91950468 kinetics body work energy impulse momentum

5/20/2018 91950468 Kinetics Body Work Energy Impulse Momentum

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ynam cs

Kinetics of rigid body:

work & ener Work & energy equation for rigid body

Impulse & momentum of rigid body

Prof. Jacek Uziak


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Rigid BodyIt is an idealized model of anobject that does not deform

or c ange s ape.

points on the body is constant.An le between an two lines

on or in the body is constant.

Because of the latter fact, whatever angular velocity a givenline has at an instant, that angular velocity is also that of all

2

.

In other words, a body only has one velocity (which wouldhave one parameter in planar motion, three in 3D motion).


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Eulers laws (generalized 2nd

Newtons Law) Laws of motion for a rigid body are known as Eulers laws. Euler gave two laws for the motion of a rigid body.

Equations of motion for system of particlesn

Gi

MaF ==

the translational motion of the

rigid body

Angular momentum of a system of particles:

GG HM &= Describes how the change of angularmomentum of the rigid body is

3(G- centre of mass)

and couples applied on the body.


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WORK & ENERGYThe principle of work and energy for a rigid body isexpressed in the form:

1 + 1 2 = 2where T1 and T2 re resent the initial and final values of thekinetic energy of the rigid body and U1 2 the work of the

external forcesacting on the rigid body.

U = Fcos ds

s2

s1

where Fis the magnitude of the force, - angle it forms with

4

the direction of motion of A, and sthe variable of integration

measuring the distance traveled by Aalong its path.


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The work of a couple of moment M applied to a rigid body

=

2

ur ng a ro a on n q o e r g o y:

1

T= mv 2 + I21

2

1

where vis the velocity of the mass centerGof the bod w the an ular velocit of

G

the body, and I its moment of inertia aboutan axis through Gperpendicular to the

v

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.


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T= mv2

+ I21 1

G

e net c energy o a r g o y nplane motion may be separated intov

(1) the kinetic ener mv 2 associated with themotion of the mass center Gof the body, and

(2) the kinetic energy I2 associated with the

ro a on o e o y a ou .

6


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Note that in the general plane motion ofeach bod , it is necessar to use thoseparameters which refer to the translation of

the mass centre G and the rotation of theo y a ou e ax s pass ng roug .

Kinetic energy o a bo y in pure trans ationor pure rotation can now be considered as a

7


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The translatin ri id bod has a mass mand

G all of its particles have a common velocity v.

Body does not have any angular velocity andv

21

2

an angular velocity :

O21

2 O

8

O

body about the fixed axis.


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Kinetic Energy: General Plane MotionKinetic energy of a body in general plane

mot on may a so e expresse n terms orotational velocity about the instantaneous.

In this approach, the general plane motioninstantaneous centre Cand the kineticener can be ex ressed as

1

92

C


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When a rigid body, or a system of rigid bodies, moves

,work and energy may be expressed in the form

T + =T +

which is referred to as theprinciple of conservation ofenergy. This principle may be used to solve problemsinvolving conservative forces such as the force of gravityor the force exerted by a spring.

The concept of power is extended to a rotating bodysubjected to a couple

Power = = =MdUdt

ddt

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where Mis the magnitude of the couple and w is theangular velocity of the body.


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A uniform disk of mass m i i h

O

vo

slip on a horizontal

surface.Determine the kineticenergy of the disk when

cen re a avelocity vo as shown

11


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The disk is in general plane motion and,

K.E. = Linear K.E. (at centre of mass) +otat ona . . a out centre o mass

or 222

1

2

1.. oo ImvEK +=

vo=21

mRIo =where or

12


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,

222

2

222

4

3

4

1

2

1

2

1

2

1

2

1.. ooo

oo mvmvmv

R

vmRmvEK =+=+=

13


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The alternative method is to consider themotion as pure rotation about thenstantaneous center o zero ve oc tywhich, in this case, is the point of contact

.approach the kinetic energy is given by the

,21.. CIEK =

where IC is the moment of inertia about

14

,

parallel axis theorem as


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2222

22 mRmRmRmRIIoC =+=+=

vo

=

Angular velocity is still

,2

222 3311 vo

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2 4222..

RC


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Example The velocit of the 8-kcylinder is 0.3 m/sat acertain instant.

What is its speed v afterdropping an additional 1.5 m?

drum is 12 kg, its centroidal=

mm, and the radius of itsgroove is ri= 200 mm.

The frictional moment at Ois constant 3 Nm.

16Answer: 3.01 m/s


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Exam le

The 15-kgwheel is released from rest and rolls on

its hubs without slipping.Calculate the velocity vof the centre Oof thewheel after it has moved a distance x = 3 mdown

e nc ne.The radius of gyration of the wheel about Ois

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.

Answer: 0.85 m/s


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Im ulse & Momentum: Ri id BodTheprinciple of impulse and momentumderivedor a sys em o par c es can e app e o e

motion of a rigid body.

Syst Momenta1 + Syst Ext Imp1 2 = Syst Momenta2

(m)v

P

I

18


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Syst Momenta1 + Syst Ext Imp1 2 = Syst Momenta2

For a rigid body the system of the momenta of the particlesforming the body is equivalent to a vector mvattached to themass cen er o e o y an a cou e .

The vector mvis associated with translation of the body withGand represents the linear momentum of the body, while thecouple Icorresponds to the rotation of the body about Gand represents the angular momentumof the body about an

.

(m)v

P mvG

I

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The principle of impulse and momentum can be expressedgrap ica y y rawing t ree iagrams representingrespectively

the s stem of initial momenta of the bod

the impulses of the external forces acting on it, and the system of the final momenta of the body.,

y components, and the moments about any given point of

the vectors shown in the figure, we obtain threeequations o motion w ic may e so ve or t e esireunknowns.

mv2Fdt

1

GG

20xO

1

xO xO

2


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. ,which is similar to that for a particle andwhich could be formulated in the

following way:

system of bodies are acting on the body,

remains constant.

n o er wor , mvG con an w ere, mis the total mass of the system, and vG is

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e ve oc y o e ma cen re .


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. momentum, which could be expressed as

Provided no external moments act on asystem, the sum of the angular

momenta of the bodies in the s stemremains constant

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e secon aw s use u n mec an caanalysis of situations such as

o ngagemen o ro a ng par s w en aclutch is released.

velocity of a body changes if its

some internal means (e.g. a gyratingdancer s ins faster b ulling

her/his arms inwards).

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Exam le The 28 g bullet has ahorizontal velocity of 500m s as it stri es t e gcompound pendulum, which

925 mm.

mm, calculate the angularvelocit of the endulum500 m/s

with its embedded bulletimmediately after the

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impact.

Answer: 0.703 rad/s


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ExampleTh h h ivelocity of 500 m/s as it strikesthe 10 k slender bar OAwhich

is suspended from point Oand isinitiall at rest.

It takes 0.001 s for the bullet

to embed itself in the bar.Calculate the time average ofthe horizontal force O exerted

by the pin on the bar at Oduringthe interaction between the

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bullet and the bar.

Answer: 3750 N


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The constant 40 Norce a e o

the 36 kg stepped.

is k = 200 mm, and it rolls on the incline withoutsli in .

If the cylinder is at rest when the force is first

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Answer: 24.18 rad/s

91950468 kinetics body work energy impulse momentum

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