Vector Mechanics for Engineers: Dynamics Chapter 13 Notes

RVCC

ENGR 133-51 - Engineering Mechanics II: Dynamics

Kinetics of Particles: Energy and Momentum Methods Chapter

13

Systems of Particles Chapter

14 Vector Mechanics for Engineers: Dynamics F. Beer & Al

Introduction • Previously, problems dealing with the motion of particles  were  solved through the fundamental equation of motion, F  ma. Current chapter introduces two additional methods of analysis.

• Method of work and energy: directly relates force, mass, velocity and displacement (F, m, v, d). • Method of impulse and momentum: directly relates force, mass, velocity, and time (F, m, v, t).

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Work of a Force  • Differential vector dr is the particle displacement. • Work of the force is

  dU  F  dr  F ds cos   ..............................

• Work is a scalar quantity, i.e., it has magnitude and sign but not direction. It can be >0, vB . • Upon impact the bodies undergo a period of deformation, at the end of which, they are in contact and moving at a common velocity u. • A period of restitution follows during which the bodies either regain their original shape or remain permanently deformed (the restitution impulse pushes the particles apart). • We wish to determine the final velocities of the two bodies. Considering a system with both particles, no external forces act. Then the total momentum of the two body system is preserved: mAvA  mB vB  mAvA  mB vB • A second relation between the final velocities is required. Apply the principle to each particle. 13/14- 77

Direct Central Impact Particle A deforms under the action of B

• Period of deformation: m Av A   Pdt  m Au

During the period of restitution (when A regain its shape) consider again the action of B on A:

• Period of restitution: It is possible to define a coefficient of restitution:

It depends on: materials, shape, velocity and size of particles

m Au   Rdt  m AvA

e  coefficient of restitution Rdt u  vA    Pdt vA  u  0  e 1 13/14- 78

Direct Central Impact

vB  u e u  vB

• A similar analysis of particle B yields

vB  vA  ev A  vB 

• Combining the relations leads to the desired second relation between the final velocities. • Perfectly plastic impact, e = 0: vB  vA  v

? .................................

• Perfectly elastic impact, e = 1: Total energy and total momentum conserved.

? .......................

Note that only in the case of perfectly elastic impact the total energy of the two particles as well as their total momentum is conserved. The principle of work and energy cannot be used for impact problems since it is not known how the internal forces of deformation and restitution vary. However, knowing the particle’s velocities before and after the collision, the energy loss during collision can be calculated as U  T  T 12

2

1

If perfectly elastic impact If plastic impact 13/14- 79

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Problem 16

Hints: •Consider four (4) different moments: 0 before the ball is released after stretching the cord; 1 when the ball hits the ceiling, 2 when the ball bounce back after the collision and finally 3 when is will be all stretched. • To apply the impulse –momentum, you need velocity before the impact, therefore conservation of energy… • Apply impulse-momentum (impact ballceiling)… • Apply equation of conservation of energy to the ball just after the collision to determine stretch. •Weight of ball is non-impulsive 13/14- 81

Oblique Central Impact • Final velocities are unknown in magnitude and direction. Four equations are required (choosing n and t axes).

Consider particles perfectly smooth and frictionless (i.e. no external impulse)

Impulse is only due to internal forces along n axis • No tangential impulse component; tangential component of momentum for each particle is conserved. • Normal component of total momentum of the two particles is conserved. • Normal components of relative velocities before and after impact are related by the coefficient of restitution.

v A t  vA t

vB t  vB t

m A v A n  mB v B n  m A vA n  mB vB n

vB n  vA n  evA n  vB n  13/14- 82

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Sample Problem 13.15

The magnitude and direction of the velocities of two identical frictionless balls before they strike each other are as shown. Assuming e = 0.9, determine the magnitude and direction of the velocity of each ball after the impact.

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Problem 17

Two smooth disks A and B, having a mass of 1 kg and 2 kg respectively, collide with the velocities shown. If the coefficient of restitution for the disks is e = 0.75, determine the velocity of each disk after the collision. Hints: • What kind of impact? • Assume the four unknown components of v after impact all in the positive directions (if come out negative means opposite directions)

Components of initial velocities (along x and y or n and t):

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Sample Problem 13.14

A ball is thrown against a frictionless, vertical wall. Immediately before the ball strikes the wall, its velocity has a magnitude v and forms angle of 30o with the horizontal. Knowing that e = 0.90, determine the magnitude and direction of the velocity of the ball as it rebounds from the wall. 13/14- 86

Sample Problem 13.17

A 30 kg block is dropped from a height of 2 m onto the the 10 kg pan of a spring scale. Assuming the impact to be perfectly plastic, determine the maximum deflection of the pan. The constant of the spring is k = 20 kN/m.

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Homework:

Solving Problems on Your Own Page ~832

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