kinetika partikel
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Kinetics of Particle
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Solution Approach
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A. direct application of Newton’s second law (called theforcemass-acceleration method)B. use of work and energy principlesC. solution by impulse and momentum methods
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SECTION A: FORCE, MASS, ANDACCELERATION
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Newton’s SecondLaw
If the resultant force acting on a particle is not zero, the particlewill have an acceleration proportional to the magnitude of theresultant and in the direction of this resultant force .
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SECTION A: FORCE, MASS, ANDACCELERATION
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Experiment: subjecting a mass particle to the action ofa single force, F1, F2, …Fn. 1. the ratios of applied force to corresponding
acceleration all equal the same number, providedthe units used for measurement are not changedin the experiments.
the constant C is a measure of some invariable
property of the particle.This property is the inertia of the particle, which isits resistance to rate of change of velocity .The mass m is used as a quantitative measure ofinertia.
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SECTION A: FORCE, MASS, ANDACCELERATION
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2. The acceleration is always in the direction ofthe applied force. Vector relation:
U.S. customary units : the units of mass (slugs)are derived from the units of force (poundsforce, lb) divided by acceleration (feet persecond squared, ft/sec2). Thus, the mass units areslugs = lb-sec2/ft.
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absolute system (SI) gravitational system (US)
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Equation of Motionand Solution of Problems
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Equation of motion:
Two Types of Dynamics Problems:1. a of the particle is either specified or can be
determined, we then determine the correspondingforces.
2. The forces F acting on the particle are specified and wemust determine the resulting motion.
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Equation of Motionand Solution of Problems
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Constrained and Unconstrained Motionunconstrained motion: the particle is free of mechanicalguides and follows a path determined by its initial motionand by the forces which are applied to it from external
sources . An airplane or rocket in flight and an electronmoving in a charged field are examples of unconstrainedmotion.constrained motion: the path of the particle is partially or
totally determined by restraining guides. A train movingalong its track and a collar sliding along a fixed shaft.
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Equation of Motionand Solution of Problems
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Degrees of freedomindependent coordinates are required to specify the positionof the particle at any instant.three degrees of freedom: 3 independent coordinates arerequired to specify the position free moving particle,airplane, rocket in free flight.two degrees of freedom : a marble sliding on the curved surfaceof a bowlone degree of freedom : If a particle is constrained to move along
a fixed linear path, as is the collar sliding along a fixed shaft.
Free-Body Diagram: every force, known and unknown, whichacts on the particle is represented and thus accounted for.
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Equation of Motionand Solution of Problems
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The general procedure1. Identify the motion2. choose the coordinate system3. draw the free-body diagram of the body.4. Obtain the appropriate force summations from this
diagram in the usual way.
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Rectilinear Motion
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moving at x direction only General case
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Example
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Free body diagram
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Example
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A small inspection car with a mass of 200 kg runs alongthe fixed overhead cable and is controlled by the attachedcable at A. Determine the acceleration of the car when thecontrol cable is horizontal and under a tension T 2.4 kN.Also find the total force P exerted by the supporting cableon the wheels.
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Example
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Free body diagram
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Curvilinear Motion
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The choice of appropriate coordinatesystem depends on the conditions of theproblems.
Rectangular coordinates
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Curvilinear Motion
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Normal and tangential coordinates
Polar coordinates
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Curvilinear Motion
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The general procedure1. Identify the motion2. Choose the coordinate system3. draw the free-body diagram of the body.4. Obtain the appropriate force summations from this
diagram in the usual way.
The free-body diagram should be complete to avoid incorrectforce summations.
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Example
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A 1500-kg car enters a section of curved road in thehorizontal plane and slows down at a uniform rate from aspeed of 100 km/h at A to a speed of 50 km/h as it passesC . The radius of curvature of the road at A is 400 m andat C is 80 m. Determine the total horizontal forceexerted by the road on the tires at positions A, B, and C .Point B is the inflection point where the curvaturechanges direction.
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Contoh
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Example
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Example
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Angel of direction of a and F?
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Work and Kinetic Energy
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These cases involve:1. integration of the forces with respect to the
displacement of the particle the equations of work andenergy
2. integration of the forces with respect to the time theyare applied the equations of impulse and momentum .
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Work and Kinetic Energy
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The work done by the force F duringthe displacement d r is defined as
dr = ds cos α dU = F.ds cos α = F cos α dsF cos α = Ft
The component Fn = F sinα
normal tothe displacement does no work.Work is positive if the workingcomponent Ft is in the direction of thedisplacement and negative if it is in theopposite direction.
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Units of WorkThe SI units of work are those of force (N) timesdisplacement (m) or This unit is given the special name
joule (J).
In the U.S. customary system, work has the units of ft-lb.
Calculation of Work
Or,
Work and Kinetic Energy
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Work and Kinetic Energy
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(2) Work Associated with a Spring ForceIf the body begins at the undeformed spring position and thenmoves to the right, the spring force is to the left; if the body beginsat x 1= 0 and moves to the left, the spring force is to the right.if the body moves from an arbitrary initial position x 1= 0 to theundeformed final position x 2=0, so the work is positive.
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(3) Work Associated with Weight A. Case g = constant constant If the altitude variation is small
The horizontal movement does not contribute to this work.If the body rises, then ( y 2> y 1) > 0 and this work is negative. If
the body falls, ( y 2 < y 1) < 0 and the work is positive.
Work and Kinetic Energy
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Work and Kinetic Energy
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(3) Work Associated with Weight(B). Case (b) g ≠ constant If large changes in altitude occur
if a body rises to a higher altitude ( r 2 > r 1), this work isnegative, if the body falls to a lower altitude ( r 2 < r 1), the workis positive.
2
21110672.6
kg m N
G
kg1098.5 24Massa bumi
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Work and Curvilinear Motion
Principle of Work and Kinetic Energy
Work and Kinetic Energy
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work-energy equation
the total work done by all forces acting on a particle as it moves from point 1 to point 2 equals the corresponding change in kinetic energy
of the particle • T selalu positif• Δ T bisa positif, negatif, atau nol
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Work and Kinetic Energy
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Daya: Kerja yang dilakukan dalam waktu tertentu
Efficiency
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Example
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Calculate the velocity v of the 50-kg crate when itreaches the bottom of the chute at B if it is given an initialvelocity of 4 m/s down the chute at A. The coefficient ofkinetic friction is 0.30.
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Solution
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Gravitational Potential EnergyThe gravitational potential energy Vg of theparticle is defined as the work m.g.h doneagainst the gravitational field to elevate theparticle a distance h above some arbitrary
reference plane (called a datum )In going from one level at h=h1 to a higherlevel at h=h2, the change in potential energy
If Vg2 > Vg1 potential energy is increasing,so the work U 1-2 is negatif . So : -m.g. ∆ hThe work is positive if potential energydecreases
Potential Energy
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Potential Energy
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Gravitational Potential EnergyWhen large changes in altitude, thegravitational force G mme/r 2 = mgR2/r 2 isno longer constant (inconstant).The work done by GPE: U = ∆ Vg
It is customary to take ( Vg )2=0 whenr 2=~, so that with this datum we have
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Potential Energy
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Elastic Potential Energy
The force exerted on the spring by the moving body isequal and opposite to the force F exerted by the springon the body SO: The work done on the spring is the negative of the work
done on the body.Therefore, we may replace the work U done by the spring onthe body by -Ve,
positive
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Potential Energy
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Work-Energy Equation
law of conservation of dynamical energy
Δ TU’ 1-2 : the work of all external forces other than
gravitational forces and spring forces
U’ 1-2 = 0
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Problem
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The 6-lb slider is released from rest at position 1 and slideswith negligible friction in a vertical plane along the circular rod.The attached spring has a stiffness of 2 lb/in. and has anunstretched length of 24 in. Determine the velocity of theslider as it passes position 2.
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Solution
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IMPULSE AND MOMENTUM
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Linear Impulse and Linear Momentum
the linear momentum of partikel:G = mv ( kg.m/s or N.s or lb-sec )
the resultant of all forces acting on a particleequals its time rate of change of linearmomentum .
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IMPULSE AND MOMENTUM
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Conservation of Linear Momentum
If ΣF = 0, So G1 = G2 Constant Conserved
Impulsive forces : very large force and has short durationNon-impulsive force : can be neglected in comparison
with impulsive forces. Example: the weight of a baseballduring its collision with a bat — the weight of the ball(about 5 oz) is small compared with the force exerted onthe ball by the bat.
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IMPULSE AND MOMENTUM
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ExampleA tennis player strikes the tennis ball with her racket whenthe ball is at the uppermost point of its trajectory asshown. The horizontal velocity of the ball just before
impact with the racket is v 1= 50 ft/sec, and just afterimpact its velocity is v 2 = 70 ft/sec directed at the 15 angleas shown. If the 2-oz (0.125 lb) ball is in contact with theracket for 0.02 sec, determine the magnitude of the
average force R exerted by the racket on the ball. Alsodetermine the angle made by R with the horizontal.
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IMPULSE AND MOMENTUM
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Solution
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IMPULSE AND MOMENTUM
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We note that the impact force Ry = 3.64 lb is considerablylarger than the 0.125-lb weight of the ball. Thus, the weightmg , a nonimpulsive force , could have been neglected assmall in comparison with Ry . Had we neglected the weight,the computed value of Ry would have been 3.52 lb.
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IMPULSE AND MOMENTUM
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Angular Impulse and Angular Momentum Angular momentum H O of P about O: The moment of thelinear momentum vector m v about the origin O
• The angular momentum is a vectorperpendicular to the plane A defined by r and v .
• The sense of H O is clearly defined by the right-hand rule for cross products .
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IMPULSE AND MOMENTUM
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Angular Impulse and Angular Momentum
mvr sin θ , is the magnitude ofthe cross product H O r mv
S1 Units: kg (m/s) m = kg m 2/s = N m s
US : [lb/(ft/sec2)][ft/sec][ft] = lb-ft-sec
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IMPULSE AND MOMENTUM
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The Angular Impulse-Momentum Principle
The product of moment and time is defined as angular
impulse,total angular impulse on m about the fixed point O equals thecorresponding change in angular momentum of m about O .
SI Nm s = kg m2/s
US lb-ft-sec
X-component
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IMPULSE AND MOMENTUM
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Conservation of Angular Momentum
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Example
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A small sphere has the position and velocity indicated inthe figure and is acted upon by the force F . Determine theangular momentum H O about point O and the timederivative Ho
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Solution
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As with moments of forces, the position vector must runfrom the reference point ( O in this case) to the line ofaction of the linear momentum mv. Here r runs directlyto the particle.
Special Application of Particle Kinetics:
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Special Application of Particle Kinetics:Direct Central Impact
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Impact refers to the collision between two bodies and ischaracterized by the generation of relatively large contactforces which act over a very short interval of time .If the velocities of the two particles are directed along the lineof impact, the impact is said to be a direct impact .If either or both particles move along a line other than the lineof impact, the impact is said to be an oblique impact (Fig.13.20b).
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Special Application of Particle Kinetics:
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Special Application of Particle Kinetics:Direct Central Impact
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Coefficient of Restitution
e = 0, Perfectly Plastic Impact/inelastic . When e = 0, so
V2’ = v1’ . There is no period of restitution, and bothparticles stay together after impact and the loss ofenergy is a maximum. If v2’ = v1’ = v’; then:
m1v1 + m2v2 = (m1 + m2)v’
e = 1 , Perfectly Elastic Impact.
IMPULSE AND MOMENTUM
Obli C l I b k i i
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Oblique Central Impact = tumbukan miring
1. Kekekalan momentum pada arah n
2. Kekekalan momentum pada arah t
3. Coefficient of restitution
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Example
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A ball is projected onto the heavy plate with a velocity of 50ft/sec at the 30 angle shown. If the effective coefficient ofrestitution is 0.5, compute the rebound velocity v and its angle.
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Solution
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Let the ball be denoted body 1 and the plate body 2.
The rebound velocity v’ and its angle θ’ are then
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Contoh
Sebuah gerbong kereta 10.000 kg yang berjalan dengan laju24 m/s menabrak gerbong lain yang sejenis yang sedangdalam keadaan diam, setelah menabrak kecepatan keduanya12 m/s, hitung berapa besar energi kinetik awal yang diubahmenjadi energi panas atau bentuk energi lainnya.
Sebelum tumbukan
Setelah tumbukan
Energi yang diubah menjadi bentuk lain :2,88 x 10 6 J – 1,44 x 10 6 J = 1,44 x 10 6 J
J x smkg vm 62211 1088,2/24000.1021
2
1
J x smkg 62 1044,1/12000.2021