2Linear Impulse and Linear Momentum Consider the general curvilinear motion in space of a particle of mass m, where the particle is located by its position vector measured from a fixed origin O.The velocity of the particle is is tangent to its path. The resultant force of all forces on m is in the direction of its accelerationWe may write the basic equation of motion for the particle, asorWhere the product of the mass and velocity is defined as the linear momentum of the particle. This equation states that the resultant of all forces acting on a particle equals its time rate of change of linear momentum.
3These equations may be applied independently of one another. Principle of Linear Impulse and MomentumIn SI, the units of linear momentum are seen to be kg.m/s, which also equals N.s.Linear momentum equation is one of the most useful and important relationships in dynamics, and it is valid as long as mass m of the particle is not changing with time.We now write the three scalar components of linear momentum equation asThese equations may be applied independently of one another.
4The Linear Impulse-Momentum Principle Principle of Linear Impulse and MomentumAll that we have done so far is to rewrite Newton’s second law in an alternative form in terms of momentum. But we may describe the effect of the resultant force on the linear momentum of the particle over a finite period of time simply by integrating the linear momentum equation with respect to time t. Multiplying the equation by dt gives , which we integrate from time t1 to time t2 to obtainHere 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 this equation states that the total linear impulse on m equals the corresponding change in linear momentum of m.
5Alternatively, we may write Principle of Linear Impulse and MomentumAlternatively, we may writewhich says that the initial linear momentum of the body plus the linear impulse applied to it equals its final linear momentum.mv1+=
6Principle of Linear Impulse and Momentum The impulse integral is a vector which, in general, we may involve changes in both magnitude and direction during the time interval. Under these conditions, it will be necessary to express and in component form and then combine the integrated components. The components become the scalar equations, which are independent of one another.
7Principle of Linear Impulse and Momentum There are cases where a force acting on a particle changes with the time in a manner determined by experimental measurements or by other approximate means. In this case, a graphical or numerical integration must be performed. If, for example, a force acting on a particle in a given direction changes with the time as indicated in the figure, the impulse, , of this force from t1 to t2 is the shaded area under the curve.
8Conservation of Linear Momentum Principle of Linear Impulse and MomentumConservation of Linear MomentumIf the resultant force on a particle is zero during an interval of time, its linear momentum G remains constant. In this case, the linear momentum of the particle is said to be conserved. Linear momentum may be conserved in one direction, such as x, but not necessarily in the y- or z- direction.This equation expresses the principle of conservation of linear momentum.
9PROBLEMSPrinciple of Linear Impulse and Momentum1. The 200-kg lunar lander is descending onto the moon’s surface with a velocity of 6 m/s when its retro-engine is fired. If the engine produces a thrust T for 4 s which varies with the time as shown and then cuts off, calculate the velocity of the lander when t=5 s, assuming that it has not yet landed. Gravitational acceleration at the moon’s surface is 1.62 m/s2.
11PROBLEMSPrinciple of Linear Impulse and Momentum2. The 9-kg block is moving to the right with a velocity of 0.6 m/s on a horizontal surface when a force P is applied to it at time t=0. Calculate the velocity v of the block when t=0.4 s. The kinetic coefficient of friction is mk=0.3.
13PROBLEMSPrinciple of Linear Impulse and Momentum3. A tennis player strikes the tennis ball with her racket while the ball is still rising. The ball speed before impact with the racket is v1=15 m/s and after impact its speed is v2=22 m/s, with directions as shown in the figure. If the 60-g ball is in contact with the racket for 0.05 s, determine the magnitude of the average force R exerted by the racket on the ball. Find the angle b made by R with the horizontal.
14W=mg Rx R Ry R Ry Rx SOLUTION 20° 10° b y x in x direction in y directionRxRyRb
15PROBLEMSPrinciple of Linear Impulse and Momentum4. The 40-kg boy has taken a running jump from the upper surface and lands on his 5-kg skateboard with a velocity of 5 m/s in the plane of the figure as shown. If his impact with the skateboard has a time duration of 0.05 s, determine the final speed v along the horizontal surface and the total normal force N exerted by the surface on the skateboard wheels during the impact.
16PROBLEMS (mB+mS)g y x N Linear momentum is conserved in x-direction; Principle of Linear Impulse and Momentum(mB+mS)gyxNLinear momentum is conserved in x-direction;in y direction
17Angular Impulse and Angular Momentum In addition to the equations of linear impulse and linear momentum, there exists a parallel set of equations for angular impulse and angular momentum. First, we define the term angular momentum. Figure shows a particle P of mass m moving along a curve in space. The particle is located by its position vector with respect to a convenient origin O of fixed coordinates x-y-z.y
18Angular Impulse and Angular Momentum The velocity of the particle is , and its linear momentum is The moment of the linear momentum vector about the origin O is defined as the angular momentum of P about O and is given by the cross-product relation for the moment of a vectorThe angular momentum is a vector perpendicular to the plane A defined by and . The sense of is clearly defined by the right-hand rule for cross products.
19In SI units, angular momentum has the units kg.m2/s =N.m.s. Angular Impulse and Angular MomentumThe scalar components of angular momentum may be obtained from the expansionso thatIn SI units, angular momentum has the units kg.m2/s =N.m.s.
20Rate of Change of Angular Momentum Angular Impulse and Angular MomentumIf represents the resultant of all forces acting on the particle P, the moment about the origin O is the vector cross productWe now differentiate with time, using the rule for the differentiation of a cross product and obtainThe term is zero since the cross product of parallel vectors is zero.
21Substitution into the expression for moment about O gives Angular Impulse and Angular MomentumSubstitution into the expression for moment about O givesThe scalar components of this equation is
22The Angular Impulse-Momentum Principle Angular Impulse and Angular MomentumThe Angular Impulse-Momentum PrincipleTo obtain the effect of the moment on the angular momentum of the particle over a finite period of time, we integrate from time t1 to t2.orThe total angular impulse on m about the fixed point O equals the corresponding change in angular momentum of m about O.Alternatively, we may write
23Plane-Motion Application Angular Impulse and Angular MomentumPlane-Motion ApplicationMost of the applications can be analyzed as plane-motion problems where moments are taken about a single axis normal to the plane motion. In this case, the angular momentum may change magnitude and sense, but the direction of the vector remains unaltered.
24Conservation of Angular Momentum Angular Impulse and Angular MomentumConservation of Angular MomentumIf the resultant moment about a fixed point O of all forces acting on a particle is zero during an interval of time, its angular momentum remains constant. In this case, the angular momentum of the particle is said to be conserved. Angular momentum may be conserved about one axis but not about another axis.This equation expresses the principle of conservation of angular momentum.
25PROBLEMSAngular Impulse and Angular Momentum1. The assembly starts from rest and reaches an angular speed of 150 rev/min under the action of a 20 N force T applied to the string for t seconds. Determine t. Neglect friction and all masses except those of the four 3-kg spheres, which may be treated as particles.
27PROBLEMSAngular Impulse and Angular Momentum2. A pendulum consists of two 3.2 kg concentrated masses positioned as shown on a light but rigid bar. The pendulum is swinging through the vertical position with a clockwise angular velocity w=6 rad/s when a 50-g bullet traveling with velocity v=300 m/s in the direction shown strikes the lower mass and becomes embedded in it. Calculate the angular velocity w which the pendulum has immediately after impact and find the maximum deflection q of the pendulum.
28SOLUTION (2) Angular momentum is conserved during impact; (1) 1 2 q w Angular Impulse and Angular Momentum(2)Angular momentum is conserved during impact;(1)12qww´v1v1´v2´v2O
29SOLUTIONAngular Impulse and Angular Momentum12qww´v1v1´v2´v2OEnergy considerations after impact;(Datum at O)