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11 - 1 Impulse and Momentum Some more extremely useful relations are the concepts known as impulse and momentum. Conservation of momentum underlies several.

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Presentation on theme: "11 - 1 Impulse and Momentum Some more extremely useful relations are the concepts known as impulse and momentum. Conservation of momentum underlies several."— Presentation transcript:

1 11 - 1 Impulse and Momentum Some more extremely useful relations are the concepts known as impulse and momentum. Conservation of momentum underlies several important physical laws in a variety of scientific disciplines from quantum electrodynamics to fluid mechanics. A derivation for this fundamental quantity is shown below: We now define a quantity known as momentum, as, and the change in momentum over a discrete time interval is defined as the impulse, : Note: In comparison with potential and kinetic energy, which are scalar quantities, impulse and momentum are vector quantities and must be added vectorially.

2 11 - 2 Starting from a generalized expression for momentum We can take the derivative of both sides of the equation: And we arrive at another expression for Newton’s Second law. The force is equal to the rate of change of momentum. Thus, when the net force on an object is zero, then the momentum must be constant. This is the famous law of conservation of momentum. Conservation of Linear Momentum

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4 11 - 4 Initial conditions: Linear Momentum prior to impactLinear Momentum after impact Conservation of Linear Momentum After impact:

5 11 - 5 Given the following information: Leads to 2 equations and 2 unknowns. These can be solved simultaneously, leading to the following result:

6 11 - 6 Relation between Force and Impulse As shown in previous slides, the impulse can be written as: If the force is constant over a discrete time interval, then it can be taken out of the integral and the impulse is re-written as: This equation is very useful for modeling collisions, because the force of impact is not usually known (or measurable), but the initial and final velocities are known and/or measurable.

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9 11 - 9 Angular Momentum Like linear momentum, there is also an expression for angular momentum, which describes how much inertia an object carries in rotating about a coordinate center. While its derivation will be reserved for later in this course, the mathematical expression is given here as: Path of particle The cross product denotes that the angular momentum points in an entirely different direction from the instantaneous position vector,, and the instantaneous velocity vector. The direction can be found according to the right-hand rule. The O means that the angular momentum is taken with respect to the origin.

10 11 - 10 Conservation of Angular Momentum The rate of change of angular momentum can be derived as follows: (i.e. the cross product of a vector with itself is always zero (i.e. they point in the same direction) We define this new vector,, which is the rate of change of angular moment, and it is called the moment of force about a pivot. Note: If the net force (or moment in this case) is zero, then it leads directly to the law of conservation of angular momentum.

11 11 - 11 Momentum according to Isaac Asimov Isaac Asimov wrote in "Understanding Physics": “This tendency for motion (or for rest) to maintain itself steadily unless made to do otherwise by some interfering force can be viewed as a kind of "laziness," a kind of unwillingness to make a change. And indeed, Newton's first law of motion is referred to as the principle of inertia, from a Latin word meaning "idleness" or "laziness." He added a footnote: "In Aristotle's time the earth was considered a motionless body fixed at the center of the universe; the notion of 'rest' therefore had a literal meaning. What we ordinarily consider 'rest' nowadays is a state of being motionless with respect to the surface of the earth. But we know (and Newton did, too) that the earth itself is in motion about the sun and about its own axis. A body resting on the surface of the earth is therefore not really in a state of rest at all."

12 11 - 12 Central Forces In many forces, such as gravitational, electrical forces, etc., the magnitude of the force only depends on the distance away from the source (i.e. ), and the direction is always towards or away from the source. Using these assumptions, we can derive an expression for the rotation frequency:

13 11 - 13 Using this result, we can arrive at an expression for radial force written purely in terms of the radial distance Now the question becomes, what function f(r) satisfies the solution to this differential form? Or alternatively, what is the solution to this equation:

14 11 - 14 For example, consider the dot product of the velocity vector with itself. What is the time derivative of this quantity? The trick is to write it in differential form:

15 11 - 15 If you now multiply this function by ½ m, you will find that: The result is that the equation for radial motion is now reduced to: The next step is to write f(r) in a form that allows for solution of this differential equation.

16 11 - 16 Consider for the moment that the function f(r) is actually derivative of another function. After some manipulation, this formula becomes: So, now we can re-write the equation on the previous slide as:

17 11 - 17 This equation is easily solvable. It is simply a constant. But we also know that the expression for the square of velocity. This equation then becomes upon multiplication by ½ m…

18 11 - 18 This is the kinetic energy stored in the linear velocity This is the kinetic energy stored in the angular momentum This is the potential energy term When we make the substitution… The above equation becomes: Linear momentum Energy term Angular momentum Energy term

19 11 - 19 Problem 1 A 25- g steel-jacket bullet is fired horizontally with a velocity of 600 m/s and ricochets off a steel plate along the path CD with a velocity of 400 m/s. Knowing that the bullet leaves a 10- mm scratch on the plate and assuming that its average speed is 500 m/s while it is in contact with the plate, determine the magnitude and direction of the average impulsive force exerted by the bullet on the plate. 10 mm 15 o 20 o B C D A

20 11 - 20 Problem 1 10 mm 15 o 20 o B C D A A 25- g steel-jacket bullet is fired horizontally with a velocity of 600 m/s and ricochets off a steel plate along the path CD with a velocity of 400 m/s. Knowing that the bullet leaves a 10- mm scratch on the plate and assuming that its average speed is 500 m/s while it is in contact with the plate, determine the magnitude and direction of the average impulsive force exerted by the bullet on the plate. 1. Draw a momentum impulse diagram: The diagram shows the particle, its momentum at t 1 and at t 2, and the impulses of the forces exerted on the particle during the time interval t 1 to t 2.

21 11 - 21 Problem 1 10 mm 15 o 20 o B C D A A 25- g steel-jacket bullet is fired horizontally with a velocity of 600 m/s and ricochets off a steel plate along the path CD with a velocity of 400 m/s. Knowing that the bullet leaves a 10- mm scratch on the plate and assuming that its average speed is 500 m/s while it is in contact with the plate, determine the magnitude and direction of the average impulsive force exerted by the bullet on the plate. 2. Apply the principle of impulse and momentum: The final momentum mv 2 of the particle is obtained by adding its initial momentum mv 1 and the impulse of the forces F acting on the particle during the time interval considered. mv 1 +  F  t = mv 2  F is sum of the impulsive forces (the forces that are large enough to produce a definite change in momentum).

22 11 - 22 Problem 1 Solution 10 mm 15 o 20 o B C D A Draw a momentum impulse diagram. += x y F x  t F y  t m v1m v1 x y 15 o x y m v2m v2 20 o Since the bullet leaves a 10-mm scratch and its average speed is 500 m/s, the time of contact  t is:  t = (0.010 m) / (500 m/s) = 2 x 10 -5 s

23 11 - 23 Problem 1 Solution += x y F x  t F y  t m v1m v1 x y 15 o x y m v2m v2 20 o Apply the principle of impulse and momentum. mv 1 +  F  t = mv 2 (0.025 kg)(600 m/s)cos15 o +F x  2 x 10 -5 s   = (0.025 kg)(400 m/s)cos20 o F x = - 254.6 kN -(0.025 kg)(600 m/s)sin15 o +F y  2 x 10 -5 s  =(0.025 kg)(400 m/s) sin20 o F y = 365.1 kN + x components: + y components:

24 11 - 24 Problem 1 Solution += x y F x  t F y  t m v1m v1 x y 15 o x y m v2m v2 20 o F x = - 254.6 kN, F y = 365.1 kN F = ( -254.6 kN ) 2 + ( 365.12 kN ) 2 = 445 kN F = 445 kN 40.1 o

25 11 - 25 140 kg 650 kg 1.2 m Problem 2 The 650- kg hammer of a drop-hammer pile driver falls from a height of 1.2 m onto the top of a 140- kg pile, driving it 110 mm into the ground. Assuming perfectly plastic impact ( e = 0 ), determine the average resistance of the ground to penetration.

26 11 - 26 Problem 2 140 kg 650 kg 1.2 m The 650- kg hammer of a drop-hammer pile driver falls from a height of 1.2 m onto the top of a 140- kg pile, driving it 110 mm into the ground. Assuming perfectly plastic impact ( e = 0 ), determine the average resistance of the ground to penetration. 1. Apply conservation of energy principle : When a particle moves under the action of a conservative force, the sum of the kinetic and potential energies of the particle remains constant. T 1 + V 1 = T 2 + V 2 where 1 and 2 are two positions of the particle. 1a. Kinetic energy: The kinetic energy at each point on the path is given by: T = m v 2 1 2

27 11 - 27 Problem 2 140 kg 650 kg 1.2 m 1b. Potential energy: The potential energy of a weight W close to the surface of the earth at a height y above a given datum is given by: V g = W y The 650- kg hammer of a drop-hammer pile driver falls from a height of 1.2 m onto the top of a 140- kg pile, driving it 110 mm into the ground. Assuming perfectly plastic impact ( e = 0 ), determine the average resistance of the ground to penetration.

28 11 - 28 Problem 2 140 kg 650 kg 1.2 m 2. Apply conservation of momentum principle: During an impact of two bodies A and B, the total momentum of A and B is conserved if no impulsive external force is applied. m A v A + m B v B = m A v’ A + m B v’ B where v A and v B denote the velocities of the bodies before the impact and v’ A and v’ B denote their velocities after the impact. For perfectly plastic impact ( e = 0 ), v’ A = v’ B = v’ and m A v A + m B v B = (m A + m B ) v’ The 650- kg hammer of a drop-hammer pile driver falls from a height of 1.2 m onto the top of a 140- kg pile, driving it 110 mm into the ground. Assuming perfectly plastic impact ( e = 0 ), determine the average resistance of the ground to penetration.

29 11 - 29 Problem 2 140 kg 650 kg 1.2 m 3. Apply principle of work and energy: When a particle moves from position 1 to position 2 under the action of a force F, the work of the force F is equal to to the change in the kinetic energy of the particle. T 1 + U 1 2 = T 2 where T 1 = m v 1 2, T 2 = m v 2 2 and U 1 2 = F. dr 1 2 1 2 The 650- kg hammer of a drop-hammer pile driver falls from a height of 1.2 m onto the top of a 140- kg pile, driving it 110 mm into the ground. Assuming perfectly plastic impact ( e = 0 ), determine the average resistance of the ground to penetration.

30 11 - 30 Problem 2 Solution 650 kg 1.2 m y Apply conservation of energy principle. Motion of the hammer during the drop just before impact. Position 1 T 1 + V 1 = T 2 + V 2 0 + mg (1.2 m) = m v 2 2 + 0 v 2 2 = 2 ( 9.81 m/s 2 )(1.2 m) v 2 = 4.85 m/s 1 2 v 1 = 0650 kg 1.2 m Position 2 v2v2

31 11 - 31 Problem 2 Solution Impact process: Apply conservation of momentum principle. mH vHmH vH m P v P = 0 m H v’ m P v’ Before impact: After impact: m H v H + m P v P = ( m H + m P ) v’ (650 kg)( 4.85 m/s) + 0 = (650 kg + 140 kg) v’ v’ = 3.99 m/s

32 11 - 32 Problem 2 Solution v’ v = 0 W R 110 mm Apply principle of work and energy. Hammer and pile move against ground resistance. Position 1Position 2 Work T 1 + U 1 2 = T 2 ( m H + m P ) v’ 2 + (W H + W P - R) y = 0 (650 + 140) (3.99 m/s) 2 + [(650 + 140)(9.81) - R](0.110) = 0 R = 65,000 N 1 2 1 2

33 11 - 33 Problem 3 A small sphere B of mass m is attached to an inextensible cord of length 2a, which passes around the fixed peg A and is attached to a fixed support at O. The sphere is held close to the support at O and released with no initial velocity. It drops freely to point C, where the cord becomes taut, and swings in a vertical plane, first about A and then about O. Determine the vertical distance from line OD to the highest point C’’ that the sphere will reach. 45 o O B C C’ C’’ D A a

34 11 - 34 Problem 3 A small sphere B of mass m is attached to an inextensible cord of length 2a, which passes around the fixed peg A and is attached to a fixed support at O. The sphere is held close to the support at O and released with no initial velocity. It drops freely to point C, where the cord becomes taut, and swings in a vertical plane, first about A and then about O. Determine the vertical distance from line OD to the highest point C’’ that the sphere will reach. 1. Apply conservation of energy principle : When a particle moves under the action of a conservative force, the sum of the kinetic and potential energies of the particle remains constant. T 1 + V 1 = T 2 + V 2 where 1 and 2 are two positions of the particle. 45 o O B C C’ C’’ D A a

35 11 - 35 1a. Kinetic energy: The kinetic energy at each end of the path is given by: T = m v 2 Problem 3 A small sphere B of mass m is attached to an inextensible cord of length 2a, which passes around the fixed peg A and is attached to a fixed support at O. The sphere is held close to the support at O and released with no initial velocity. It drops freely to point C, where the cord becomes taut, and swings in a vertical plane, first about A and then about O. Determine the vertical distance from line OD to the highest point C’’ that the sphere will reach. 1 2 45 o O B C C’ C’’ D A a

36 11 - 36 Problem 3 A small sphere B of mass m is attached to an inextensible cord of length 2a, which passes around the fixed peg A and is attached to a fixed support at O. The sphere is held close to the support at O and released with no initial velocity. It drops freely to point C, where the cord becomes taut, and swings in a vertical plane, first about A and then about O. Determine the vertical distance from line OD to the highest point C’’ that the sphere will reach. 1b. Potential energy: The potential energy of a weight W close to the surface of the earth at a height y above a given datum is given by: V g = W y 45 o O B C C’ C’’ D A a

37 11 - 37 Problem 3 A small sphere B of mass m is attached to an inextensible cord of length 2a, which passes around the fixed peg A and is attached to a fixed support at O. The sphere is held close to the support at O and released with no initial velocity. It drops freely to point C, where the cord becomes taut, and swings in a vertical plane, first about A and then about O. Determine the vertical distance from line OD to the highest point C’’ that the sphere will reach. 45 o O B C C’ C’’ D A a 2. Apply the principle of impulse and momentum: The final momentum mv 2 of the particle is obtained by adding its initial momentum mv 1 and the impulse of the forces F acting on the particle during the time interval considered. mv 1 +  F  t = mv 2  F is sum of the impulsive forces (the forces that are large enough to produce a definite change in momentum).

38 11 - 38 Problem 3 Solution Apply conservation of energy principle. Motion of the sphere from point B to point C (just before the cord is taut). 45 o O B D A a v B = 0 Position 1 45 o O B C D A a vCvC Position 2 y T 1 + V 1 = T 2 + V 2 0 + 0 = m v C 2 - m g (2 a sin 45 o ) v C = 1.682 g a 1 2

39 11 - 39 Problem 3 Solution Consider the sphere at point C as the cord becomes taut and the velocity of the sphere changes to be in direction normal to the cord. Apply impulse and momentum principle. 45 o O B C D A a vCvC O B C D A a v’ C 45 o t t Momentum is conserved in the tangential direction since the external impulse (the cord on the sphere) is in the normal direction. m v C cos 45 o = m v’ C v’ C = v C cos 45 o = 1.682 g a cos 45 o v’ C = 1.1892 g a

40 11 - 40 Problem 3 Solution Motion of the sphere from point C to point C’’. 45 o O B C D A a v’ C = 1.1892 g a Position 2 45 o O B C’’ D A a y d 2 a sin45 o Position 3 T 2 + V 2 = T 3 + V 3 m (v’ C ) 2 - m g (2 a sin 45 o ) = 0 - m g d m (1.1892) 2 g a - m g (2 a sin 45 o ) = 0 - m g d d = 0.707 a v C’’ = 0 1 2 1 2 Apply conservation of energy principle.

41 11 - 41 AA BB lAlA lBlB A B C D Problem 4 A small sphere A attached to a cord AC is released from rest in the position shown and hits an identical sphere B hanging from a vertical cord BD. If the maximum angle  B formed by cord BD with the vertical in the subsequent motion of sphere B is to be equal to the angle  A, determine the required value of the ratio l B / l A of the lengths of the two cords in terms of the coefficient of restitution e between the two spheres.

42 11 - 42 Problem 4 AA BB lAlA lBlB A B C D A small sphere A attached to a cord AC is released from rest in the position shown and hits an identical sphere B hanging from a vertical cord BD. If the maximum angle  B formed by cord BD with the vertical in the subsequent motion of sphere B is to be equal to the angle  A, determine the required value of the ratio l B / l A of the lengths of the two cords in terms of the coefficient of restitution e between the two spheres. 1. Apply principle of conservation of energy: When a particle moves under the action of a conservative force, the sum of the kinetic and potential energies of the particle remains constant. T 1 + V 1 = T 2 + V 2 where 1 and 2 are two positions of the particle.

43 11 - 43 Problem 4 AA BB lAlA lBlB A B C D A small sphere A attached to a cord AC is released from rest in the position shown and hits an identical sphere B hanging from a vertical cord BD. If the maximum angle  B formed by cord BD with the vertical in the subsequent motion of sphere B is to be equal to the angle  A, determine the required value of the ratio l B / l A of the lengths of the two cords in terms of the coefficient of restitution e between the two spheres. 1a. Kinetic energy: The kinetic energy at each point on the path is given by: T = m v 2 1b. Potential energy: The potential energy of a weight W close to the surface of the earth at a height y above a given datum is given by: V g = W y 1 2

44 11 - 44 Problem 4 AA BB lAlA lBlB A B C D A small sphere A attached to a cord AC is released from rest in the position shown and hits an identical sphere B hanging from a vertical cord BD. If the maximum angle  B formed by cord BD with the vertical in the subsequent motion of sphere B is to be equal to the angle  A, determine the required value of the ratio l B / l A of the lengths of the two cords in terms of the coefficient of restitution e between the two spheres. 2. Apply conservation of momentum principle: During an impact of two bodies A and B, the total momentum of A and B is conserved if no impulsive external force is applied. m A v A + m B v B = m A v’ A + m B v’ B where v A and v B denote the velocities of the bodies before the impact and v’ A and v’ B denote their velocities after the impact.

45 11 - 45 Problem 4 AA BB lAlA lBlB A B C D A small sphere A attached to a cord AC is released from rest in the position shown and hits an identical sphere B hanging from a vertical cord BD. If the maximum angle  B formed by cord BD with the vertical in the subsequent motion of sphere B is to be equal to the angle  A, determine the required value of the ratio l B / l A of the lengths of the two cords in terms of the coefficient of restitution e between the two spheres. 3. Apply the relationship for the coefficient of restitution: For impact of two particles A and B : v’ B - v’ A = e (v A - v B ) where v’ B - v’ A and v A - v B are the relative velocities, normal to the impact plane, after and before the impact, respectively, and e is the coefficient of restitution.

46 11 - 46 Problem 4 Solution AA BB lAlA lBlB A B C D Motion of sphere A from its release until it hits sphere B. Position 1 Position 2 AA lAlA A C vA2vA2 v A1 = 0 l A ( 1 - cos  A ) y

47 11 - 47 Problem 4 Solution Apply principle of conservation of energy. Position 1 Position 2 AA lAlA A C vA2vA2 v A1 = 0 l A ( 1 - cos  A ) y T 1 + V 1 = T 2 + V 2 0 + m g l A ( 1 - cos  A ) = m (v A2 ) 2 + 0 v A2 = 2 g l A ( 1 - cos  A ) 1 2 Motion of sphere A from its release until it hits sphere B.

48 11 - 48 Problem 4 Solution AA BB lAlA lBlB A B C D Collision of balls A and B. B C D vA2vA2 v B2 = 0 Before impact B C D v’ A2 v’ B2 After impact

49 11 - 49 Problem 4 Solution B C D vA2vA2 v B2 = 0 Before impact B C D v’ A2 v’ B2 After impact Apply conservation of momentum principle. m v A2 = m v’ A2 + m v’ B2 v A2 = v’ A2 + v’ B2 (1) Apply the relationship for the coefficient of restitution. ( v’ B2 - v’ A2 ) = e ( v A2 )(2) Eliminating v’ A2 from equations (1) and (2) gives: v’ B2 = ( 1+ e ) 2 vA2vA2

50 11 - 50 Problem 4 Solution AA BB lAlA lBlB A B C D Motion of sphere B following the collision. BB lBlB B D l B ( 1 - cos  B ) Position 2 v’ B2 vB3vB3 Position 3

51 11 - 51 Problem 4 Solution Motion of sphere B following the collision. BB lBlB B D l B ( 1 - cos  B ) Position 2 v’ B2 vB3vB3 Position 3 y Apply principle of conservation of energy. T 2 + V 2 = T 3 + V 3 m ( v’ B2 ) 2 + 0 = 0 + m g l B ( 1 - cos  B ) 1 2 Substituting v’ B2 = ( 1+ e ) 2 vA2vA2 and v A2 = 2 g l A ( 1 - cos  A ) and  B =  A gives: lBlB lAlA = ( ) 1 + e 2 2


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