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2. Motion Motion or rest is relative. In describing movement we use the notion of a reference frame, i.e. the physical object to which a coordinate system is attached. Linear, right-angle coordinate system is named the Cartesian system. 2.1. Position and path Position of a particle P is given by the position vector r î , ˆj , k̂ constant unit vectors r î x ˆjy k̂z Motion can be described by giving the vector equation r r t or equivalently three scalar equations 1 x xt y y t z z t parametric equations of the path (trajectory). Eliminating time t one obtains the equation of the path in implicit form y f1 x z f 2 y or y = f(x) in two dimensions. Example: Uniform circular motion in the plane xy (θ = ωt) Parametric equation: ( 1 ) x r cos t ( 2 ) y r sin t equivalent to the vector equation r ( t ) r( î cos t ˆj sin t ) Eliminating time from (1) and (2) one obtains the equation of the path in the form of a circle x2 y2 r 2 2 2.2. Displacement and velocity When the position vector changes during a certain time interval, then the displacement of a particle is r r2 r1 r ( t2 ) r ( t1 ) The average velocity is a vector and is defined as r v avg t The instantaneous velocity at some instant (simply velocity - prędkość) is the limit of average velocity when the time interval tends to zero. r d r t 0 t dt v lim Velocity vector is tangent to the path at the particle’s position (arc secant transforms to tangent when Δt tends to zero). 3 Velocity, cont. Velocity vector can be rewritten as r s r s lim lim t 0 s t t 0 s t 0 t v lim v v The path length Δs is the length of the arc AB . - unit vector tangent to the path at a given position of the particle v - magnitude of velocity vector which is called speed (szybkość). The scalar components of velocity vector along the appropriate axes can be calculated by differentiating the components of the position vector d i x j y k z dr i dx j dy k dz i v j v k v V x y z dt dt dt dt dt vx, vy, vz – scalar components of v . 4 Sample problem A particle moves along an x axis and its position changes with time as follows x 2t 2 10t 6 where x is given in meters and t in seconds (the coefficients also have appropriate dimensions). What is its velocity at t = 2 s? dx 4t 10 dt v( 2 ) 8 10 2m / s v Minus sign indicates that the particle at t = 2 s moves in the negative direction of x with a speed of 2 m/s. The velocity is changing with time. 5 2.3. Acceleration The average acceleration is defined as the change in velocity in a time interval Δt v 2 v1 v aavg t t When Δt tends to zero, one obtains in the limit the instantaneous acceleration (or acceleration) v d v t 0 t dt a lim Acceleration is a time derivative of the velocity vector (both magnitude and direction of the velocity are important). The components of acceleration vector can be calculated by differentiating the scalar components of velocity 6 Acceleration, cont. a d v dv x dv y dv z i j k i a x j ay k az dt dt dt dt Then ax dv x , dt ay dv y dt , az dv z dt Alternatively an acceleration can be rewritten using components connected to the variation of the magnitude and the direction of velocity vector because and 1 Following the discussion on the right, one gets s then d v n dt d d ds dt ds dt d v d dv a v dt dt dt ρ - radius of curvature d d 1 n n ds ds 7 Acceleration, cont. Finally one obtains dv a n an a dt v2 an a v2 dv dt The component called centripetal acceleration, responsible for the variation of velocity direction (directed to the center of curvature) The component connected with the variation of velocity magnitude (directed tangent to the path) 8 Projectile motion (oblique projection) g The motion of an object in a vertical plane under the influence of gravitational force is known as “projectile motion.” The projectile is launched with an initial velocity vo at an angle θ0. The horizontal and vertical velocity components are: v 0 x v 0 cos 0 v 0 y v 0 sin 0 Motion along the x-axis has zero acceleration. Motion along the y-axis has uniform acceleration ay = -g. In this case v x v 0 x const v y v 0 y gt 9 Projectile motion, cont. The x component of the position is then (for x0 = 0 and y0 = 0) x v 0 cos 0 dt ( v 0 cos 0 ) t (1) Calculation of the y component gives y v y dt ( v 0 y gt 2 gt ) dt v 0 y t 2 (2) Eliminating t from equations (1) and (2) one obtains the path of the motion. gx 2 y ( tgθ0 )x 2( v 0 cos θ0 )2 (3) This is the equation of a parabola. 10 Projectile motion, cont. The equation of the path can be used for calculation of path parameters, e.g. the horizontal range and the maximum height. Solving eq.(3) for y = 0 one gets II in vacuum I in air gx 2 ( tg θ0 )x 0 2 2( v 0 cos θ0 ) x1 0 v 02 sin 2 x2 R g H A O t R From the symmetry analysis it follows, that the maximum height corresponds to y value at x = R/2. In this case we have v 02 sin 2 y( R / 2 ) H 2g 11 Uniform circular motion A particle moves with constant speed v along a circular path with radius ρ. The speed is constant but the velocity vector is not. The fact that the velocity changes means that the acceleration is not zero. The centripetal acceleration is given by the equation v2 an As v = const then a dv 0 dt a an The time of a complete full revolution is called the period and is given by the equation T 2 r 1 v f where f is the frequency (number of revolutions per second). 12