Principle of Linear Impulse And Momentum

Principle of Linear Impulse And Momentum (2 of 4) The principle of linear impulse and momentum is obtained by integrating the equation of motion with respect to time. The equation of motion can be written  F  m a  m(dv / dt ) Separating variables and integrating between the limits v  v1 at t  t1 and v  v2 at t  t2 results in t2 v2   F dt  m  dv  mv t1 2  mv1 v1 This equation represents the principle of linear impulse and momentum. It relates the particle’s final velocity (v1 ) (v2 ) and initial velocity and the forces acting on the particle as a function of time. Principle of Linear Impulse And Momentum (3 of 4) Linear momentum: The vector mv is called the linear momentum, denoted as L. This vector has the same direction as v. 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 force during its time interval of action. I acts in the same direction as F and has units of N ·s or lb·s. The impulse may be determined by direct integration. Graphically, it can be represented by the area under the force versus time curve. If F is constant, then I  F (t 2  t1 ) Principle of Linear Impulse And Momentum (4 of 4) The principle of linear impulse and momentum in vector form is written as t2 mv1    F dt  mv 2 t1 The particle’s initial momentum plus the sum of all the impulses applied from t1 to t2 is equal to the particle’s final momentum. The two momentum diagrams indicate direction and magnitude of the particle’s initial and final momentum, mv1 and mv2 . The impulse diagram is similar to a free body diagram, but includes the time duration of the forces acting on the particle. Impulse And Momentum: Scalar Equations Since the principle of linear impulse and momentum is a vector equation, it can be resolved into its x, y, z component scalar equations: t2 m(vx )    Fx dt  m(v x ) 2 t1 t2 m(v y )1    Fy dt  m(v y ) 2 t1 t2 m(vz )1    Fz dt  m(v z ) 2 t1 The scalar equations provide a convenient means for applying the principle of linear impulse and momentum once the velocity and 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 the direction of the particle’s initial and final velocities, drawing the impulse and momentum diagrams for the particle. Show the 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 principle of linear impulse and momentum using its scalar form. Forces as functions of time must be integrated to obtain impulses. If a force is constant, its impulse is the product of the force’s magnitude and time interval over which it acts. Example I (1 of 3) Given: A 0.5 kilogram ball strikes the rough ground and rebounds with the velocities shown. Neglect the ball’s weight during the time it impacts the ground. Find: The magnitude of impulsive force exerted on the ball. Plan: 1) Draw momentum and impulse diagrams of the ball as it hits the surface. 2) Apply the principle of impulse and momentum to determine the impulsive force. Example I (2 of 3) Solution: 1) Impulse and momentum diagrams can be drawn as: The impulse caused by the ball’s weight and the normal force N can be neglected because their magnitudes are very small as compared to the impulse from the ground. Example I (3 of 3) 2) The principle of impulse and momentum can be applied along the direction of motion: t2 mv1    F dt  mv2 t1 t2  0.5(25 cos 45 i  25 sin 45 j )    F dt   t1  0.5(10 cos 30 i  10 sin 30 j )   The impulsive force vector is t2 I    Fdt  (4.509i  11.34 j ) N .s t1 Magnitude: I  4.5092  11.34 2  12.2 N .s Example II (1 of 3) Given: The wheels of the 1500 kilogram car generate a traction force F described by the graph. Find: The speed of the car when t = 6 s if the car starts from rest. Plan: 1) Draw the momentum and impulse diagrams of the car. 2) Apply the principle of impulse and momentum to determine the speed. Example II (2 of 3) Solution: 1) The impulse and momentum diagrams can be drawn as: Why is the traction force F drawn acting to the right? The impulse caused by the weight W and the normal force N can be canceled. Why? Example II (3 of 3) 2) The principle of impulse and momentum can be applied along the direction of motion: t2   mv1    F dt  mv2 t1 (1500 kg )(0 m/s)  0.5(6000 N)(2 s)  (6000 N)(4 s)  (1500 kg)(v 2 m / s ) v2  20 m / s Principle of linear impulse and momentum for a system of particles For the system of particles shown, the internal forces fi between particles always occur in pairs with equal magnitude and opposite directions. Thus the internal impulses sum to zero. The linear impulse and momentum equation for this system only includes the impulse of external forces. t2  m i v i 1    F i dt   m i v i 2 t1 Motion of the Center of Mass For a system of particles, we can define a “fictitious” center of mass of an aggregate particle of mass where m is the sum  mi  of all the particles. This system of particles then has an aggregate velocity of vG   mi vi  / mtot mtot , tot The motion of this fictitious mass is based on motion of the center of mass for the system. The position vector r   m r  / m describes the G i i tot motion of the center of mass. Conservation of linear momentum for a system of particles When the sum of external impulses acting on a system of objects is zero, the linear impulsemomentum equation simplifies to  m i v i 1   m i v i 2 This equation is referred to as the conservation of linear momentum. Conservation of linear momentum is often applied when particles collide or interact. When particles impact, only impulsive forces cause a change of linear momentum. The sledgehammer applies an impulsive force to the stake. The weight of the stake is considered negligible, or non-impulsive, as compared to the force of the sledgehammer. Also, provided the stake is driven into soft ground with little resistance, the impulse of the ground acting on the stake is considered non-impulsive. Example 1 (1 of 2) Given: Spring constant k  10 kN/m m A  15 kg, vA  0 m/s, mB  10 kg, vB  15 m/s The blocks couple together at impact. Find: The maximum compression of the spring. Plan: 1) We can consider both blocks as a single system and apply the conservation of linear momentum to find the velocity after impact, but before the spring compresses. 2) Then use the energy conservation to find the compression of the spring. Example 1 (2 of 2) Solution: 1) Conservation of linear momentum    mi vi 0   mi vi 1 10  15i   15  10  vi  v  6m/s  6 m/s  2) Energy conservation equation T V  T  V 1 0.5 15 + 10 1 −6 2 2 2 + 0 = 0 + 0.5 10000 (𝑠2 So the maximum compression of the spring is ) smax  0.3m. Example 2 (1 of 2) Given: Two rail cars with masses of m A  20 Mg and mB  15Mg and velocities as shown. Find: The speed of the car A after collision if the cars collide and rebound such that B moves to the right with a speed of 2 m/s. Also find the average impulsive force between the cars if the collision place in 0.5 s. Plan: Use conservation of linear momentum to find the velocity of the car A after collision (all internal impulses cancel). Then use the principle of impulse and momentum to find the impulsive force by looking at only one car. Example 2 (2 of 2) Solution: Conservation of linear momentum (x-dir): mA v A1   mB vB1   mA v A 2   mB vB 2  20,0003  15,000 1.5  20,000v A 2  15,0002 v A 2  0.375m / s Impulse and momentum on car A (x-dir): m A v A1    Fdt  m A v A 2  20,0003   Fdt  20,0000.375  Fdt  52,000 N .s The average force is  Fdt  52,500 N .s  Favg 0.5 sec ;  105kN