INTRODUCTION. The three general approaches to the solution of kinetics. a) Direct application of Newton s law (called the forcemass-acceleration

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2 INTRODUCTION According to Newton s law, a particle will accelerate when it is subjected to unbalanced force. Kinetics is the study of the relations between unbalanced forces and resulting changes in motion. The three general approaches to the solution of kinetics problems are: a) Direct application of Newton s law (called the forcemass-acceleration method) b) Work and energy principles c) Impulse and momentum methods

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4 The basic relation between force and acceleration is found in Newton s second law, the verification of which is entirely experimental. Newton s second law can be stated as follows: If the resultant force acting on a particle is not zero, the particle will have an acceleration proportional to the magnitude of the resultant and in the direction of this resultant force.

5 We subject a particle to the action of a single force F 1 and we measure the acceleration a 1 of the particle. The ratio F 1 /a 1 of the magnitudes of the force and the acceleration will be some number C 1. We then repeat the experiment by subjecting the same particle to a different force F and measuring the corresponding acceleration a. The ratio F /a of the magnitudes will again produce a number C. The experiment is repeated as many times as desired. We draw two important conclusions from the results of these experiments. First, the ratios of applied force to corresponding acceleration all equal the same number. Thus, F1 F F = =... = n = C, a constant a a a 1 n

6 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. For a particle of high inertia (large C), the acceleration will be small for a given force F. On the other hand, if the inertia is small, the acceleration will be large. The mass m is used as a quantitative measure of inertia, and therefore, we may write the expression F C = = km a where k is a constant introduced to account for the units used. Thus, we may express the relation obtained from the experiments as F =kma

7 where F is the magnitude of the resultant force acting on the particle of mass m, and a is the magnitude of the resulting acceleration of the particle. The second conclusion is that the acceleration is always in the direction of the applied force. r F = r kma (Equation of Motion) In SI unit system, k=1.

8 Primary Inertial System (Birincil (Temel) Eylemsizlik Sistemi) Although the results of ideal experiment are obtained for measurements made relative to the fixed primary inertial system, they are equally valid for measurements made with respect to any nonrotating reference system which translates with a constant velocity with respect to the primary system. Newton s second law holds equally well in a nonaccelerating system, so that we may define an inertial system as any system in which equation of motion is valid.

9 If the ideal experiment described were performed on the surface of the earth and all measurements were made relative to a reference system attached to the earth, the measured results would show a slight discrepancy from those predicted by the equation of motion, because the measured acceleration would not be the correct absolute acceleration. These discrepancy would dissappear when we introduced the corrections due to the acceleration components of the earth.

10 These corrections are negligible for most engineering problems which involve the motions of structures and machines on the surface of the earth. An increasing number of problem occur, particularly in the fields of rocket and spacecraft design, where the acceleration components of the earth are of primary concern.

11 The concept of time, considered an absolute quantity in Newtonian theory, received a basically different interpretation in the theory of relativity announced by Einstein. Although the difference between the mechanics of Newton and Einstein is basic, there is a practical difference in the results given by the two theories only when velocities of the order of the speed of light (300x10 6 m/s) are encountered.

12 Solution of Problems We encounter two types of problems. 1) The acceleration is either specified or can be determined directly from known kinematic conditions. We then determine the corresponding forces which act on the particle by direct substitution into the equation of motion. r F r = ma ) The forces acting on the particle are specified and we must determine the resulting motion. If the forces are constant, the acceleration is also constant and is easily found from the equation of motion. When the forces are functions of time, position or velocity, the equation of motion becomes a differential equation which must be integrated to determine the velocity and displacement.

13 Constrained and Unconstrained Motion (Serbest ve Kısıtlanmış Hareket) (Degree of Freedom-Serbestlik Derecesi) There are two physically distict types of motion. The first type is unconstrained motion where the particle is free of mechanical guides and follows a path determined by initial motion and by the forces which are applied to it from external sources. An airplane or rocket in flight and an electron moving in a charged field are examples of unconstrained motion.

14 The second type is constrained motion where the path of the particle is partially or totally determined by restraining guides. A marble is partially constrained to move in the horizontal plane. A train moving along its track and a collar sliding along a fixed shaft are examples of more fully constrained motion.

15 The choice of an appropriate coordinate system is frequently indicated by the number and geometry of the constraints. Thus, if a particle is free to move in space, the particle is said to have three degrees of freedom since three independent coordinates are required to specify its position at any instant.

16 The marble sliding on the surface has two degrees of freedom. Collar sliding along a fixed shaft has only one degree of freedom.

17 Free-Body Diagram (FBD) Serbest Cisim Diyagramı (SCD) When applying any of the force-mass-acceleration equations of motion, we must account correctly for all forces acting on the particle. The best way to do this is to draw the particle s free body diagram (FBD).

18 In statics the resultant equals zero F r =0 whereas in dynamics it is equated to the r F r ma product of mass and acceleration =.

19 If we choose the x-direction, for example, as the direction of the rectilinear motion of a particle, the acceleration in the y- and z-direction will be zero. ΣF x = ma ΣF ΣF y z = = 0 0 x

20 1) F x = ma x ) F y = ma y ax = x v& =& x& ay = y v& = & y ( F ) ( ) x F = y F + a = a x + a y 1) F = ) a t t ma t F n = man = v& =& s& v ( s& ) a n = ρ = ρ ( F ) ( ) t F = n F + a = a t + a n

21 1) F = ) a r r ma r = & r rθ& a F = θ maθ θ = r& θ+ r&& θ F ( F ) + ( ) = F r θ a r = a + a θ

22 1) x ma x ) y ma y 3) z F = ax = x v& =& x& F = ay = y v& = & y F z = ma = v& =& z& az z ( F ) + ( F ) ( F ) = x y z F + x y a = a + a + a z 1) F = ) a r r ma r = & r rθ& a Fθ = maθ θ = r& θ+ r&& θ 3) F z = ma z az = z v& =& z& ( F) ( ) ( ) r + F Fz F = + θ = ar + aθ a + a z

23 F = 1) R ma R ) F = 3) a R a θ = R& Rcos φ & θ θ ma R & φ = Rcosφθ& + R& cosφθ& Rsinφφθ && a = R& φ + R& & φ+ Rcosφsinφ & θ φ θ F φ = maφ ( F ) + ( F ) ( ) R θ F F + = φ a= ar + a θ + a φ

24 Rectilinear Motion 1. The block shown is observed to have a velocity v 1 =0 m/s as it passes point A and a velocity v =10 m/s as it passes point B on the incline. Calculate the coefficient of kinetic friction µ k between the block and incline if x=75 m and θ=15 o.

25 Rectilinear Motion. A force P is applied to the initially stationary cart. Determine the velocity and displacement at time t=5 s for each of the force histories P 1 and P. Neglect friction.

26 Rectilinear Motion 3. The sliders A and B are connected by a light rigid bar and move with negligible friction in the slots, both of which lie in a horizontal plane. For the position shown, the velocity of A is 0.4 m/s to the right. Determine the acceleration of each slider and the force in the bar at this instant.

27 Curvilinear Motion 4. A 0.8-kg slider is propelled upward at A along the fixed curved bar which lies in a vertical plane. If the slider is observed to have a speed of 4 m/s as it passes position B, determine, (a) the magnitude N of the force exerted by the fixed rod on the slider and (b) the rate at which the speed of the slider is increasing. Assume that friction is negligible.

28 Curvilinear Motion 5. The collar has a mass of 5 kg and is confined to move along the smooth circular rod which lies in the horizontal plane. The attached spring has an unstretched length of 00 mm. If, at the instant β=30 o, the collar has a speed v= m/s, determine the magnitude of normal force of the rod on the collar and the collar s acceleration.

29 Curvilinear Motion 6. The 1 kg collar slides along the smooth parabolic rod in the vertical plane towards point O. The spring whose stiffness is k=600 N/m has an unstretched length of 1 m. If, at the position shown in the figure, velocity of the collar is 3.5 m/s, determine the force acting on the collar by the parabolic rod for this instant. Neglect the friction. y B k y= 3 9 x 1 m m m x O 3/4 m

30 Curvilinear Motion 7. The slotted arm revolves in the horizontal plane about the fixed vertical axis through point O. The kg slider C is drawn toward O at the constant rate of 50 mm/s by pulling the cord S. At the instant for which r=5 mm, the arm has a counterclockwise angular velocity ω=6 rad/s and is slowing down at the rate of rad/s. For this instant, determine the tension T in the cord and the magnitude N of the force exerted on the slider by the sides of the smooth radial slot. Indicate which side, A or B, of the slot contacts the slider.

31 Curvilinear Motion 8. The slotted arm OB rotates in a horizontal plane about point O of the fixed circular cam with constant angular velocity 15 rad/s. The spring has a stiffness of 5 kn/m and is uncompressed when θ=0. the smooth roller A has a mass of 0.5 kg. determine the normal force N which cam exerts on A and also the force R exerted on A by the sides of the slot whenθ=45 o. All surfaces are smooth. Neglect the small diameter of the roller.

32 Curvilinear Motion 9. Pin B has a weight of 1. N and is moving both in slotted arm OC and circular slot DE. Determine the radial and transverse forces acting on B. Take & θ = 15 rad/s, & θ = 50 rad/s, θ=0 o. Neglect the friction. r D B C O θ b E