week 12 ---impulse+momentum+impact
DESCRIPTION
UNSW CVEN1300TRANSCRIPT
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ImpulseMomentum
Impact
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Impulse & Momentum
Analysis velocity of objects subjected to external forces and impacts.
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Principle of Linear Impulse & Momentum
Consider the curvilinear motion of a particle of mass m.
The velocity is in a tangential direction to the path.
rv
The resultant force will be in the direction of acceleration.
dtdvmvmma F
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Principle of Impulse & Momentum
dtdG
dtmvd
dtdvm )(F
mvG
GF The resultant force acting on a particle equals its rate of change of linear momentum.
The product of the mass m and the velocity v is defined as the linear momentum.
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LINEAR MOMENTUM AND IMPULSE
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.
Linear momentum: The vector mv is called the linear momentum, denoted as G. This vector has the same direction as v. The linear momentum vector has units of (kg·m)/s.
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Principle of Impulse & Momentum
dtdG
dtmvd
dtdvm )(F G Fdt
12
2
1
GGFdtt
t
12
2
1
mvmvFdtt
t
21
2
1
GFdtGt
t
Initial linear momentum + Linear impulse
= Final momentum
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Principle of Impulse & Momentum
xxxx
t
tx mvmvGGdtF 1212
2
1
The three components in x, y and z are
yyyy
t
ty mvmvGGdtF 1212
2
1
zzzz
t
tz mvmvGGdtF 1212
2
1
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Principle of Impulse & Momentum
x
t
txx mvdtFmv 21
2
1
Initial linear momentum + Linear impulse
= Final momentum
y
t
tyy mvdtFmv 21
2
1
z
t
tzz mvdtFmv 21
2
1
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Principle of Impulse & Momentum
• It states that the impulse applied to an object during an interval of time is equal to the change in the object’s linear momentum.
12
2
1
mvmvFdtt
t
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Procedure for Analysis
• 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 momentum and force impulse vectors.
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Procedure for Analysis
• 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.
• 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.
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EXAMPLE 1The 100-kg stone is originally at rest on the smoothhorizontally surface. If a towing force of 200 N,acting at an angle of 45°, is applied to the stone for10 s, determine the final velocity and the normalforce which the surface exerts on the stone duringthe time interval.
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EXAMPLE 1Free Body Diagram
Impulse )( 12 ttFI c
Velocity: horizontal and positive when pointing to the right
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EXAMPLE 1
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EXAMPLE 1
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EXAMPLE 2The 250-N crate is acted upon by a force having avariable magnitude P = (100t) N. Determine thecrate’s velocity 2 s after P has been applied. Theinitial velocity is v1 = 1 m/s down the plane, and thecoefficient of kinetic friction between the crate andthe plane is μk = 0.3.
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EXAMPLE 2Free Body Diagram
Impulse
--- For varying force P:integrating P = 100t overthe 2-s time interval.
---For constant forces(Weight, normal forceand frictional force): )( 12 ttFI c
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EXAMPLE 2
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EXAMPLE 2
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EXAMPLE 3Given: A 40 g golf ball is hit over
a time interval of 3 ms by a driver. The ball leaves with a velocity of 35 m/s, at an angle of 40°. Neglect the ball’s weight while it is struck.
Find: The average impulsive force exerted on the ball and the momentum of the ball 1 s after it leaves the club face.
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EXAMPLE 3Plan: 1) Draw the momentum and impulsive
diagrams of the ball as it is struck.2) Apply the principle of impulse and
momentum to determine the average impulsive force.
3) Use kinematic relations to determine the velocity of the ball after 1 s. Then calculate the linear momentum.
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Solution:1) The impulse and momentum diagrams can be
drawn:
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 of the club. Since the initial velocity (vO) is zero, the impulse from the driver must be in the direction of the final velocity (v1).
mvO = 0
+ =
W dt 0
N dt 0 F dt
mv1
40°
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2) The principle of impulse and momentum can be applied along the direction of motion:
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3) After impact, the ball acts as a projectile undergoing free-flight motion. Using the constant acceleration equations for projectile motion:
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Considering two particles interacting during a time interval t, if the interaction forces are F one particle.
Conservation of Linear Momentum
Then it will be F for the other particle.
FtG 1
FtG 2
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Conservation of Linear Momentum
Therefore the total change in linear momentum for the system is
0)( FtFt
21 GG
0G
or
This is known as the principle of conservation of linear momentum.
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For the system of particles shown, the internal forces fibetween particles always occur in pairs with equal magnitude and opposite directions. Thus the internal impulses sum to zero.
Principle of Linear Impulse & Momentum for a System of Particles
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The internal forces acting between particles do not appear with this summation, since by Newton’s third law they occur in equal but opposite collinear pairs and therefore cancel out.
Principle of Linear Impulse & Momentum for a System of Particles
dtdvmvmamF i
iiiiii
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The linear impulse and momentum equation for this system only includes the impulse of external forces.
Principle of Linear Impulse & Momentum for a System of Particles
21
2
1
ii
t
tiii vmdtFvm
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EXAMPLE 4Given:Two rail cars with masses of mA = 15 Mg and
mB = 12 Mg and velocities as shown.
Find: The speed of the cars after they meet and connect. Also find the average impulsive force between the cars if the coupling takes place in 0.8 s.
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Plan: Use conservation of linear momentum to find the velocity of the two cars after connection (all internal impulses cancel). Then use the principle of impulse and momentum to find the impulsive force by looking at only one car.
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Solution:
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Impact
Impact occurs whentwo bodies collidewith each otherduring a very shortperiod of time,causing relativelylarge (impulsive)forces to be exertedbetween the bodies.
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ImpactThe line of impact passes through the mass centers of the particles.Central impact: thedirection of motion of themass centers of the twocolliding particles isalong the line of impact.
Oblique impact: one orboth of the particles ismoving at an angle withthe line of impact.
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Central Impact• Consider two smooth particles A and B with, theinitial momentum as shown• Provided , collision will Eventually occur.
11 )()( BA vv
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The particles are deformable or non-rigid duringthe collision.
• An equal but opposite deformation impulse ∫ Pdtis exerted on each other.
Central Impact
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At the instant of maximum deformation• Their relative motion is zero• Both particles move with a common velocity v.
Central Impact
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• Afterward a period of restitution occurs (theparticles will either return to their original shapeor remain permanently deformed).• The equal but opposite restitution impulse ∫R dtpushes the particle apart from one another.
Central Impact
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Just after the separation the particles will havethe final momentum, where 22 )()( AB vv
Central Impact
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Conservation of linear momentum
Central Impact
2)( AvWith two unknown velocities 2)( Bvand
2211 )()()()( BBAABBAA vmvmvmvm
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The coefficient of restitution.
Central Impact
11
22
)()()()(
BA
AB
vvvve
22 )()( AB vv
Relative velocity just after impact
11 )()( BA vv
Relative velocity just before impact
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Coefficient of restitution e has a value between 0and 1 depending on the material property of the particles
Central Impact
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Find the final velocities just after central impact.
Procedure for Analysis-Central Impact
Given:• Coefficient of restitution,• Mass of each particle• Initial velocity of each particle just before impact
Equations:• The conservation of momentum: Σmv1 = Σmv2• coefficient of restitution relating the relativevelocities before and just after collision.
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EXAMPLE 5The bag A, having a mass of 6 kg is released fromrest at the position θ = 0°. After falling to θ = 90°, isstrikes an 18 kg box B. If the coefficient ofrestitution between the bag and the box is e = 0.5,determine the velocities of the bag and box justafter impact and the loss of energy during collision.
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Solution
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Coefficient of Restitution
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Oblique ImpactParticles travel at an angle with the line of impact.
The equations for centric impact are still valid for oblique impact. We need to work out
2)( Av 2)( Bv 2 2 or 2)( Axv 2)( Ayv 2)( Bxv 2)( Byv
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Choose the line of impact as the x axis.
Procedure for Analysis-Oblique Impact
The impulsive forces of deformation and restitution act only in the x direction.
2)( Axv 2)( Ayv 2)( Bxv 2)( Byv
Resolving the velocities into components along x and yaxes.
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Conservation of momentum along x axis.
Procedure for Analysis-Oblique Impact
Coefficient of restitution
21 )()( xx vv
11
22
)()()()(
BxAx
AxBx
vvvve
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Conservation of momentum along y axis (no impulse on A and B)
Procedure for Analysis-Oblique Impact
21 )()( ByBByB vmvm
21 )()( AyAAyA vmvm
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EXAMPLE 6Two smooth disks A and B, having mass of 1 kgand 2 kg respectively, collide with the velocitiesshown. If the coefficient of restitution for the disksis e = 0.75, determine the x and y components ofthe final velocity of each disk just after collision.
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Solution
Resolving each of the initial velocities into x and ycomponents, we have
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Conservation of “x” Momentum
The four unknown velocity components after collision are assumed to act in the positive directions.
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Coefficient of (x) Restitution. Both disks are assumed to have components of velocity in the +x direction after collision
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Conservation of “y” Momentum