10/12/2015 1 3. Kinetics of Particles 3.1 Force, Mass and Acceleration 3.2 Work and Energy 3.3 Impulse and Momentum 3.4 Impact 1 3.1 Force, Mass and Acceleration • We draw two important conclusions from the results of the experiments. First, the ratios of applied force to corresponding acceleration all equal the same number, provided the units used for measurement are not changed in the experiments. Thus, 1 1 = 2 2 =⋯= = , constant • We conclude that the constant C is a measure of some invariable property of the particle. This property is the inertia of the particle, which is its resistance to rate of change of velocity. 2
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
10/12/2015
1
3. Kinetics of Particles
3.1 Force, Mass and Acceleration
3.2 Work and Energy
3.3 Impulse and Momentum
3.4 Impact
1
3.1 Force, Mass and Acceleration
• We draw two important conclusions from the results of the experiments. First, the ratios of applied force to corresponding acceleration all equal the same number, provided the units used for measurement are not changed in the experiments. Thus,
𝐹1𝑎1
=𝐹2𝑎2
= ⋯ =𝐹
𝑎= 𝐶, 𝑎 constant
• We conclude that the constant C is a measure of some invariable property of the particle. This property is the inertia of the particle, which is its resistance to rate of change of velocity.
2
10/12/2015
2
3.1 Force, Mass and Acceleration (contd.)
• When a particle of mass m is subjected to the action of forces
𝐅1, 𝐅2, 𝐅3, . . . whose vector sum is ∑𝐅, and equation is,
∑𝐅 = 𝑚𝐚
Rectilinear Motion
• If we choose the x-direction, for example, as the direction of the rectilinear motion of a particle of mass m, the acceleration in the y- and z-directions will be zero and the scalar components of equation become,
∑𝐹𝑥 = 𝑚𝑎𝑥
∑𝐹𝑦 = 0
∑𝐹𝑧 = 0
3
3.1 Force, Mass and Acceleration Sample Problem (1)
A 75-kg man stands on a spring scale in an elevator. During the first 3 seconds of motion from rest, the tension T in the pulling cable is 8300 N. Find the reading R of the scale in newtons during this interval and the upward velocity v of the elevator at the end of the 3 seconds. The total mass of the elevator, man, and scale is 750 kg.
4
10/12/2015
3
Solution
5
3.1 Force, Mass and Acceleration Sample Problem (2)
6
The 125-kg concrete block A is released from rest in the position shown and pulls the 200-kg log up the 30° ramp. If the coefficient of kinetic friction between the log and the ramp is 0.5, determine the velocity of the block as it hits the ground at B.
10/12/2015
4
Solution
7
3.1 Force, Mass and Acceleration Sample Problem (3)
8
The sliders A and B are connected by a light rigid bar of length 𝑙 = 0.5 m and move with negligible friction in the slots, both of which lie in a horizontal plane. For the position where 𝑥𝐴 = 0.4 m, the velocity of A is 𝑣𝐴 = 0.9 m/s to the right. Determine the acceleration of each slider and the force in the bar at this instant.
10/12/2015
5
Solution
9
3.1 Force, Mass and Acceleration Curvilinear Motion
10
• We now rewrite the equation in three ways, the choice of which depends on which coordinate system is most appropriate.
∑𝐹𝑥 = 𝑚𝑎𝑥
∑𝐹𝑦 = 𝑚𝑎𝑦
𝑎𝑥 = 𝑥 and 𝑎𝑦 = 𝑦
10/12/2015
6
3.1 Force, Mass and Acceleration Normal and Tangential Coordinates
11
∑𝐹𝑛 = 𝑚𝑎𝑛
∑𝐹𝑡 = 𝑚𝑎𝑡
𝑎𝑛 = 𝜌𝛽 2 =𝑣2
𝜌= 𝑣𝛽
𝑎𝑡 = 𝑣
𝑣 = 𝜌𝛽
3.1 Force, Mass and Acceleration Sample Problem (4)
12
Determine the maximum speed v which the sliding block may have as it passes point A without losing contact with the surface.
10/12/2015
7
Solution
13
3.1 Force, Mass and Acceleration Sample Problem (5)
14
Small objects are released from rest at A and slide down the smooth circular surface of radius R to a conveyor B. Determine the expression for the normal contact force N between the guide and each object in terms of 𝜃 and specify the correct angular velocity 𝜔 of the conveyor pulley of radius r to prevent any sliding on the belt as the objects transfer to the conveyor.
10/12/2015
8
Solution
15
3.1 Force, Mass and Acceleration Polar Coordinates
16
∑𝐹𝑟 = 𝑚𝑎𝑟
∑𝐹𝜃 = 𝑚𝑎𝜃
𝑎𝑟 = 𝑟 − 𝑟𝜃 2
𝑎𝜃 = 𝑟𝜃 + 2𝑟 𝜃
10/12/2015
9
3.1 Force, Mass and Acceleration Sample Problem (6)
17
The small 180 g slider A moves without appreciable friction in the hollow tube, which rotates in a horizontal plane with a constant angular speed Ω = 7 rad/s . The slider is launched with an initial speed 𝑟0 = 20 m/s relative to the tube at the inertial coordinates 𝑥 = 150 mm and 𝑦 = 0 . Determine the magnitude P of the horizontal force exerted on the slider by the tube just before the slider exits the tube.
Solution
18
10/12/2015
10
3.2 Work and Energy Definition of Work
The work done by the force F during the displacement dr is defined as
𝑑𝑈 = 𝐅. 𝑑𝐫
𝑑𝑈 = 𝐹 𝑑𝑠 𝑐𝑜𝑠𝛼 • 𝛼: the angle between 𝐅 and 𝑑𝐫 • 𝑑𝑠: the magnitude of the 𝑑𝐫
19
3.2 Work and Energy Definition of Work (contd.)
• Work is positive if the working component 𝐹𝑡 is in the direction of the displacement and negative if it is in the opposite direction.
• The SI units of work are those of force (N) times displacement (m) or N.m. This unit is given the special name joule (J), which is defined as the work done by a force of 1 N acting through a distance of 1 m in the direction of the force.
20
10/12/2015
11
3.2 Work and Energy Calculation of Work
• During a finite movement of the point of application of a force does an amount of work equal to
𝑈 = 𝐹. 𝑑𝑟 = (𝐹𝑥𝑑𝑥 + 𝐹𝑦𝑑𝑦 + 𝐹𝑧𝑑𝑧)
𝑈 = 𝐹𝑡𝑑𝑠
21
3.2 Work and Energy Calculation of Work (contd.)
• We now consider the work done on a particle of mass 𝑚, moving along a curved path under the action of the force 𝐅, which stands for the resultant ∑𝐅 of all forces acting on the particle. The position of 𝑚 is specified by the position vector 𝐫, and its displacement along its path during the time 𝑑𝑡 is represented by the change 𝑑𝐫 in its position vector.
22
10/12/2015
12
3.2 Work and Energy Kinetic Energy (contd.)
𝑈1−2 = 𝐅. 𝑑𝐫 = 𝐹𝑡𝑑𝑠𝑠2
𝑠1
2
1
𝑈1−2 = 𝐅. 𝑑𝐫 = 𝑚𝐚. 𝑑𝐫2
1
2
1
But 𝐚. 𝑑𝐫 = 𝑎𝑡𝑑𝑠, where 𝑎𝑡 is the tangential component of the acceleration of 𝑚, where 𝑎𝑡𝑑𝑠 = 𝑣𝑑𝑣. Thus, the espression for the work of 𝐅 becomes
𝑈1−2 = 𝐅. 𝑑𝐫 = 𝑚𝑣 𝑑𝑣 =1
2𝑚(𝑣2
2 − 𝑣12)
𝑣2
𝑣1
2
1
23
3.2 Work and Energy Kinetic Energy
• The kinetic energy 𝑇 of the particle is defined as
𝑇 =1
2𝑚𝑣2
and is the total work which must be done on the particle to
bring it from a state of rest to a velocity 𝑣.
24
10/12/2015
13
3.2 Work and Energy Power
• The capacity of a machine is measured by the time rate at which it can do work or deliver energy. The total work or energy output is not a measure of this capacity since a motor, no matter how small, can deliver a large amount of energy if given sufficient time.
• Accordingly, the power 𝑃 developed by a force 𝐅 which does an amount of work 𝑈 is 𝑃 = 𝑑𝑈/𝑑𝑡 = 𝐅. 𝑑𝐫/𝑑𝑡. Because 𝑑𝐫/𝑑𝑡 is the velocity 𝐯 of the point of application of the force, we have
𝑃 = 𝐅. 𝐯
25
3.2 Work and Energy Power (contd.)
• Power is clearly a scalar quantity, and in SI it has the units of N.m/s = J/s. The special unit for power is the watt (W), which equals one joule per second (J/s). In U.S. customary units, the unit for mechanical power is the horsepower (hp). These units and their numerical equivalences are
1 W = 1 J/s
1 hp = 500 ft_lb/sec = 33,000ft_lb/min
1 hp = 746 W = 0.746 kW
26
10/12/2015
14
3.2 Work and Energy Efficiency
• The ratio of the work done by a machine to the work done on
the machine during the same time interval is called the
mechanical efficiency 𝑒𝑚 of the machine.
• Efficiency is always less than unity since every device operates with some loss of energy and since energy cannot be created within the machine.
𝑒𝑚 =𝑃𝑜𝑢𝑡𝑝𝑢𝑡𝑃𝑖𝑛𝑝𝑢𝑡
27
3.2 Work and Energy Potential Energy
Gravitational Potential Energy
• The gravitational potential energy 𝑉𝑔 of the particle is defined
as the work 𝑚𝑔ℎ done against the gravitational field to elevate the particle a distance ℎ above some arbitrary reference plane (called a datum), where 𝑉𝑔 is taken to be zero.
Thus, we write the potential energy as
𝑉𝑔 = 𝑚𝑔ℎ
28
10/12/2015
15
3.2 Work and Energy Potential Energy (contd.)
Gravitational Potential Energy (contd.)
• This work is called potential energy because it may be converted into energy if the particle is allowed to do work on a supporting body while it returns to its lower original datum plane. In going from one level at ℎ = ℎ1 to a higher level at ℎ = ℎ2, the change in potential energy becomes
∆𝑉𝑔 = 𝑚𝑔 ℎ2 − ℎ1 = 𝑚𝑔∆ℎ
• The corresponding work done by the gravitational force on the particle is −𝑚𝑔∆ℎ. Thus, the work done by the gravitational force is the negative of the change in potential energy.
29
3.2 Work and Energy Potential Energy (contd.)
Elastic Potential Energy
• The work which is done on the spring to deform it is stored in the spring and is called its elastic potential energy 𝑉𝑒.
• The force supported by the spring at any deformation 𝑥, tensile or compressive, from its undeformed position is 𝐹 = 𝑘𝑥. Thus, we define the elastic potential energy of the spring as the work done on it to deform it an amount 𝑥, and we have
𝑉𝑒 = 𝑘𝑥 𝑑𝑥 =1
2𝑘𝑥2
𝑥
0
30
10/12/2015
16
3.2 Work and Energy Potential Energy (contd.)
Elastic Potential Energy
• If the deformation, either tensile or compressive, of a spring increases from 𝑥1 to 𝑥2 during the motion, then the change in potential energy of the spring is its final value minus its initial value or
∆𝑉𝑒 =1
2𝑘(𝑥2
2 − 𝑥12)
31
3.2 Work and Energy Work-Energy Equation
𝑈1_2′ = ∆𝑇 + ∆𝑉𝑔 + ∆𝑉𝑒
Note that equation may be rewritten in the equivalent form
𝑇1 + 𝑉𝑔1 + 𝑉𝑒1 + 𝑈1−2 ′ = 𝑇2 + 𝑉𝑔2 + 𝑉𝑒2
We may rewrite the alternative work-energy relation, for a
particle-and-spring system as
𝑈1_2′ = ∆ 𝑇 + 𝑉𝑔 + 𝑉𝑒 = ∆𝐸
Where 𝐸 = 𝑇 + 𝑉𝑔 + 𝑉𝑒 is the total mechanical energy of the
particle and its attached linear spring.
32
10/12/2015
17
3.2 Work and Energy Work-Energy Equation (contd.)
• For problems where the only forces are garvitational, elastic, and nonworking constraint forces, the 𝑈′-termis zero, and the energy equation becomes merely
∆𝐸 = 0 or 𝐸 = constant
Where 𝐸 is constant, we see that transfers of energy between kinetic and potential may take place as long as the total mechnaical energy 𝑇 + 𝑉𝑔 + 𝑉𝑒 does not change.
• ∆𝐸 = 0 or 𝐸 = constant equation express the law of conservation of dynamical energy.
33
3.2 Work and Energy Sample Problem (7)
The 10-kg slider moves with negligible friction up the inclined guide. The attached spring has a stiffness of 60 N/m and is stretched 0.6 m in position A, where the slider is released from rest. The 250-N force is constant and the pulley offers negligible resistance to the motion of the cord. Calculate the velocity 𝑣 of the slider as it passes point C.
34
10/12/2015
18
Solution
35
3.2 Work and Energy Sample Problem (8)
The 3-kg slider is released from rest at position 1 and slides with negligible friction in a vertical plane along the circular rod. The attached spring has a stiffness of 350 N/m and has an unstretched length of 0.6 m. Determine the velocity of the slider as it passes position 2.
36
10/12/2015
19
Solution
37
3.2 Work and Energy Sample Problem (9)
The mechanism is released from rest with 𝜃 = 180° where the uncompressed spring of stiffness 𝑘 = 900 N/m is just touching the underside of the 4-kg collar. Determine the angle 𝜃 corresponding to the maximum compression of the spring. Motion is in the vertical plane, and the mass of the links may be neglected.
38
10/12/2015
20
Solution
39
3.3 Impulse and Momentum Linear Impulse and Linear Momentum
∑𝐅 = 𝑚𝐯 =𝑑
𝑑𝑡𝑚𝐯 or ∑𝐅 = 𝐆
where the product of the mass and velocity is defined as the linear momentum 𝐆 = 𝑚𝐯 of the particle. The above equation states that the resultant of all forces acting on a particle equals its time rate of change of linear momentum.
40
10/12/2015
21
3.3 Impulse and Momentum Linear Impulse and Linear Momentum (contd.)
• We now write three scalar components of the equation ∑𝐅 = 𝐆 as
∑F𝑥 = G 𝑥
∑F𝑦 = G 𝑦
∑F𝑧 = G 𝑧
• These equations may be applied independently of one another.
41
3.3 Impulse and Momentum The Linear Impulse-Momentum Principle
• Multiplying the equation ∑𝐅 = 𝐆 by 𝑑𝑡 gives ∑𝐅 𝑑𝑡 = 𝑑𝐆, which we integrate from time 𝑡1to time 𝑡2 to obtain
∑𝐅𝑡2
𝑡1
𝑑𝑡 = 𝐆𝟐 − 𝐆𝟏 = ∆𝐆
• Here the linear momentum at time 𝑡2 is 𝐆𝟐 = 𝑚𝐯𝟐 and the linear momentum at time 𝑡1 is 𝐆𝟏 = 𝑚𝐯𝟏. The product of force and time is defined as the linear impulse of the force, and the above equation states that the total linear impulse on 𝑚 equals the corresponding change in linear momentum of 𝑚.
42
10/12/2015
22
3.3 Impulse and Momentum The Linear Impulse-Momentum Principle (contd.)
• The components of ∑𝐅𝑡2𝑡1
𝑑𝑡 = 𝐆𝟐 − 𝐆𝟏 = ∆𝐆 equation
become the scalar equations
∑F𝑥
𝑡2
𝑡1
𝑑𝑡 = 𝑚𝑣𝑥 2 − 𝑚𝑣𝑥 1
∑F𝑦
𝑡2
𝑡1
𝑑𝑡 = 𝑚𝑣𝑦 2− 𝑚𝑣𝑦 1
∑F𝑧
𝑡2
𝑡1
𝑑𝑡 = 𝑚𝑣𝑧 2 − 𝑚𝑣𝑧 1
43
3.3 Impulse and Momentum Conservation of Linear Momentum
• If the resultant force on a particle is zero during an interval of time, we see that equation ∑𝐅 = 𝐆 requires that its linear momentum 𝐆 remain constant. In this case, the linear momentum of the particle is said to be conserved.
• If there are two or more particles that interact during an interval of time, the momentum of these particles due to interactive forces cancel out each other. Therefore the total linear momentum of the ‘system of particles’ is also conserved if there is no external force acting on the system.
44
10/12/2015
23
3.3 Impulse and Momentum Sample Problem (10)
The loaded 150-kg skip is rolling down the incline at 4 m/s when a force P is applied to the cable as shown at time t 0. The force P is increased uniformly with the time until it reaches 600 N at t 4 s, after which time it remains constant at this value. Calculate
(a) the time at which the skip reverses its direction and
(b) the velocity v of the skip at t 8 s. Treat the skip as a particle.
45
Solution
46
10/12/2015
24
3.3 Impulse and Momentum Sample Problem (11)
The 50-g bullet traveling at 600 m/s strikes the 4-kg block centrally and is embedded within it. If the block slides on a smooth horizontal plane with a velocity of 12 m/s in the direction shown prior to impact, determine the velocity v of the block and embedded bullet immediately after impact.
47
Solution
48
10/12/2015
25
3.3 Impulse and Momentum Sample Problem (12)
Car B weighing 1500 kg and traveling west at 48 km/h collides with car A weighing 1600 kg and traveling north at 32 km/h as shown. If the two cars become entangled and move together as a unit after the crash, compute the magnitude v of their common velocity immediately after the impact and the angle 𝜃 made by the velocity vector with the north direction.
49
Solution
50
10/12/2015
26
3.3 Impulse and Momentum Angular Impulse and Angular Momentum
• The moment of the linear momentum vector 𝑚𝐯 about the origin 𝑂 is defined as the angular momentum 𝐇𝑂 of 𝑃 about 𝑂 and is given by the cross-product relation for the moment of a vector
𝐇𝑂 = 𝐫 ×𝑚𝐯
• The angular momentum then is a vector perpendicular to the plane 𝐴 defined by 𝐫 and 𝐯.
51
3.3 Impulse and Momentum Rate of Change of Angular Momentum
• The moment about origin O is the vector cross product
∑𝐌𝑂 = 𝐫 × ∑𝐅 = 𝐫 ×𝑚𝐯
𝐇 𝑂 = 𝐫 × 𝑚𝐯 + 𝐫 ×𝑚𝐯 = 𝐯 ×𝑚𝐯+ 𝐫 ×𝑚𝐯
• The term 𝐯 × 𝑚𝐯 is zero since the cross product of parallel vectors is identically zero.
∑𝐌𝑂 = 𝐇 𝑂
• This equation states that the moment about the fixed point O of all forces acting on m equals the time rate change of angular momentum of m about O.
52
10/12/2015
27
3.3 Impulse and Momentum Rate of Change of Angular Momentum (contd.)
• Equation ∑𝐌𝑂 = 𝐇 𝑂 is a vector equation with scalar components
∑M𝑂𝑥 = H 𝑂𝑥
∑M𝑂𝑦 = H 𝑂𝑦
∑M𝑂𝑧 = H 𝑂𝑧
53
3.3 Impulse and Momentum The Angular Impulse-Momentum Principle
∑𝐌O
𝑡2
𝑡1
𝑑𝑡 = 𝐇𝑂2 −𝐇𝑂1 = ∆𝐇𝑂
Where 𝐇𝑂2 = 𝐫𝟐 ×𝑚𝐯𝟐 and 𝐇𝑂1 = 𝐫𝟏 ×𝑚𝐯𝟏 . The product of moment and time is defined as angular impulse, and the above equation states that the total angular impulse on m about the fixed point O equals the corresponding change in angular momentum of m about O.
54
10/12/2015
28
3.3 Impulse and Momentum Plane-Motion Applications
• For a particle of mass 𝑚 moving along a curved path in the 𝑥 − 𝑦 plane, the angular momenta about 𝑂 at points 1 and 2 have the magnitudes 𝐻𝑂1 = 𝐫𝟏 ×𝑚𝐯𝟏 = 𝑚𝑣1𝑑1 and
𝐻𝑂2 = 𝐫𝟐 ×𝑚𝐯𝟐 = 𝑚𝑣2𝑑2, respectively. In the illustration
both 𝐻𝑂1 and 𝐻𝑂2 are represented in the counter clockwise
sense in accord with the direction of the moment of the linear momentum.
55
3.3 Impulse and Momentum Conservation of Angular Momentum
• If the resultant moment about a fixed point O of all forces acting on a particle is zero during an interval of time, ∑𝐌𝑂 = 𝐇 𝑂 requires that its angular momentum 𝐇𝑂 about that point remain constant. In this case, the angular momentum of the particle is said to be conserved.
∆𝐇𝑂 = 𝟎 or 𝐇𝑂1 = 𝐇𝑂2
The above equations express the principle of conservation of angular momentum.
56
10/12/2015
29
• Consider now the motion of two particles a and b which interact during an interval of time. If the interactive forces 𝐅 and −𝐅 between them are the only unbalanced forces acting on the particles during the interval, it follows that the moments of the equal and opposite forces about any fixed point O not on their line of action are equal and opposite.
• Therefore the total angular momentum of the ‘system of particles’ is also conserved if there is no external force acting on the system.
57
3.3 Impulse and Momentum Conservation of Angular Momentum
3.3 Impulse and Momentum Sample Problem (13)
A small mass particle is given an initial velocity 𝐯0 tangent to the horizontal rim of a smooth hemispherical bowl at a radius 𝑟0 from the vertical centerline, as shown at point 𝐴. As the particle slides past point 𝐵, a distance ℎ below 𝐴 and a distance 𝑟 from the vertical centerline, its velocity 𝐯 makes an angle 𝜃 with the horizontal tangent to the bowl through B. Determine 𝜃.
58
10/12/2015
30
Solution
59
3.3 Impulse and Momentum Sample Problem (14)
The two spheres of equal mass 𝑚 are able to slide along the horizontal rotating rod. If they are initially latched in position a distance 𝑟 from the rotating axis with the assembly rotating freely with an angular velocity 𝜔0, determine the new angular velocity 𝜔 after the spheres are released and finally assume positions at the ends of the rod at a radial distance of 2𝑟. Also find the fraction 𝑛 of the initial kinetic energy of the system which is lost. Neglect the small mass of the rod and shaft.
60
10/12/2015
31
Solution
61
3.4 Impact
• Impact refers to the collision between two bodies and is characterized by the generation of relatively large contact forces which act over a very short interval of time.
Direct Central Impact
• The collinear motion of two spheres of masses 𝑚1 and 𝑚2 traveling with velocities 𝑣1 and 𝑣2. If 𝑣1 is greater than 𝑣2, collision occurs with the contact forces directed along the line of centers.
62
10/12/2015
32
3.4 Impact
Direct Central Impact (contd.)
• Because the contact forces are equal and opposite during
impact, the linear momentum of the system remains unchanged. Thus, we apply the law of conservation of linear momentum and write
𝑚1𝑣1 +𝑚2𝑣2 = 𝑚1𝑣1
′ +𝑚2𝑣2′
63
3.4 Impact Coefficient of Restitution
• 𝑒 is the magnitude of the restoration impulse to the magnitude of the deformation impulse. This ratio is called the coefficient of restitution.
64
10/12/2015
33
3.4 Impact Coefficient of Restitution (contd.)
𝑒 = 𝐹𝑟𝑑𝑡𝑡
𝑡0
𝐹𝑑𝑑𝑡𝑡00
𝑒 =𝑣2′ − 𝑣1
′
𝑣1 − 𝑣2=
relative velocity of seperation
relative velocity of approach
• If the two initial velocities 𝑣1 and 𝑣2 and the coefficient of restitution 𝑒 are known, then equations 𝑚1𝑣1 +𝑚2𝑣2 =
𝑚1𝑣1′ +𝑚2𝑣2
′ and 𝑒 =𝑣2′−𝑣1
′
𝑣1−𝑣2 give us two equations in the two
unknown final velocities 𝑣1′ and 𝑣2
′ .
65
3.4 Impact Energy Loss During Impact
• According to this classical theory of impact, the value 𝑒 =1 means that the capacity of the two particles to recover equals their tendency to deform. This condition is one of elastic impact with no energy loss. The value 𝑒 = 0 , on the other hand, describes inelastic or plastic impact where the particles cling together after collision and the loss of energy is a maximum. All impact conditions lie somewhere between these two extremes.
66
10/12/2015
34
3.4 Impact Sample Problem (15)
The ram of a pile driver has a mass of 800 kg and is released from rest 2 m above the top of the 2400-kg pile. If the ram rebounds to a height of 0.1 m after impact with the pile, calculate
(a) the velocity 𝑣𝑝′ of the pile immediately after impact,
(b) the coefficient of restitution e, and
(c) the percentage loss of energy due to the impact.
67
Solution
68
10/12/2015
35
3.4 Impact Oblique Central Impact
(1) Momentum of the system is conserved in the𝑛-direction. This gives
𝑚1 𝑣1 𝑛 +𝑚2 𝑣2 𝑛 = 𝑚1 𝑣1
′𝑛 +𝑚2 𝑣2
′𝑛
69
3.4 Impact Oblique Central Impact (contd.)
(2) and (3) The momentum for each particle is conserved in the 𝑡 -direction since there is no impulse on either particle in the 𝑡 -direction. Thus,
𝑚1 𝑣1 𝑡 = 𝑚1 𝑣1′𝑡
𝑚2 𝑣2 𝑡 = 𝑚2 𝑣2
′𝑡
(4) The coefficient of restitution, as in the case of direct central impact, is the positive ratio of the recovery impulse to the
deformation impulse. 𝑒 =𝑣2′−𝑣1
′
𝑣1−𝑣2 applies, then, to the velocity
components in the 𝑛 -direction.
𝑒 =𝑣2′𝑛 − 𝑣1
′𝑛
𝑣1 𝑛 − 𝑣2 𝑛
70
10/12/2015
36
3.4 Impact Sample Problem (16)
A ball is projected onto the heavy plate with a velocity of 16 m/s at the 30° angle shown. If the effective coefficient of restitution is 0.5, compute the rebound velocity 𝑣′ and its angle 𝜃′.
71
Solution
72
Related Documents
