Physics 211 Space - time & space-space diagrams Kinetic Equations of Motion Projectile motion Uniform circular motion Moving coordinate systems Relative motion Galilean Transformation of coordinates 3: Two Dimensional Motion
r(t 1 ) r(t 2 ) x y space-space diagram
at 1 dvt 1 dt d 2 xt 1 2 i d 2 yt 1 2 j at 1 dvt 1 a x t 1 i a y t 1 j speed= vt vt dxt dt 2 dyt dt 2 acceleration = at at d 2 xt dt 2 2 d 2 yt 2 2 average quantities= final value-initial value time taken
acceleration due force of gravity near the earths surface is approximately constant Neglect air resistance Neglect rotation of earth Then we can use kinetic equations of motion for projectile motion
i j y x
a g 9.81j m s 2 r0 0 rt 1 2 gt 2 j v x 0 t i v y 0 t j v x 0 t i v y 0 t j xt v x 0 t yt v y 0 t y x Horizontal and vertical positions
velocity in x direction x t = dx dt v x 0 velocity in y direction y t = dy dt v y 0 9.81t when projectile reaches highest point v y (t)=0 v y 0 9.81t 0 t high v y 0 9.81 xt high v x 0 v y 0 9.81 ; yt high v y 0 v y 0 9. 2 = 1 2 v y 0 2 9. = v y 0 projectile hits ground when yt 0 t v y 0 t 0 t 0 or t 2 vy0vy0 9.81 x max 2 v x 0 v y 0 9.81 v v
trajectory angle tan slope of tangent to path at t 0 slope of velocity vector at t 0 v y 0 v x 0 dy dt dx dt dy dx tan 1 v y 0 v x 0 y x
initial speed= v0 v x 0 2 v y 0 2 v x 0 v0 cos v y 0 v0 sin xt high v0 2 sin cos g v0 2 sin2 2g yt high v0 2 sin 2 2g and x range 2 v x 0 v y 0 9.81 2v0 2 sin cos g v0 2 sin2 g Maximun height and range can be expressed in terms of v0 and
y x v(0) v0 2 sin 2 2g v0 2 sin (2 g
Uniform circular motion r v vt v=constant; rt r position vector rt = xt ,yt rcos t ,rsin t rcos t ,rsin t rcos t,rsin t t t d t dt constant is angular speed angular acceleration d t dt 0 distance travelled s = r r t linear speed=v= ds dt r
rt rcos t i rsin t j r ˆ rt vt drt dt r sin t i r cos t j r sin t i cos t j r ˆ vt at dvt dt r 2 cos t i 2 sin t j r 2 cos t i sin t j r 2 ˆ rt at r 2 ˆ rt v 2 r ˆ rt at v 2 r constant
Relative motion observer 1 observer 2 u(t)=u=constant r 2 (t) r 1 (t) r(t)=ut
r 1 t position vector of object in coordinate system r 2 t r 1 t r 2 t ut r 2 t r 1 t ut v 2 t v 1 t u a 2 t a 1 t Galilean Transformation 1 2 1 2
Nonuniform curvilinear motion
rt rt cos t i sin t j vt drt dt v t ˆ vt find unit vector, ˆ ct perpendicular to ˆ vt points to center of curvature of path at this point ˆ vt ˆ vt 1 d dt ˆ vt ˆ vt 0 d ˆ vt ˆ vt ˆ vt d ˆ vt 0 ˆ vt d ˆ vt 0 thus ˆ ct d ˆ vt dt d ˆ vt
a t dvt d v t ˆ vt v t d ˆ vt at a t t ˆ vt +a r t ˆ ct a r t v t 2 rt radial(centripetal) acceleration a t t rt t tangential acceleration rt distance to center of curvature