Chapter I: KINEMATICS OF PARTICLES POSITION, VELOCITY, ACCELERATION A particle is a body whose physical dimensions are very small compared with the radius of curvature of its path. The figure shows a particle moving along some general curvilinear path z in k space and is located at path P point P at a certain instant. O z i x y j x y The position vector of the particle at any time t can be described by fixed rectangular coordinates measured from the z origin O to point P k rP / O path as P rP / O = xi + yj + zk O z i x y j x y Let us construct a coordinate system x'y'z', parallel to the xyz coordinate system and let the origin of this system be point O'. z' The position vector of z point P with respect O' rO′ / O to O' is rP / O′ = x′i + y ′j + z ′k x' x rP / O′ P path rP / O O y' y According to the parallelogram law of addition, the relation between the position vectors is: rP / O = rO′ / O + rP / O′ z' z O' rO′ / O x' x rP / O′ P path rP / O O y' y rQ / O = rO′ / O + rQ / O′ For point Q r In both equations O′ / O is a fixed vector. The difference between the position vectors of a body at two different instants is named “displacement” (yer değiştirme). At time t if the body is at point P and at time t+δt it is at point Q, then the displacement δr is, z' z δr = rQ / O − rP / O O' rO′ / O O rP / O′ rP / O t rQ / O P rQ / O′ δr path t + δt x' x or δr = rQ/O′ − rP/O′ Q y' y This displacement is independent of the origin of the coordinate system since δr = (r/O′/ O + rQ / O′ ) - (r/O′/ O + rP / O′ )= rQ / O′ - rP / O′ rP /O rQ /O For an instantaneous z' z time difference the direction of δr will O' rO′ / O be tangent to the path. O rP / O′ rP / O t rQ / O P rQ / O′ δr path Q t + δt x' x y' y The velocity of a body is defined as the time rate of change of its position. drP / O δ r = rP / O vP = lim = δt →0 δt dt Since the displacement δr is independent of the origin of the coordinate system, the velocity vector v P will also be independent of the origin of the coordinate system. Also the direction of v P will be the same as the direction of δr , that is tangent to the path of the body. The velocity vector in terms of its components, v P = v x i + v y j + v z k = xi + yj + zk dx x = = vx dt dy y = = vy dt dz z = = vz dt The acceleration of a body is defined as the time rate of change of its velocity, dv p dt = ap The acceleration vector in terms of its components is given as aP = a x i + a y j + a z k = vx i + v y j + vz k aP = xi + yj + zk 2 d x dv x ax = 2 = dt dt d 2 y dv y ay = 2 = dt dt 2 d z dv z az = 2 = dt dt If the position, velocity and acceleration of a body can be described by only their x components, the y and z components are zero, such a motion is named as “rectilinear motion” (doğrusal hareket) . In this case, the x axis may be taken as the axis of the motion and the body may move along a straight line with varying velocity and acceleration. If only the z components of the position, velocity and acceleration vectors are zero, x and y components are different from zero, such a motion is named as “ plane curvilinear motion” (düzlemde eğrisel hareket) . If all the components of the position, velocity and acceleration vectors related to the motion of a body are different from zero, such a motion is named as “general curvilinear motion” or “space curvilinear motion” (genel eğrisel hareket veya uzayda eğrisel hareket). Consider a particle moving along a straight line which is at point P at time t. The position of the particle at time t is s measured from some convenient reference point O fixed on the line. t t=0 t+∆t s=0 v=v0 rP / O = s , v P = v = s, a P = a = v = s At time t+∆t the particle has moved to P’ and its coordinate becomes s+∆s. The change in the position coordinate during the interval ∆t is called the displacement ∆s. The displacement would be negative if the particle moved in the negative s-direction. t t=0 t+∆t s=0 v=v0 rP / O = s , v P = v = s, a P = a = v = s The average velocity of the particle during the interval ∆t is the displacement divided by the time interval or vav ∆s = ∆t As ∆t becomes smaller and approaches zero in the limit, the average velocity approaches the instantaneous velocity of the particle ∆s ds v = lim = = s ∆t →0 ∆t dt Thus, the velocity is the time rate of change of the position coordinate s. The velocity is positive or negative depending on whether the corresponding displacement is positive or negative. v is (+) if the particle is moving right and v is (-) if the particle is moving left. t=0 s=0 v=v0 t t+∆t The average acceleration of the particle during the interval ∆t is the change in its velocity divided by the time interval or a av ∆v = ∆t As ∆t becomes smaller and approaches zero in the limit, the average acceleration approaches the instantaneous acceleration of the particle ∆v dv d 2 s a = lim = = 2 = v = s ∆t →0 ∆t dt dt The acceleration is positive or negative depending on whether the velocity is increasing or decreasing. If the signs of velocity and acceleration are the same then velocity is increasing and it is said that the particle is accelerating (pozitif ivmelenme, hızlanma). If the signs of the velocity and acceleration are opposite, then velocity is decreasing and it is said that the particle is decelerating (negatif ivmelenme, yavaşlama). Note that the acceleration would be positive if the particle had a negative velocity which becoming less negative. By eliminating the time dt between velocity and acceleration equations, we obtain a differential equation relating, displacement, velocity and acceleration. ds dt = v ads = vdv dv dv = ⋅v a= dt ds , or sds = sds The basic relationships between displacement, velocity, acceleration and time can be summarized as: 2 ds v = = s dt dv d s a = = 2 = v = s dt dt ads = vdv or sds = sds Although position, velocity and acceleration are vector expressions, for rectilinear motion, where the direction of the motion is that of the given straight-line path, they may be expressed by their magnitudes indicating the sense of the vectors along the path described by a plus (+) or minus (-) sign. The displacement of the particle may not always be equal to the path taken. Displacement is the vector difference between the initial and final positions of the particle along the path. If the particle completes its motion at exactly the same point it starts its motion then the displacement is zero. But the path taken will be the distance it has traveled along the path and will not be equal to zero. 1 2 start 3 final position If the position coordinate s is known for all values of the time t, successive mathematical and graphical differentiation with respect to t gives the velocity and acceleration. In many problems, however, the relationship between position coordinate and time is unknown, and it must be determined by successive integration from the acceleration. Depending on the nature of the forces, the acceleration may be a function of time, velocity or position coordinate, or a combined function of these quantities. The procedure for integrating the differential equation in each case is indicated as follows: For all cases let the initial conditions be assumed as, t=0, s=s0 , v=v0. dv =a dt ∗ v − v0 = at ⇒ ds =v dt ∗ v t v0 0 ∫ dv = a ∫ dt ⇒ v = v0 + at ⇒ s t s0 0 ∫ ds = ∫ vdt t s − s0 = ∫ (v0 + at )dt ⇒ s = s 0 + v0 t + 0 ∗ v ∫ v0 1 2 at 2 vdv = ads s ∫ vdv = a ds ⇒ s0 ( ) 1 2 v − v02 = a (s − s0 ) ⇒ 2 v 2 = v02 + 2a (s − s0 ) dv = a = f (t ) dt ∗ ∫ f (t )dt t v0 0 ∫ dv = ∫ f (t )dt ⇒ t v − v0 = v t ⇒ v = v0 + ∫ f (t )dt 0 0 v = g (t ) ∗ ds =v dt s ⇒ t ∫ ds = ∫ g (t )dt s0 0 t ⇒ s = s0 + ∫ g (t )dt 0 dv = a = f (v ) dt ∗ v ⇒ t= ∫ ⇒ dv = dt f(v) dv f(v) v0 This result gives t as a function of v. Another approach is vdv = ads = f (v)ds v ⇒ s = s0 + ∫ v0 vdv f(v) ⇒ vdv = ds f(v) v ⇒ ∫ v0 vdv = f(v) s ∫ ds s0 ∗ vdv = ads = f ( s )ds ⇒ v s v0 s0 ∫ vdv = ∫ f (s)ds s ⇒ v 2 − v02 = 2 ∫ f (s)ds s0 s v = 2 v02 +2 ∫ f (s)ds 0 s v= g (s) ∗ ds v= = g ( s) dt s ⇒ t= ∫ s0 ds g(s) ⇒ ds = dt g ( s) s ⇒ ∫ s0 ds = g(s) t ∫ dt 0 Figure a is a schematic plot of the variation of s with t from time t1 to time t2 for some given rectilinear motion. By constructing the tangent to the curve at any time t, we obtain the slope, which is the velocity v=ds/dt (Figure b). ds =v dt Similarly, the slope dv/dt of the v-t curve at any instant gives the acceleration at that instant (Figure c). dv =a dt dx =v dt We see from Figure b that the area under the v-t curve during time dt is vdt, which is the displacement ds. The net displacement of the particle during the interval from t1 to t2 is the corresponding area under the curve, which is s2 t2 ∫ ds = ∫ vdt s1 ⇒ s2 − s1 = (area under v-t curve) t1 Similarly, from Figure c we see that the area under the a-t curve during time dt is adt, which is the velocity dv. The net change in velocity between t1 and t2 is the corresponding area under the curve, which is v2 t2 ∫ dv = ∫ adt v1 t1 ⇒ v2 − v1 = (area under a-t curve) dx =v dt When the acceleration a is plotted as a function of the position coordinate s, the area under the curve during a displacement ds is ads is v2 ∫ v1 s2 ∫ vdv = ads s1 ⇒ ( ) 1 2 v2 − v12 = (area under a-s curve) 2 Example: The slope of the v-t graphic gives the magnitude of the acceleration for that instant. dv =a dt If the magnitude of the acceleration is (+) then either the velocity is increasing along the +s direction or the velocity is decreasing along –s direction. If acceleration is (–) either the velocity is decreasing along the +s direction or it is increasing along the –s direction. The slope of the x-t graphic gives the magnitude of the velocity for that instant. dx dt =v If v is + the particle moves in +s direction, it v is – the particle moves in –s direction.
