plane kinetics of rigid bodies - iit guwahatiiitg.ac.in/kd/lecture notes/me101-lecture34-kd.pdf ·...
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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
1ME101 - Division III Kaustubh Dasgupta
:: Earlier discussion on rectilinear/curvilinear motion
- 2 equations of motion
:: Plane kinetics of rigid bodies
- Additional equation of motion
- Account for the rigid body rotation
Plane Kinetics of Rigid Bodies
yy
xx
maF
maF
2ME101 - Division III Kaustubh Dasgupta
Plane Kinetics of Rigid BodiesGeneral Equations of Motion
G is the mass center of the body
ActionDynamic
Response
3ME101 - Division III Kaustubh Dasgupta
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
4ME101 - Division III Kaustubh Dasgupta
Rigid Body Kinetics :: Force/Mass/Acc Plane Motion Equationsω = ωk ; α = αk ; α = ω
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:
5ME101 - Division III
Rigid Body Kinetics :: Force/Mass/Acc
Alternative DerivationUsing mass center G as the reference point:
Accln of mi is vector sum of three terms:
a, 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
ma = translational dynamic response
Ῑ α = rotational dynamic response
ME101 - Division III Kaustubh Dasgupta
Rigid Body Kinetics :: Force/Mass/Acc
Alternative Moment EquationsMoment @ any arbitrary point P:
ρ is the vector from P to mass center G,
and a is the mass center accln.
ρ x ma = moment of magnitude of ma @ P ma 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)
7ME101 - Division III Kaustubh Dasgupta
Rigid Body Kinetics :: Force/Mass/AccConstrained and Unconstrained Motion
:: Motion of a rigid body may be constrained or unconstrained
a x, a y, and α can be
determined independently
using plane motion eqns
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.
8ME101 - Division III Kaustubh Dasgupta
Rigid Body Kinetics :: Force/Mass/AccSystems of Interconnected Bodies
:: If motion of connected rigid bodies are related kinematically
analyze the bodies as an entire 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
Two rigid bodies hinged at A
Forces in the connection A are
internal to the system
9ME101 - Division III Kaustubh Dasgupta
Rigid Body Kinetics :: Force/Mass/AccApplication to three cases of rigid body motion:
TranslationNo angular motion of body (ω and α will be zero)
Mass moment of inertia will not be effective
10ME101 - Division III Kaustubh Dasgupta
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
11ME101 - Division III Kaustubh Dasgupta
Example on TranslationDraw 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
12ME101 - Division III Kaustubh Dasgupta
Rigid Body Kinetics :: Force/Mass/Acc
Rotation @ a Fixed AxisMass 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:
Units of Mass moment of inertia: kg m2
ρ = constant throughout the body
13ME101 - Division III Kaustubh Dasgupta
Plane Kinetics of Rigid BodiesMass 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
I and k are the values @ an axis
through mass center
14ME101 - Division III Kaustubh Dasgupta
Plane Kinetics of Rigid BodiesMass Moments of Inertia
Plane Motion:
Mass MI of the plate (with motion in x-y plane)
@ z-axis through O:
3-D Motion:
o o
15ME101 - Division III Kaustubh Dasgupta
Plane Kinetics of Rigid BodiesMass Moments of InertiaThin 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:
16ME101 - Division III Kaustubh Dasgupta
Plane Kinetics of Rigid BodiesMass Moments of InertiaProducts 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
17ME101 - Division III Kaustubh Dasgupta
Plane Kinetics of Rigid BodiesFixed 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: an = r ω2 and at = r α
Two scalar comp of force eqns:
ΣFn = m r ω2 and ΣFt = m r α
Moment of the resultants @ rotn axis O:
Using parallel axis theorem:
18ME101 - Division III Kaustubh Dasgupta
Plane Kinetics of Rigid BodiesFixed Axis Rotation
• If the body rotates @ G a = 0 ΣF = 0
Resultant of the applied forces will only be couple I α
Center of Percussion
• The resultant-force comp (ma t = m r α) and the resultant
couple I α can be combined to form an equivalent system
with the force m r α acting at a point Q along OG.
Point Q can be located by:
Using parallel axis theorem: Io = I + m r 2
and radius of gyration @ O: ko = √(Io /m) Io = ko2m
Location of Point Q: q = ko2/ r
• Point Q is known as Center of Percussion
• Resultant of all forces applied to the body must
pass through Q
ΣMQ = 0
19ME101 - Division III Kaustubh Dasgupta
Plane Kinetics of Rigid Bodies
Fixed Axis RotationExample: 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
20ME101 - Division III Kaustubh Dasgupta
Plane Kinetics of Rigid Bodies
Force, Mass, and AccelerationFixed Axis RotationExample:
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 I α = Io α
T will come into picture
more calculations
T can be eliminated by drawing FBD
of the entire system.
21ME101 - Division III Kaustubh Dasgupta
Plane Kinetics of Rigid Bodies
Force, Mass, and AccelerationFixed Axis RotationExample:
Solution:
KD: Resultant of the force system will be the
couple I α plus moment due to ma of the block
I = k2m I = 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
22ME101 - Division III Kaustubh Dasgupta