LECTUR~14_15.1 - PRINCIPLE OF LINEAR IMPULSE AND MOMENTUM...

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Unformatted text preview: PRINCIPLE OF LINEAR IMPULSE AND MOMENTUM Today’s Objectives: Students will be able to: 1.  Calculate the linear momentum of a particle and linear impulse of In-Class Activities: a force. 2.  Apply the principle of linear •  Linear Momentum and Impulse impulse and momentum. •  Principle of Linear Impulse and Momentum •  Concept Quiz •  Group Problem Solving •  Attention Quiz READING QUIZ 1. 2. The linear impulse and momentum equation is obtained by integrating the ______ with respect to time. A) friction force B) equation of motion C) kinetic energy D) potential energy Which parameter is not involved in the linear impulse and momentum equation? A) Velocity B) Displacement C) Time D) Force APPLICATIONS A dent in an trailer fender can be removed using an impulse tool, which delivers a force over a very short time interval. To do so, the weight is gripped and jerked upwards, striking the stop ring. How can we determine the magnitude of the linear impulse applied to the fender? Could you analyze a carpenter’s hammer striking a nail in the same fashion? Sure! APPLICATIONS (continued) A good example of impulse is the action of hitting a ball with a bat. The impulse is the average force exerted by the bat multiplied by the time the bat and ball are in contact. Is the impulse a vector? Is the impulse pointing in the same direction as the force being applied? Given the situation of hitting a ball, how can we predict the resultant motion of the ball? APPLICATIONS (continued) When a stake is struck by a sledgehammer, a large impulse force is delivered to the stake and drives it into the ground. If we know the initial speed of the sledgehammer and the duration of impact, how can we determine the magnitude of the impulsive force delivered to the stake? PRINCIPLE OF LINEAR IMPULSE AND MOMENTUM (Section 15.1) The next method we will consider for solving particle kinetics problems is obtained by integrating the equation of motion with respect to time. The result is referred to as the principle of impulse and momentum. It can be applied to problems involving both linear and angular motion. This principle is useful for solving problems that involve force, velocity, and time. It can also be used to analyze the mechanics of impact (taken up in a later section). PRINCIPLE OF LINEAR IMPULSE AND MOMENTUM (continued) The principle of linear impulse and momentum is obtained by integrating the equation of motion with respect to time. The equation of motion can be written ∑F = m a = m (dv/dt) Separating variables and integrating between the limits v = v1 at t = t1 and v = v2 at t = t2 results in t2 v2 ∑ ∫ F dt = m ∫ dv = mv2 – mv1 t1 v1 This equation represents the principle of linear impulse and momentum. It relates the particle’s Dinal velocity (v2) and initial velocity (v1) and the forces acting on the particle as a function of time. PRINCIPLE OF LINEAR IMPULSE AND MOMENTUM (continued) Linear momentum: The vector mv is called the linear momentum, denoted as L. This vector has the same direction as v. The linear momentum vector has units of (kg·m)/s or (slug·ft)/s. Linear impulse: The integral ∫F dt is the linear impulse, denoted I. It is a vector quantity measuring the effect of a force during its time interval of action. I acts in the same direction as F and has units of N·s or lb·s. The impulse may be determined by direct integration. Graphically, it can be represented by the area under the force versus time curve. If F is constant, then I = F (t2 – t1) . PRINCIPLE OF LINEAR IMPULSE AND MOMENTUM (continued) The principle of linear impulse and momentum in vector form is written as t2 mv1 + ∑ ∫t F dt = mv2 1 The particle’s initial momentum plus the sum of all the impulses applied from t1 to t2 is equal to the particle’s Dinal momentum. The two momentum diagrams indicate direction and magnitude of the particle’s initial and Dinal momentum, mv1 and mv2. The impulse diagram is similar to a free-­‐body diagram, but includes the time duration of the forces acting on the particle. IMPULSE AND MOMENTUM: SCALAR EQUATIONS Since the principle of linear impulse and momentum is a vector equation, it can be resolved into its x, y, z component scalar equations: t2 m(vx)1 + ∑ Fx dt = m(vx)2 t1 t2 m(vy)1 + ∑ Fy dt = m(vy)2 t1 t2 m(vz)1 + ∑ Fz dt = m(vz)2 ∫ ∫ ∫ t 1 The scalar equations provide a convenient means for applying the principle of linear impulse and momentum once the velocity and force vectors have been resolved into x, y, z components. PROBLEM SOLVING •  Establish the x, y, z coordinate system. •  Draw the particle’s free-­‐body diagram and establish the direction of the particle’s initial and Dinal velocities, drawing the impulse and momentum diagrams for the particle. Show the linear momenta and force impulse vectors. •  Resolve the force and velocity (or impulse and momentum) vectors into their x, y, z components, and apply the principle of linear impulse and momentum using its scalar form. •  Forces as functions of time must be integrated to obtain impulses. If a force is constant, its impulse is the product of the force’s magnitude and time interval over which it acts. EXAMPLE Given: A 0.5-­‐kg ball strikes the rough ground and rebounds with the velocities shown. Neglect the ball’s weight during the time it impacts the ground. Find: The magnitude of impulsive force exerted on the ball. Plan: 1) 2) Draw the momentum and impulse diagrams of the ball as it hits the surface. Apply the principle of impulse and momentum to determine the impulsive force. EXAMPLE (continued) Solution: 1) The impulse and momentum diagrams can be drawn as: ∫ W dt ≈ 0 = + 45° mv1 mv2 ∫ F dt 30° ∫ N dt ≈ 0 The impulse caused by the ball’s weight and the normal force N can be neglected because their magnitudes are very small as compared to the impulse from the ground. EXAMPLE (continued) 2) The principle of impulse and momentum can be applied along the direction of motion: t2 mv1 + ∑ F dt = mv2 ∫ t 1 ⇒ t2 ∫t 0.5 (25 cos 45° i − 25 sin 45° j) + ∑ F dt = 0.5 (10 cos 30° i + 10 sin 30° j) 1 The impulsive force vector is t2 I = ∑ F dt = (4.509 i + 11.34 j ) N⋅s ∫ t 1 Magnitude: I = √ 4.5092 + 11.342 = 12.2 N⋅s CONCEPT QUIZ 1. Calculate the impulse due to the force. F 10 N Force curve A) 20 kg·m/s B) 10 kg·m/s C) 5 N·s D) 15 N·s 2s 2. A constant force F is applied for 2 s to change the particle’s velocity from v1 to v2. Determine the force F if the particle’s mass is 2 kg. A) (17.3 j) N B) (–10 i +17.3 j) N C) (20 i +17.3 j) N D) (10 i +17.3 j) N v2=20 m/s 60° v1=10 m/s t GROUP PROBLEM SOLVING Given: A 0.05-kg golf ball is struck by the club and travels trajectory shown, θ = 30°, R = 150 m. Assume the club maintains contact with the ball for 0.5 ms. along the Find: The average impulsive force exerted on the ball. Plan: 1) Find v using the kinematics equations. 2) Draw the momentum and impulse diagrams of the ball. 3) Apply the principle of impulse and momentum to determine the impulsive force. GROUP PROBLEM SOLVING (continued) Solution: 1)  Kinematics : horizontal motion equation : x = x0 + vx t 150 = 0 + v (cos 30) t ⇒ t = 150 / (v cos 30) Solving for v: v = 41.2 m/s GROUP PROBLEM SOLVING (continued) 2) Draw the momentum and impulse diagrams are ∫ W dt ≈ 0 mv = + ∫ F dt 30° ∫ N dt ≈ 0 The impulse generated by the weight of the golf ball is very small compared to that generated by the force of the impact. Hence, it and the resultant normal force can be neglected. GROUP PROBLEM SOLVING (continued) 3) Now, apply the principle of impulse and momentum to determine the impulsive force. m (0) + ∑ ∫ F dt = mv, where v = 41.2 m/s ⇒ Favg (0.5) 10-3 = (0.05) (41.2 cos 30 i + 41.2 sin 30 j) The average impulsive force is Favg = 4120 (cos 30° i + sin 30° j) = (3568 i + 2060 j) N ∫ W dt ≈ 0 mv = + ∫ F dt ∫ N dt ≈ 0 30° ATTENTION QUIZ 1. Jet engines on the 100-­‐Mg VTOL aircraft exert a constant vertical force of 981 kN as it hovers. Determine the net impulse on the aircraft over t = 10 s. A) -­‐981 kN·s B) 0 kN·s C) 981 kN·s D) 9810 kN·s 2. A 100-­‐N(≈10-­‐kg) cabinet is placed on a smooth surface. If a force of a 100 N is applied for 2 s, determine the net impulse on the cabinet during this time interval. A) 0 N·s B) 100 N·s C) 200 N·s D) 300 N·s 30° ...
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