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Learn about the conservation of linear momentum equation, impulse, and average force in impacts, including central and oblique impacts. Understand the coefficient of restitution and apply the principles to analyze oblique and central impacts. Additionally, explore angular momentum and angular impulse principles.
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King Fahd University of Petroleum & Minerals Mechanical Engineering Dynamics ME 201 BY Dr. Meyassar N. Al-Haddad Lecture # 22
Principle of Linear Impulse and Momentum Initial momentum + Sum of all Impulse = Final momentum
0 Conservation of Linear Momentum for a system of Particles Conservation of linear momentum equation
Impulse Impulse & Average Force
Impact • Impact occurs when two bodies collide with each other during a very short period of time. • Types of impact: • Central impact • Oblique impact Line of impact Plane of impact
Coefficient of restitution e Coefficient of restitution e is defined as the ratio of the restitution impulse to the deformation impulse. Coefficient of restitution e is defined as the ratio of relative velocity after impact to the relative velocity before impact Coefficient of restitution e is according to the impact velocity, material, size and shape of the colliding body, Coefficient of restitution e: range between 0-1 Coefficient of restitution e:is defined along the line of impact only Elastic impact e = 1(re-bounce with same velocity) Plastic impact e = 0 (couple or stick together and move with common velocity)
Procedure for Analysis • Identify the intial velocity ““you may use” • T1+ V1 = T2+ V2 • Apply the conservation of momentum along the line of impact, you will get one equation with two unknown velocity • Use the coefficient of restitution to obtain a second equation • Solve both equation for final velocities after the impact
Four unknowns Oblique Impact Central Impact : one Dimension Oblique Impact : Two Dimension
Procedure for Analysis • Establish x-axis as line of impact • Establish y-axis as plane of impact • Resolve the velocity components • along x, and y as • Apply the conservation of momentum along the line of impact • Use the coefficient of restitution to obtain a second equation • Solve both equation for final velocities along x-axis after the impact • The momentum is conserved along the plane of impact; so
ANGULAR MOMENTUM • For a point object the angular momentum is v m Units - kg.m2/s or sl.ft2/s It is a vector. Here the vector is pointing toward you. Using right-hand rule r
Example 15-3 W=50 Ib P=(20t) Ib v2=? t=2 sec. v1=3 ft/s mk=0.3
Example 15-3 From rest vB=? t=6 sec. Block A Block B
Problem m=12 Mg Fy=150 kN V=? h=? t=6 s Start from rest
Problem m = 28 Mg At rest V = ? t = 4 s F = 4 - 0.01 t2
Example 15-4 mA=15 Mg mB=12 Mg Couple together V2=? After coupling Favg = ? In 0.8 s
Example 15-5 mC =1200-Ib mp = 8-Ib vp=1500ft/s t = 0.03 s. vc2= ? Favg= ? Favg
Example 15-7 mp = 800 kg mH = 300 kg From rest Impulse = ? Couple together Impulse Impulse =
Momentum of particle A,B is conserved along the y axis, since no impulse acts on particle A,B
Problem m = 400 g v1= 2 m/s M = 0.6 N.m v2=? t = 3 s