Principle Of Linear Impulse And Momentum
PRINCIPLE OF LINEAR IMPULSE AND MOMENTUMToday’s Objectives:Students will be able to:1. Calculate the linear momentumof a particle and linear impulseof a force.2. Apply the principle of linearimpulse and momentum.In-Class Activities: Check Homework Reading Quiz Applications Linear Momentum And Impulse Principle of Linear Impulse AndMomentum Concept Quiz Group Problem Solving Attention Quiz
READING QUIZ1. The linear impulse and momentum equation is obtained byintegrating the with respect to time.A) friction forceB) equation of motionC) kinetic energyD) potential energy2. Which parameter is not involved in the linear impulse andmomentum equation?A) VelocityB) ForceC) TimeD) Acceleration
APPLICATIONSA dent in an automotive fendercan be removed using an impulsetool, which delivers a force over avery short time interval.How can we determine themagnitude of the linear impulseapplied to the fender?Could you analyze a carpenter’shammer striking a nail in thesame fashion?
APPLICATIONS(continued)Sure! When a stake is struck by asledgehammer, a large impulsiveforce is delivered to the stake anddrives it into the ground.If we know the initial speed of thesledgehammer and the duration ofimpact, how can we determine themagnitude of the impulsive forcedelivered to the stake?
PRINCIPLE OF LINEAR IMPULSE AND MOMENTUM(Section 15.1)Linear momentum: The vector mv is called the linearmomentum, denoted as L. This vector has the same direction asv. The linear momentum vector has units of (kg·m)/s or(slug·ft)/s.Linear impulse: The integral F dt is the linear impulse,denoted I. It is a vector quantity measuring the effect of a forceduring its time interval of action. I acts in the same direction asF and has units of N·s or lb·s.The impulse may be determined bydirect integration. Graphically, itcan be represented by the area underthe force versus time curve. If F isconstant, thenI F (t2 – t1) .
PRINCIPLE OF LINEAR IMPULSE AND MOMENTUM(continued)The next method we will consider for solving particlekinetics problems is obtained by integrating the equation ofmotion with respect to time.The result is referred to as the principle of impulse andmomentum. It can be applied to problems involving bothlinear and angular motion.This principle is useful for solving problems that involveforce, velocity, and time. It can also be used to analyze themechanics of impact (taken up in a later section).
PRINCIPLE OF LINEAR IMPULSE AND MOMENTUM(continued)The principle of linear impulse and momentum is obtainedby integrating the equation of motion with respect to time.The equation of motion can be writtenF m a m (dv/dt)Separating variables and integrating between the limits v v1at t t1 and v v2 at t t2 results int2v2F dt m dv mv2 – mv1t1v1This equation represents the principle of linear impulseand momentum. It relates the particle’s final velocity, v2,and initial velocity (v1) and the forces acting on theparticle as a function of time.
PRINCIPLE OF LINEAR IMPULSE AND MOMENTUM(continued)The principle of linear impulse and momentum intvector form is written as 2F dt mv2mv1 t1The particle’s initial momentum plus the sum of all theimpulses applied from t1 to t2 is equal to the particle’sfinal momentum.The two momentum diagrams indicate directionand magnitude of the particle’s initial and finalmomentum, mv1 and mv2. The impulse diagram issimilar to a free body diagram, but includes thetime duration of the forces acting on the particle.
IMPULSE AND MOMENTUM: SCALAR EQUATIONSSince the principle of linear impulse and momentum is avector equation, it can be resolved into its x, y, z componentt2scalar equations:m(vx)1 Fx dt m(vx)2t1t2m(vy)1 Fy dt m(vy)2t1t2m(vz)1 Fz dt m(vz)2t1The scalar equations provide a convenient means for applyingthe principle of linear impulse and momentum once the velocityand force vectors have been resolved into x, y, z components.
PROBLEM SOLVING Establish the x, y, z coordinate system. Draw the particle’s free body diagram and establish thedirection of the particle’s initial and final velocities, drawingthe impulse and momentum diagrams for the particle. Showthe linear momenta and force impulse vectors. Resolve the force and velocity (or impulse and momentum)vectors into their x, y, z components, and apply the principleof linear impulse and momentum using its scalar form. Forces as functions of time must be integrated to obtainimpulses. If a force is constant, its impulse is the product ofthe force’s magnitude and time interval over which it acts.
EXAMPLEGiven: A 40 g golf ball is hit over a timeinterval of 3 ms by a driver. Theball leaves with a velocity of 35 m/s,at an angle of 40 . Neglect theball’s weight while it is struck.Find: The average impulsive force exerted on the ball and themomentum of the ball 1 s after it leaves the club face.Plan: 1) Draw the momentum and impulsive diagrams of theball as it is struck.2) Apply the principle of impulse and momentum todetermine the average impulsive force.3) Use kinematic relations to determine the velocity ofthe ball after 1 s. Then calculate the linearmomentum.
EXAMPLE(continued)Solution:1) The impulse and momentum diagrams can be drawn:W dt0mv1 mvO 0F dt40N dt 0The impulse caused by the ball’s weight and the normalforce N can be neglected because their magnitudes arevery small as compared to the impulse of the club. Sincethe initial velocity (vO) is zero, the impulse from the drivermust be in the direction of the final velocity (v1).
EXAMPLE(continued)2) The principle of impulse and momentum can be applied alongthe direction of motion:t140mvO F dt mv1t0The average impulsive force can be treated as a constantvalue over the duration of impact. Using vO 0,0.0030 Favg dt mv10Favg(0.003 – 0) mv1(0.003) Favg (0.04)(35)Favg 467 N40
EXAMPLE(continued)3) After impact, the ball acts as a projectile undergoing freeflight motion. Using the constant acceleration equations forprojectile motion:v2x v1x v1 cos 40 35 cos 40 26.81 m/sv2y v1y – gt 35 sin 40 – (9.81)(1) 12.69 m/s v2 (26.81 i 12.69 j) m/sThe linear momentum is calculated as L m v .L2 mv2 (0.04)(26.81 i 12.69 j) (kg·m)/sL2 (1.07 i 0.508 j) (kg·m)/sL2 1.18 (kg·m)/s25.4
CONCEPT QUIZF1. Calculate the impulse due to the force.A) 20 kg·m/sB) 10 kg·m/sC) 5 N·sD) 15 N·s10 N2s2. A constant force F is applied for 2 s to change the particle’svelocity from v1 to v2. Determine the force F if the particle’smass is 2 kg.A) (17.3 j) NB) (–10 i 17.3 j) NC) (20 i 17.3 j) ND) ( 10 i 17.3 j) Nv2 20 m/s60 v1 10 m/st
GROUP PROBLEM SOLVINGGiven: The 500 kg log rests onthe ground (coefficientsof static and kineticfriction are s 0.5 andk 0.4). The winchdelivers a towing force Tto its cable at A as shown.Find: The speed of the log when t 5 s.Plan: 1) Draw the FBD of the log.2) Determine the force needed to begin moving the log, andthe time to generate this force.3) After the log starts moving, apply the principle ofimpulse and momentum to determine the speed of the logat t 5 s.
GROUP PROBLEM SOLVING(continued)Solution:1) Draw the FBD of the log:yFy 0 leads to the result thatWxN W mg (500)(9.81) 4905 N.T Before the log starts moving, uses. After the log is moving, use k.NN2) The log begins moving when the towing force T exceeds thefriction force sN. Solve for the force, then the time.T sN (0.5)(4905) 2452.5 NT 400 t2 2452.5 Nt 2.476 sSince t 4 s, the log starts moving before the towing forcereaches its maximum value.
GROUP PROBLEM SOLVING(continued)3) Apply the principle of impulse and momentum in the xdirection from the time the log starts moving at t1 2.476 s tot2t2 5 s. mv1 F dt mv2 where v1 0 at t1 2.476 st10 45400t2 dt 2.476455T dt -2.4765kNdt mv22.4766400 dt - (0.4)(4905) dt (500)v242.476 (6400)(5 - 4) – (0.4)(4905)(5 – 2.476) (500)v2 v2 15.9 m/s2.476The kinetic coefficient of friction was used since the log ismoving.(400/3)t3
ATTENTION QUIZ1. Jet engines on the 100 Mg VTOL aircraft exert a constantvertical force of 981 kN as it hovers. Determine the netimpulse on the aircraft over t 10 s.A) -981 kN·sB) 0 kN·sC) 981 kN·sD) 9810 kN·sF 981 kN2. A 100 lb cabinet is placed on a smoothsurface. If a force of a 100 lb is appliedfor 2 s, determine the impulse from theforce on the cabinet.A) 0 lb·sB) 100 lb·sC) 200 lb·sD) 300 lb·s30
This equation represents the principle of linear impulse and momentum. It relates the particle's final velocity, v 2, and initial velocity (v 1) and the forces acting on the particle as a function of time. The principle of linear impulse and momentum is obtained by integrating the equation of motion with respect to time.
Impulse and momentum The impulse of a force is the product of the force and the time interval during which it acts. On a graph of F x versus time, the impulse is equal to the area under the curve, as shown in figure to the right. Impulse-momentum theorem: The change in momentum of a particle during a time interval
Mesa Oversized Rectifier. cab impulse response pack featuring 500 Mesa Oversized guitar cabinet IRs. . and then you will need a convolution plug-in to load the IR . IR MESA Guitar Impulse Response Pack WAV IR MESA is the primary impulse response pack of a sequence. It contains over 30 impulse responses recorded in .
Momentum Impulse Newton Graphing Impulse The simple definition for impulse is that it is a change in momentum. If an object’s velocity changes, there is a change in momentum, so there must be an impulse We assume that you are not going to change the mass of an object. Δp m Δv Δp impulse (change
The Impulse system calls allow user applications to use remapping functionality provided by the Impulse Adaptive Memory System to remap their data structures. Impulse supports several remapping algorithms. User applications choose the desired remapping algorithms by calling the right Impulse system calls. This note uses detailed examples to .
1. Impulse and Momentum: You should understand impulse and linear momentum so you can: a. Relate mass, velocity, and linear momentum for a moving body, and calculate the total linear momentum of a system of bodies. Just use the good old momentum equation. b. Relate impulse to the change in linear momentum and the average force acting on a body.
This principle is the basis for describing a system through one or more of its impulse responses. Basic impulse response measurement system. In fact, judging the room's acoustical response to clapping hands is a particular case of an impulse response based measurement model, in the sense that the hands as a source, the ears
Lecture L10 - Angular Impulse and Momentum for a Particle In addition to the equations of linear impulse and momentum considered in the previous lecture, there is a parallel set of equations that relate the angular impulse and momentum. Angular Momentum We consider a particle of mass m,
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