Transcript Chapter 4
Chapter 4 Motion in two and three dimensions Two principles for 2D and 3D motions: 1) The principle of independence of force 2) The principle of superposition of motion F1 F3 F2 Section 4-1 Motion in three dimensions with constant acceleration Now we consider a particle move in three dimensions with constant acceleration. We can represent the acceleration as a vector: a ax i a y j az k The particle starts at t=0 with initial position r0 x 0 i y 0 j z o k and an initial velocity v0 v0 x i v0 y j v0 z k . a , a constant ax , a y , az , all constants v v a t v v a t x 0x y 0y v v z 0z x y az t v v0 a t （4-1） In a similar way: 1 2 x x0 v0 x t a x t 2 1 2 y y v0 y t a y t 0 2 1 2 z z 0 v0 z t a z t 2 1 2 r r0 v0 t a t 2 （4-2） Section 4-2 Newton’s laws in three dimensional vector form F ma (4-3) Which includes the three component equations F F x max z maz F y ma y (4-4) Sample problem 1. A crate of mass m=62 kg is sliding without friction with an initial velocity of v0=6.4 m/s along the floor. In an attempt to move it in a different direction, Tom pushes opposite to its initial motion with a constant force of a magnitude F1=81N, while Jane pushes in a perpendicular direction with a constant force of magnitude F2=105N. If they each push for 3.0s, in what direction is the crate moving when they stop pushing? Section 4-3 Projectile motion Figure 4-4 shows the initial motion of a projectile at the instant of launch. Its initial velocity is vo , directed at an angle from the horizontal. y vo 0 We choose suitable coordinate system to make: x0 0 y 0 0 o mg 0 x Fig 4-4 A particle is launched with initial velocity The components of the initial velocity are vox v0 cos 0 v oy v 0 sin 0 (4-6) Gravity is the only force acting on the particle, so the components of the net force are Fx 0 Fy mg (4-7) (4-8) ay g ax 0 v x v0 x v y v 0 y gt (4-9) Position components: 1 2 x v0 x t y v0 y t gt (4-10) 2 From Eqs. (4-10), we can eliminate t and obtain the relationship between x and y (after considering Eqs. (4-6)): g 2 y tan 0 x x (4-13) 2(v0 cos 0 ) 2 which is the equation of a trajectory (轨线) of the projectile, the equation of a parabola. Hence the trajectory of a projectile is parabolic. Fig 4-5 trajectory of a projectile y v v 0y v0 v0 x y v0 x v 0x o v v 0x 0x v0 y R x The “horizontal range R” of the projectile is defined as the distance along the horizontal where the projectile return to the level from which it was launched. Let y=0 in Eq(4-13), we obtain the range R: 2 2 2v0 v0 R sin 0 cos 0 sin 20 (4-14) g g 当子弹从枪口射出时，椰子刚好从树上由静止 自由下落 . 试说明为什么子弹总可以射中椰子 ? Sample problem 4-3. In a contest to drop a package on a target, one contestant’s plane is flying at a constant horizontal velocity of 155km/h at an elevation (海拔) of 225m toward a point directly above the target. At what angle of sight should the package be released to strike the target? Section 4-4 Drag forces and the effects on motions Drag force is a frictional force which experienced by any object that moves through a fluid medium, such as air or water. Drag force must be taken into account in the design of aircraft and seacraft. Drag forces prevent the velocity from increasing without limit in the nature. Falling motion with drag force We assume that the magnitude of the drag force D depends linearly on the speed: (4-17) D bv We choose the y axis to be vertical and the positive direction to be downward. Fy mg bv y and mg bv y ma y (4-18) dv y b ay g vy dt m dv y dt bv y g m (4-20) (4-19) With v y 0 at time t=0, we integrate two sides of Eq.(4-20) vy dv y 0 bv y g m then we obtain ln( mg bv y mg t dt 0 b ) t m mg vy (1 e b b t m ) (4-21) (4-22) • For large t, e vy b t m 0 mg b The magnitude of the terminal speed approaches a constant value, not increasing without limit. • For small t, b t 1 m So e b t m b 1 t m mg b v y (t ) [1 (1 t )] gt (small t) (4-23) b m At the beginning of the motion, it is nearly a freely falling motion. How is D changing with time in general? D D mg mg D mg (for cats) mg It is found that a cat is much safer when it falls from higher place. WHY? Projectile motion with drag force Y(m) When the drag force is considered, D bv the range R and the maximum height H will be reduced. With drag force Without drag force 60 0 o 60 0 79 R - X (m) The trajectory is also no longer symmetric about the maximum; the descending motion is much steeper than the ascending motion. 4-5 Uniform Circular Motion In uniform circular motion, the particle moves at constant speed in a circular path. Since the direction of velocity changes in the motion, it is an acceleration motion. How to find acceleration from the constant speed for uniform circular motion? v r O Fig 4-16 Find acceleration for the motion v1x v cos v1 y v sin y v p1 v v1 y p p2 1x r v2 y v2 x v x O v2 x v cos (4-25) v 2 y v sin As the particle moves along the arc from p1 to p 2 ,it covers a distance of r 2 , and a time interval t 2r . v Acceleration: Fig 4-16 ay v2 y v1 y t v2 x v1x ax 0 (4-27) t v sin v sin v 2 sin (4-28) (2r / v) r In order to find the instantaneous acceleration, we take t approaches zero, (then angle goes to zero) so that p1 and p 2 both approach p , which gives v 2 sin v2 sin v2 a y lim [ ( )] lim (4-29) 0 r r 0 r • Point p is an arbitrary point on the circle, so Eq.(4-29) is a general result for the motion. • The minus sign indicating that the acceleration at p points toward the center of the circle. • The acceleration a y is called centripetal acceleration or radial acceleration. The corresponding force is called centripetal force or seeking center force. 4-6 Relative Motion If we have two inertial frames s and s’, how about the relationship between the motions in them? S’ y’ s, s’: inertial frame P s y rps' r ps 0’ x’ rs 's 0 x rps rps' rs ' s Fig 4-17 (4-31) Take the derivative with respect to time of Eq.(4-31), we have (4-32) v ps v ps' vs 's Eq.(4-32) is a law of the transformation of velocities. It is often called the Galilean form of the law of transformation of velocities. It permits us to transform a velocity of one frame of reference (s’) to another frame of reference. Differentiate Eq.(4-32), we have d v ps d v ps ' d vs ' s dt dt dt (4-33) The last term of eq.(4-33) vanishes, because the relative velocity of two reference frames must be a constant. Thus d v ps dt d v ps' dt or a ps a ps' (4-34) The accelerations of P measured by two observers are identical. Eq.(4-34) indicates directly Newton’s second law can be equally well applied in any inertial frames. Sample problem 1. The compass in an airplane indicates that it is headed due east; its air speed indicator reads 215 km/h. A steady wind of 65 km/h is blowing due north. (a) What is the velocity of the plane with respect to the ground? (b) If the pilot wishes to fly due east, what must be the heading? That is, what must the compass read?