Division III
Kaustubh Dasgupta
12
Rigid Body Kinetics :: Force/Mass/Acc
Rotation @ a Fixed Axis
Mass Moments of Inertia
• Required in rotational acceleration of any body
• Mass m of a body is a measure of resistance to translational
acceleration
• Area moment of inertia is a measure of the distribution of area @
the axis
• Mass Moment of Inertia I is a measure of resistance to rotational
acceleration of the body
Mass moment of inertia of the body @ O-O:
ρ = constant throughout the body
Units of Mass moment of inertia: kg m2
ME101 - Division III
Kaustubh Dasgupta
13
Plane Kinetics of Rigid Bodies
Mass Moments of Inertia
Radius of Gyration (k)
•about an axis for which I is defined:
Parallel Axis Theorem (Transfer of Axes)
•Mass moment of inertia about any axis parallel to the axis passing
through mass center G:
•Radius of gyration @ an axis through C
Ī and k̄ are the values @ an axis
through mass center
ME101 - Division III
Kaustubh Dasgupta
14
Plane Kinetics of Rigid Bodies
Mass Moments of Inertia
Plane Motion:
Mass MI of the plate (with motion in x-y plane)
@ z-axis through O:
o
o
3-D Motion:
ME101 - Division III
Kaustubh Dasgupta
15
Plane Kinetics of Rigid Bodies
Mass Moments of Inertia
Thin Plates
Relationship between mass moments of inertia and area
moments of inertia exists in case of flat plates.
t = constant thickness of the plate,
ρ = constant mass density of the plate
Mass MI Izz of the plate @ z-axis normal to the plate:
Mass MI @ z-axis = mass per unit area (ρt) x Polar MI of the plate area @ z-axis
If t is much less as compared to the dimensions of the plate in its plane:
Mass MI Ixx and Iyy of the plate @ x- and y-axes are closely approximated by:
ME101 - Division III
Kaustubh Dasgupta
16
Plane Kinetics of Rigid Bodies
Mass Moments of Inertia
Products of Inertia: used in the expression for
angular momentum of rigid bodies under 3-D motion
Parallel Axis Theorem
 extremely useful while determining mass MI @ any axis OM
Direction cosines of OM: l, m, n
Unit vector along OM: λ = li + mj + nk
ME101 - Division III
Kaustubh Dasgupta
17
Plane Kinetics of Rigid Bodies
Fixed Axis Rotation
• All points in body move in a circular path @ rotation axis
• All lines of the body have the same ω and α
Accln comp of mass center: ān = r̅ ω2 and āt = r̅ α
Two scalar comp of force eqns:
ΣFn = m r̅ ω2 and ΣFt = m r̅ α
Moment of the resultants @ rotn axis O:
Using parallel axis theorem:

ME101 - Division III
Kaustubh Dasgupta
18
Plane Kinetics of Rigid Bodies
Fixed Axis Rotation
• If the body rotates @ G  a̅ = 0  ΣF = 0
 Resultant of the applied forces will only be couple Ī α
Center of Percussion
• The resultant-force comp (māt = m r̅ α) and the resultant
couple Ī α can be combined to form an equivalent system
with the force m r̅ α acting at a point Q along OG.
 Point Q can be located by:
Using parallel axis theorem: Io = Ī + m r̅ 2
and radius of gyration @ O: ko = √(Io /m)  Io = ko2m
 Location of Point Q: q = ko2/ r̅
• Point Q is known as Center of Percussion
• Resultant of all forces applied to the body must
pass through Q
 ΣMQ = 0
ME101 - Division III
Kaustubh Dasgupta
19
Plane Kinetics of Rigid Bodies
Fixed Axis Rotation
Example: A concrete block is lifted by hoisting mechanism in which the cables are
securely wrapped around the respective drums. The drums are fastened together
and rotate as a single unit @ their mass center at O. Combined mass of drum is
150 kg, and radius of gyration @ O is 450 mm. A constant tension of 1.8 kN is
maintained in the cable by the power unit at A. Determine the vertical accln of the
block and the resultant force on the bearing at O.
Solution:
Draw the FBD and Kinetic Diagrams
ME101 - Division III
Kaustubh Dasgupta
20
Plane Kinetics of Rigid Bodies
Force, Mass, and Acceleration
Fixed Axis Rotation
Example:
Solution: Two ways to draw the FBD and KD
KD: Resultant of the force system on
the drums for centroidal rotation will
be the couple Ī α = Io α
T can be eliminated by drawing FBD
of the entire system.
T will come into picture
 more calculations
ME101 - Division III
Kaustubh Dasgupta
21
Plane Kinetics of Rigid Bodies
Force, Mass, and Acceleration
Fixed Axis Rotation
Example:
Solution:
KD: Resultant of the force system will be the
couple Ī α plus moment due to ma of the block
I = k2m  Ī = Io = (0.45)2150 = 30.4 kgm2
1800(0.6) – 300(9.81)(0.3) = 30.4α + 300a(0.3)
We know a = r α  α = a/0.3
 a = 1.031 m/s2 (and α = 3.44 rad/s2)
Oy - 150(9.81) - 300(9.81) - 1800sin45 =
150(0) + 300(1.031)  Oy = 6000 N
Ox – 1800cos45 = 0  Ox = 1273 N
ME101 - Division III
Kaustubh Dasgupta
22