Kinematics (2) - Motion in Three Dimensions

Transcription

1 Kinematics (2) - Motion in Three Dimensions 1. Introduction Kinematics is a branch of mechanics which describes the motion of objects without consideration of the circumstances leading to the motion. 2. Position vector The position vector r or OM of a particle M is the vector that is originated from the origin of the system of coordinates and extends to the point M. Figure 1. Position vector in a Cartesian system of coordinates In a Cartesian system of coordinates: r = OM = x i + y j + z k (1) The magnitude of the position vector: r = OM = x 2 + y 2 + z 2 (2) When the particle M is moving, its position is changing with time, so we should consider x, y and z components as functions of time t. 1

2 3. Velocity 3.1. Average velocity: The average velocity of the particle M, is the rate of change of the position vector over a time interval t: r r (t + t) r (t) v av = = (3) t t Figure 2. Position vector at instant t and at instant t + t 3.2. Instantaneous velocity: The instantaneous velocity of the particle M, is the rate of change of the position vector over an extremely small time interval t (i.e. as t tends to zero): r v inst = v = lim t 0 t = d r (4) So the instantaneous velocity is the derivative of the position vector. From now on, instead of writing v inst, we will simply write v. Also, for simplicity, we will use the word velocity for instantaneous velocity. In Cartesian coordinates, the instantaneous velocity: v = d r = d ) (x i + y j + z k (5) = dx dy i + dz j + k (6) = v x i + vy j + vz k (7) The magnitude of v represents the speed of the particle M: (dx ) 2 ( ) dy 2 ( ) dz 2 v = v 2 x + v 2 y + v 2 z = + + (8) 2

3 3.3. Direction of the velocity: From equation 4 r v = lim t 0 t Therefore, v should be in the direction of r (see Figure 2). But as t 0, r becomes along the tangent to the curve of motion at M. Hence v is along the tangent to the curve of motion at M Unit vector: Let u t be a unit vector tangent to the trajectory at any point, so we can write the velocity vector as: v = v u t (9) 3.5. Another expression of v: We have from equation 4: r v = lim t 0 t Let s be the incremental change in the length of the arc during the time interval t, then: r r v = lim t 0 t = lim t 0 s s r t = lim t 0 s lim s t 0 t (10) But as t 0, s 0. Then: r v = lim s 0 s lim s t 0 t = d r ds ds (11) Note that ds is equal to the magnitude of the vector d r, because we are dealing with an extremely small arc. So: And d r ds = Which is the speed v of M. Hence: d r d r = u t (12) ds = ṡ (13) v = ds d r ds = ṡ u t = v u t (14) 3

4 4. Acceleration: 4.1. Average acceleration: The average acceleration of the particle M, is the rate of change of the velocity vector over a time interval t: a av = v t 4.2. Instantaneous acceleration: = v (t + t) v (t) t (15) The instantaneous acceleration of the particle M, is the rate of change of the velocity vector over an extremely small time interval t (i.e. as t tends to zero): v a inst = lim t 0 t = d v (16) So the instantaneous acceleration is the derivative of the velocity vector. From now on, instead of writing a inst, we will simply write a. Also, for simplicity, we will use the word acceleration for instantaneous acceleration. In Cartesian coordinates the instantaneous acceleration: a = d v = d (v x i + vy j + vz k ) (17) = dv x i + dv y j + dv z k (18) = d2 x 2 i + d 2 y 2 j + d 2 z 2 k (19) The magnitude of a: a = a = a 2 x + a 2 y + a 2 z (20) (dvx ) 2 ( ) 2 ( ) 2 dvy dvz = + + (21) = (d 2 x 2 ) 2 + ( d 2 ) 2 y 2 + ( d 2 z 2 ) 2 (22) In many cases, it is useful to project the acceleration vector along two directions, one in the direction of the tangent to the trajectory of motion (i.e. direction of u t ), and the other is normal to the trajectory, i.e. along the normal direction to u t, say, along the direction of a unit vector perpendicular to u t, named u n (radial or normal unit vector) Another expression of a: u t is at any point of the trajectory. Referring to equation 14, which states that: v = v u t = ṡ u t 4

5 a = d v = s u t + ṡ d u t (23) We can show that d u t ds is perpendicular to u t, using the following method: We know that: u t u t = u t 2 = 1 (24) Derive by ds both sides, we get: d u t ds u t + u t d u t ds = 0 (25) 2 u t d u t ds = 0 (26) u t d u t ds = 0 (27) So d u t ds is perpendicular to u t since their scalar product is zero. This implies that d u t ds is parallel to u n. Hence we can write: d u t ds = k u n (28) Now: Therefore d u t = d u t ds ds = k u n ṡ = ṡ k u n (29) a = s u t + ṡ 2 k u n (30) a = s u t + ṡ2 ρ u n (31) where ρ = 1 k is the radius of curvature. 5. Motion with constant acceleration: In this type of motion the acceleration vector a is a constant vector. a = d v ˆ ˆ d v = a (32) Since a = const., then integration results: Where k is a constant vector. At t = 0: v = v 0 k = v0 v = a t + k (33) Moreover: v = a t + v 0 (34) 5

6 v = d r ˆ ˆ d r = ˆ v = ( a t + v 0 ) (35) r = 1 2 a t2 + v 0 t + r 0 (36) So, for a motion with a constant acceleration, we obtain: a = const. (37) v = a t + v 0 (38) r = 1 2 a t2 + v 0 t + r 0 (39) In Cartesian coordinates, the components of the acceleration: where C 1, C 2, and C 3 are constants. The components of the velocity: a x = C 1 (40) a y = C 2 (41) a z = C 3 (42) v x = a x t + v x0 (43) v y = a y t + v y0 (44) v z = a z t + v z0 (45) where v x0, v y0, and v z0 are the initial velocity on x, y, and z-axes respectively (i.e. at t = 0). The components of the position vector: x = 1 2 a x t 2 + v 0x t + x 0 (46) y = 1 2 a y t 2 + v 0y t + y 0 (47) z = 1 2 a z t 2 + v 0z t + z 0 (48) where x 0, y 0, and z 0 are the initial positions on the x, y, and z-axes respectively (i.e. at t = 0). 6. Circular motion: In circular motion the particle moves on a circle. During this motion the position vector r changes its direction but not its magnitude. So the magnitude of the position vector r or r is constant and is equal to the radius of the circle. In this case, the radius of curvature ρ = r = R = const. 6

7 Figure 3. Circular motion The position vector: The velocity vector: OM = r = r u r (49) The acceleration vector: v = v u t = ṡ u t (50) a = dv u t + v2 ρ u n (51) = s u t + ṡ2 ρ u n (52) For a uniform circular motion: = a t + a n (53) v = const. dv = s = 0 (54) and a t = 0 (55) If the motion is circular but not uniform, then: a n = ṡ2 ρ u n = 0 (56) and a t = 0 (57) 7

8 a n = 0 (58) Physics Zone by Farid Minawi 8