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MAE 242 Dynamics – Section I Dr. Kostas Sierros. Problem 1. Problem 2. Problem 3. Problem 4. Team building exercise 1. 12.9, 12.22, 12.26, 12.42, 12.53, 12.65, 12.71, 12.83, 12.84, 12.100, 12.111, 12.153 13.5, 13.15, 13.34, 13.45, 13.51 14.10, 14.15, 14.17, 14.65.

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slide1

MAE 242

Dynamics – Section I

Dr. Kostas Sierros

slide6

Team building exercise 1

12.9, 12.22, 12.26, 12.42, 12.53, 12.65, 12.71, 12.83, 12.84, 12.100, 12.111, 12.153

13.5, 13.15, 13.34, 13.45, 13.51

14.10, 14.15, 14.17, 14.65

For Tuesday, October 2nd

Groups will be asked to solve and explain problems on the board.

Group members that are not present will get minus points

Groups that will solve and explain problems correctly will get bonus points

All group members will need to contribute during the solution of the problem

Problems that will be asked may vary from group to group

slide7

Kinetics of a particle: Impulse and Momentum

Chapter 15

Chapter objectives

  • Develop the principle of linear impulse and momentum for a particle
  • Study the conservation of linear momentum for particles
  • Analyze the mechanics of impact
  • Introduce the concept of angular impulse and momentum
  • Solve problems involving steady fluid streams and propulsion with variable mass
slide8

Lecture 11

  • Kinetics of a particle:
  • Impact (Chapter 15)
  • -15.4
slide9

Material covered

  • Kinetics of a particle:
  • Impact
  • …Next lecture…
  • Angular momentum
  • Relation between moment of a force and angular momentum
  • Angular impulse and momentum principles
slide10

Also…

http://www.nasa.gov/mission_pages/deepimpact/multimedia/di-animation.html

slide11

Today’s Objectives

  • Students should be able to:
  • Understand and analyze the mechanics of impact.
  • Analyze the motion of bodies undergoing a collision, in both central and oblique cases of impact.
slide12

Applications 1

The quality of a tennis ball is measured by the height of its bounce. This can be quantified by the coefficient of restitution of the ball.

If the height from which the ball is dropped and the height of its resulting bounce are known, how can we determine the coefficient of restitution of the ball?

slide13

Applications 2

In a game of billiards, it is important to be able to predict the trajectory and speed of a ball after it is struck by another ball.

If we know the velocity of ball A before the impact, how can we determine the magnitude and direction of the velocity of ball B after the impact?

slide14

Central impactoccurs when the directions of motion of the two colliding particles are along the line of impact.

Oblique impactoccurs when the direction of motion of one or both of the particles is at an angle to the line of impact.

Impact (section 15.4)

Impactoccurs when two bodies collide during a veryshort time period, causing large impulsive forces to be exerted between the bodies. Common examples of impact are a hammer striking a nail or a bat striking a ball. Theline of impactis a line through themass centersof the colliding particles. In general, there aretwotypes of impact:

slide15

vA

vB

Line of impact

Central impact

Central impacthappens when the velocities of the two objects are along the line of impact (recall that the line of impact is a line through the particles’ mass centers).

Once the particles contact, they may deform if they are non-rigid. In any case, energy is transferred between the two particles.

There are two primary equations used when solving impact problems. The textbook provides extensive detail on their derivation.

slide16

Central impact (continued)

In most problems, the initial velocities of the particles, (vA)1 and (vB)1, are known, and it is necessary to determine the final velocities, (vA)2 and (vB)2. So thefirst equation used is theconservation of linear momentum, applied along the line of impact.

(mA vA)1 + (mB vB)1 = (mA vA)2 + (mB vB)2

This provides one equation, but there are usually two unknowns, (vA)2 and (vB)2. So another equation is needed. Theprinciple of impulse and momentumis used to develop this equation, which involves thecoefficient of restitution, or e.

slide17

The equation defining the coefficient of restitution, e, is

(vB)2 – (vA)2

e =

(vA)1 - (vB)1

Central impact (continued)

Thecoefficient of restitution, e, is the ratio of the particles’relative separation velocityafter impact, (vB)2 – (vA)2, to the particles’relative approach velocitybefore impact, (vA)1 – (vB)1. The coefficient of restitution is also an indicator of the energy lost during the impact.

If a value for e is specified, this relation provides the second equation necessary to solve for (vA)2 and (vB)2.

slide18

Coefficient of restitution

In general, e has a value between zero and one. The two limiting conditions can be considered:

• Elastic impact (e = 1):In a perfectly elastic collision, no energy is lost and the relative separation velocity equals the relative approach velocity of the particles. In practical situations, this condition cannot be achieved.

• Plastic impact (e = 0):In a plastic impact, the relative separation velocity is zero. The particles stick together and move with a common velocity after the impact.

Some typical valuesof e are:

Steel on steel: 0.5 – 0.8 Wood on wood: 0.4 – 0.6

Lead on lead: 0.12 – 0.18 Glass on glass: 0.93 – 0.95

slide19

Impact: Energy losses

Once the particles’ velocities before and after the collision have been determined, theenergy lossduring the collision can be calculated on the basis of the difference in the particles’kinetic energy.The energy loss is

 U1-2 =  T2 -  T1 where Ti = 0.5mi(vi)2

During a collision, some of the particles’ initial kinetic energy will be lost in the form of heat, sound, or due to localized deformation.

In a plastic collision(e = 0), the energy lost is a maximum, although it does not necessarily go to zero.

slide20

Oblique impact

In anoblique impact, one or both of the particles’ motion is at an angle to the line of impact. Typically, there will be four unknowns: themagnitudes anddirectionsof the final velocities.

The four equations required to solve for the unknowns are:

Conservation of momentum and the coefficient of restitution equation are appliedalong the line of impact (x-axis):

mA(vAx)1 + mB(vBx)1 = mA(vAx)2 + mB(vBx)2

e = [(vBx)2 – (vAx)2]/[(vAx)1 – (vBx)1]

Momentum of each particle is conserved in the directionperpendicularto the line of impact (y-axis):

mA(vAy)1 = mA(vAy)2 and mB(vBy)1 = mB(vBy)2

slide21

Procedure of analysis

•In most impact problems, the initial velocities of the particles and the coefficient of restitution, e, are known, with the final velocities to be determined.

•Define the x-y axes. Typically, thex-axisis definedalongthe line of impact and they-axisis in the plane of contactperpendicularto the x-axis.

•For bothcentral and obliqueimpact problems, the following equations applyalongthe line of impact (x-dir.):

 m(vx)1 =  m(vx)2 and e = [(vBx)2 – (vAx)2]/[(vAx)1 – (vBx)1]

•Forobliqueimpact problems, the following equations are also required, appliedperpendicular to the line of impact (y-dir.):

mA(vAy)1 = mA(vAy)2 and mB(vBy)1 = mB(vBy)2

slide22

Example 1

Given:A 0.5 kg ball is ejected from the tube at A with a horizontal velocity vA = 2 m/s. The coefficient of restitution at B is e = 0.6.

Find:The horizontal distance R where the ball strikes the smooth inclined plane and the speed at which it bounces from the plane.

Plan:1) Usekinematicsto find the distance R (projectile motion).

2) The collision at B is anoblique impact, with the line of impact perpendicular to the plane.

3) Thus, the coefficient of restitution applies perpendicular to the incline and the momentum of the ball is conserved along the incline.

slide23

vx = vxo = 2 m/s ( )

vy = vyo – gt = 0 – 9.81(1.028) = -10.0886 m/s ( )

=> v = 10.285 m/s 78.8°

Example 1 (continued)

Solution:

1) Apply the equations of projectile motion to determine R. Place the origin at A (xo = yo = 0) with the initial velocity of vyo = 0, vxo = vA = 2 m/s:

x = xo + vxot => R = 0 + 2t

y = yo + vyot – 0.5gt2 => -(4 + R tan30°) = 0 + 0 – 0.5(9.81)t2

Solving these equations simultaneously yields

t = 1.028 s and R = 2.06 m

It is also necessary to calculate the velocity of the ball just before impact:

slide24

vA1

x

y

48.8°

vA2

30°

The speed is the magnitude of the velocity vector:

vA2 = ((vAx)2)2 + ((vAy)2)2 = 8.21 m/s

Example 1 (continued)

2) Solve the impact problem by using x-y axes defined along and perpendicular to the line of impact, respectively:

Denoting the ball as A and plane as B, the momentum of the ball is conserved in the y-dir:

mA(-vAy)1 = mA(-vAy)2

(vAy)2 = (vAy)1 = vA cos48.8° = 6.77 m/s

The coefficient of restitution applies in the x-dir and (vBx)1 = (vBx)2 = 0:

e = [(vBx)2 – (vAx)2]/[(vAx)1 – (vBx)1]

=> 0.6 = [0 - (vAx)2]/[-10.285 sin48.8 – 0]

=> (vAx)2 = 4.64 m/s

slide25

Example 2

Given:A 2 kg block A is released from rest, falls a distance h = 0.5 m, and strikes plate B (3 kg mass). The coefficient of restitution between A and B is e = 0.6, and the spring stiffness is k = 30 N/m.

Find:The velocity of block A just after the collision.

Plan:1) Determine the speed of the block just before the collision using projectile motion or an energy method.

2) Analyze the collision as a central impact problem.

slide26

v2 = 3.132 m/s

Example 2 (continued)

Solution:

1)Determine the speed of block A just before impact by using conservation of energy. Defining the gravitational datum at the initial position of the block (h1 = 0) and noting the block is released from rest (v1 = 0):

T1 + V1 = T2 + V2

0.5m(v1)2 + mgh1 = 0.5m(v2)2 + mgh2

0 + 0 = 0.5(2)(v2)2 + (2)(9.81)(-0.5)

This is the speed of the block just before the collision. Plate (B) is at rest, velocity of zero, before the collision.

slide27

(vA)2

(vA)1 = 3.132 m/s

A

B

+ mA(vA)1 + mB(vB)1 = mA(vA)2 + mB(vB)2

(2)(-3.132) + 0 = (2)(vA)2 + (3)(vB)2

(vB)2

(vB)1 = 0

Using the coefficient of restitution:

+ e = [(vB)2 – (vA)2]/[(vA)1 – (vB)1]

=> 0.6 = [(vB)2 – (vA)2]/[-3.132 – 0] => -1.879 = (vB)2 – (vA)2

Solving the two equations simultaneously yields

(vA)2 = -0.125 m/s , (vB)2 = -2.00 m/s

Both the block and plate will travel down after the collision.

Example 2 (continued)

2) Analyze the collision as a central impact problem.

Apply conservation of momentum to the system in the vertical direction: