Plane Kinetics of Rigid Bodies - IIT Guwahatiiitg.ac.in/kd/Lecture Notes/ME101-Lecture34-KD.pdf ·...

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


Plane Kinetics of Rigid BodiesGeneral Equations of Motion

G is the mass center of the body

ActionDynamic

Response



• 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


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


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)


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.


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


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


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


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



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


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


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


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:


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


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:


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



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



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.



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


Plane Kinetics of Rigid Bodies - IIT Guwahatiiitg.ac.in/kd/Lecture Notes/ME101-Lecture34-KD.pdf ·...

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