The branch of mechanics which deals with the motion of object is called dynamics. It is divided into two branches:

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1 M th KINEMATICS 4 CHAPTER INTRODUCTION The branch of mechanics which deals with the motion of object is called dynamics. It is divided into two branches: Kinematics: (i) Kinematics (ii) Kinetics The branch of dynamics which deals with geometry of motion of a body without any reference of the force acting on the body is called kinematics. Kinetics: The branch of dynamics which deals with geometry of motion of a body with reference to the force causing motion is called kinetics. Points to be Remember (i) (ii) (iii) The position of a particle can be specified by a vector r whose initial point is at the origin of some fixed coordinate system and the terminal point is at the particle. This vector is called position vector. If the particle is moving, the vector r changes with time. i.e. it is a function of time. The curve traced by a moving particle is called the trajectory or the path of the particle. The path of the particle can be specified by the vector equation r r (t) (i) The path of the particle can also be specified by three scalar equations x = x(t), y = y(t), z = z(t) (ii) These equations are obtained by equating the components of vectors on two sides of the equation (i). Equation gives the coordinates of the points of the path for different value s of t. We call these as parametric equations of the path.

2 Cartesian Components of Velocity & Acceleration y B Q dr = V r + δr δr P O r A Let AB be a part of the trajectory of the particle as shown in figure. Let the particle at time t be at the point P whose position vector is r. After a small time δt, let the particle reach the point Q whose position vector is r + δr. The PQ = δr is the displacement of the particle from the point P in the small time interval δt. The quotient δr δt gives the average rate of change of displacement of the particle in the interval δt. If we start decreasing the time interval δt, the displacement δr will go on deceasing and the point Q gets nearer and nearer to P. Thus δr lim δ δt can be considered as the instantaneous rate of change of displacement. This is defined as the instantaneous velocity or the simply velocity v of the particle at point P. Thus, δr dr v lim δ δt Proceeding in similar way we can see that the acceleration a (the instantaneous rate of change of velocity) at time t is given by δv a= lim δt 0δt = dv = d dr = d r In Cartesian coordinates, we can write Then r = xi + yi v = d xi + yi= dx i + dy j x

3 Thus Question 1 a = d xi + yi= d x i + d y j v x = x- component of velocity = dx v y = y- component of velocity = dy a x x- component of acceleration d x a y y- component of acceleration d y A particle is moving in such a way that it position at any time t is specified by Find the velocity and acceleration. 3 r = t 3 + t i + cost + sin tj + e t + logtk If v and a are velocity and acceleration of particle respectively. Then v dr d t3 + t i + cost + sin tj + e t + logtk = 3t + ti + sint + sintcostj + e t + 1 t k = 3t + ti + sint sintj + e t + 1 t k and a dv d 3t + ti + sint sintj + e t + 1 t k 6t + i + cost costj + e t 1 t k Question A particle P start from O at t = 0. Find tits velocity and acceleration of particle at any time t if its position at that time is given by r = at i + 4atj If v and a are velocity and acceleration of particle respectively. Then

4 4 and v dr d at i + 4atj = ati + 4aj a dv d ati + 4aj ai Question 3 At any time t, the position of a particle moving in a plane can be specified by (acoswt, asinwt) where a and w are constants. Find the component of its velocity and acceleration along the coordinates axis. Let r = acoswt i + asinwt j Differentiate w.r.t t, we get v = awsinwt i + awcoswt j Differentiate again w.r.t t, we get a = aw coswt i aw sinwt j Thus the component of velocity and acceleration are v x = awsinwt, v y = awcoswt a x = aw coswt, Question 4 a y = aw sinwt The position of particle moving along an ellipse is given by r = acost i + bsint j If a > b, find the position of the particle where velocity has maximum and minimum magnitude. As r = acost i + bsint j Differentiate w.r.t t, we get v = asint i + bcost j v = asint + bcost = a sin t + b cos t = a sin t + b 1 sin t = a sin t + b b sin t = sin ta b + b

5 5 v is maximum when sin t is maximum. i.e. sin t = 1 sint = 1 t = 90, 70 For t = 90 r = acos90 i + bsin90 j = bj For t = 70 r = acos70 i + bsin70 j = bj So the position of the particle when velocity has maximum magnitude is ± bj. v is minimum when sin t is minimum. i.e. sin t = 0 sint = 0 t = 0, 180 For t = 0 r = acos0 i + bsin0 j = ai For t = 180 r = acos180 i + bsin180 j = ai So the position of the particle when velocity has minimum magnitude is ± ai. Radial & Transverse Components of Velocity & Acceleration y s P O r θ x In polar coordinates, the position of a particle is specified by a radius vector r and the polar angle θ which are related to x and y through the relations x = rcosθ y = rsinθ provided the two coordinates frames have the same origin and the x axis and the initial line coincide. The direction of radius vector is known as radial direction and that perpendicular to it in the direction of increasing θ is called transverse direction.

6 6 Let r and s be units vectors in the radial and transverse direction respectively as shown in figure. Then r = cosθi + sinθj s = cos( θ)i + sin( θ)j = sinθi + cosθj (i) (ii) Differentiating (i) w.r.t t dr d cosθi + sinθj sinθ i dθ + cosθj dθ dθ sinθ i + cosθj dθ s By (ii) (iii) Differentiating (ii) w.r.t t We know that ds d sinθi + cosθj cosθ i dθ sinθj dθ dθ sinθ i + cosθj dθ r By (i) (iv) r = r r r = r r Now Thus, v = dr d dr dr rr. r + r dr dθ. r + r s v r = Radial component of velocity = dr = r v θ = Transverse component of velocity = r dθ = rθ Where dot denotes the differentiation with respect to time t.

7 Let a be the acceleration Then Thus, a = dv d dr dθ. r + r s d dr. r + d dθ r s = d dr = d r dr dr r + dr dr r + = d r dr r + + dr dθ + dr dθ dθ 7 s + d dθ ds rs + s + d θ ds rs + dθ r s + dr dθ s + d θ dθ r s + = d r r r dθ r + dr dθ s + d θ r s = d r r dθ r + dr dθ + d θ r s r dθ r dθ r By iii& (iv) a r = Radial component of acceleration = d r r dθ = r rθ a θ = Transverse component of acceleration = dr dθ + r d θ = rθ + rθ Question 5 A particle P moves in a plane in such away that at any time t, its distance from point O is r = at + bt and the line connecting O and P makes an angle θ = ct 3/ with a fixed line OA. Find the radial and transverse components of velocity and acceleration of particle at t = 1 Given that r = at + bt and θ = ct 3/ Differentiate w.r.t t, we get dr = a + bt and dθ = 3 ct1 Differentiate again w.r.t t, we get d r = b and d θ = 3 4 ct1

8 8 At t = 1 r = a + b and θ = c dr = a + b, dθ = 3 c, Radial component of velocity = v r = dr = a + b d r = b and d θ = 3 4 c Transverse component of velocity = v θ = r dθ = 3 ca + b Radial component of acceleration = a r = d r r dθ = b a + b 3 c = b 9 4 c a + b = 1 4 8b 9c a + b Transverse component of acceleration = a θ = dr dθ + r d θ = a + b 3 c + a + b 3 4 c Question 6 = 3 c5a + 9b 4 Find the radial and transverse components of velocity of a particle moving along the curve ax + by = 1 at any time t if the polar angle is θ = ct Given that θ = ct Differentiate w.r.t t, we get dθ = ct Also given that ax + by = 1 First we change this into polar form by putting x = rcosθ and y = rsinθ

9 9 ar cos θ + br sin θ = 1 r (acos θ + bsin θ) = 1 r acos θ + bsin θ = 1 r = acos θ + bsin θ Differentiate w.r.t t, we get dr = 1 acos θ + bsin θ 3 acosθsinθ dθ = 1 acos θ + bsin θ 3 a bsinθ dθ + bsinθcosθ dθ = 1 acos θ + bsin θ 3 a bsinθ.ct = cta bsinθ acos θ + bsin θ 3 Radial component of velocity = dr = Transverse component of velocity = r dθ = Question 7 cta bsinθ acos θ + bsin θ 3 ct acos θ + bsin θ 1 Find the radial and transverse components of acceleration of a particle moving along the circle x + y = a with constant velocity c. Given that dθ = c Differentiate w.r.t t, we get d θ = 0 Also given that x + y = a First we change this into polar form by putting x = rcosθ and y = rsinθ r cos θ + r sin θ = a r (cos θ + sin θ) = a r = a

10 10 r = a dr = 0 d r = 0 Radial component of acceleration = a r = d r r dθ = 0 ac = ac Transverse component of acceleration = a θ = dr dθ + r d θ = 0 Tangential & Normal Components of Velocity & Acceleration y B Q Tangent r + δr δr δs P r A x O Let AB be a part of the trajectory of the particle as shown in figure. Let the particle at time t be at the point P whose position vector is r. After a small time δt, let the particle reach the point Q whose position vector is r + δr. Then PQ = δr and arcpq = δs Now v = dr = dr ds. ds = v. dr ds (i) Here dr ds is a unit tangent at point P. Let t be a unit vector along the tangent at P and n unit vector along normal at the point P. Then dr t ds Using this in (i), we get

11 11 Thus, v = v t + 0.n v t = Tangential component of velocity = v v n = Normal component of velocity = 0 Hence the velocity is along the tangent to the path. y Normal n Tangent t 90 0 O Let a be the acceleration. Then a = dv = d v t = dv t + v = dv t + v dψ dψ ds ds = dv t + v dψ (Kv) ψ ds v Where dψ ds = K is called curvature and K = 1 ρ So a = dv t + v dψ. v ρ = dv t + v ρ dψ Since t and n are unit vectors along tangent and normal at P Therefore t = cosψi + sinψj n = cos( ψ)i + sin( ψ)j = sinψi + cosψj Now dψ d dψ cosψi + sinψj sinψ i + cosψ j ψ x

12 1 So Thus, n a = dv t + v ρ n Tangential component of acceleration = a t = dv Where Normal component of acceleration = a n = v ρ = Question dy 3 dx d y dx A particle is moving along the parabola x = 4ay with constant speed. Determine tangential and normal components of its acceleration when it reaches the point whose abscissa is 5a. Given that x = 4ay Differentiate w.r.t x, we get x = 4a dy dy dx dx = x a Differentiate again w.r.t x, we get d y dx 1 a Given that x = 5a therefore We know that dy dx = 5a a = dy 3 dx ρ = = = a d y 1 4 dx a Since the particle is moving with constant speed therefore dv = 0 ρ = a = a 3 3 7a 4

13 Tangential component of acceleration = a t = dv 0 Normal component of acceleration = a n = v ρ = v 7a 4 Question v 7a Find the tangential and normal component of acceleration of a point describing ellipse x y a b = 1 With uniform speed v when the particle is at (0, b). Given that x y a b = 1 x b + y a = a b Differentiate w.r.t x, we get b x + a y dy dx = 0 dy dx = b x a y Differentiate again w.r.t x, we get At (0, b) and d y dx b a dy x y dx y b y x b x a y a y b a 1 y + x b a y 3 dy dx = b 0 a b = 0 d y b dx a 1 b + 0.b a b 3 b a

14 14 We know that ρ = 1 + dy dx d y dx = b a 3 = a b Since the particle is moving with uniform speed therefore dv = 0 Thus, Tangential component of acceleration = a t = dv 0 Normal component of acceleration = a n = v ρ = v a b bv a \ Question 10 A particle is moving with uniform speed along the curve x y = a x + a 5 Show that acceleration has maximum value 10v 9a Given that x y = a x + a 5 y = a + a3 5 x Differentiate w.r.t x, we get dy a3 dx 5 x3 Differentiate again w.r.t x, we get d y 6a3 dx We know that 5 x4

15 ρ = = We know that 1 + dy 3 dx d y dx a3 5 x3 6a 3 5 x4 = a6 5x 6 6a 3 5x 4 a = dv t v ρ n Since the particle is moving with constant speed therefore dv = 0 a = v ρ n a = v v n n = 1 ρ ρ a will maximum when ρ is minimum. Differentiate (i) w.r.t x, we get 3 = 5x4 + 4a 6 6a 3 5x6 5x 6 3 dρ dx 30a3x5 5x6 + 4a x 5 5x 6 + 4a a 3 x 4 30a 3 x 5 5x6 + 4a a 6 x 10 30a 3 x 5 45x 5 5x 6 + 4a 6 150a 3 x 4 5x6 + 4a a 3 x 6 45x 6 55x 6 + 4a 6 5x6 + 4a a 3 x 6 45x 6 5x 6 0a 6 5x6 + 4a a 3 x 6 0x 6 0a 6 05x6 + 4a a 3 x 6 x 6 a 6 05x6 + 4a a 3 x 6 x a x 4 + x a + a 4 Putting dρ = 0, we get dx Since x = ± a 3 = 5x6 + 4a a 3 x 5 (i)

16 16 dρ dρ < 0 before x = a and dx dx > 0 Therefore ρ is minimum when x = a after x = a Thus ρ min = 5a6 + 4a a 3 a 5 Maximum value of acceleration = = 9a6 3 30a 8 = 7a 30 = 9 10 a v ρ min = v 9 10 a = 10v 9a %%%%% End of The Chapter # 4 %%%%%

Chapter 4. Motion in Two Dimensions. Professor Wa el Salah

Chapter 4. Motion in Two Dimensions. Professor Wa el Salah Chapter 4 Motion in Two Dimensions Kinematics in Two Dimensions Will study the vector nature of position, velocity and acceleration in greater detail. Will treat projectile motion and uniform circular