{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

9-15-08LectPHY2048

9-15-08LectPHY2048 - Summary of 3d kinematics An object...

Info iconThis preview shows pages 1–8. Sign up to view the full content.

View Full Document Right Arrow Icon
Summary of 3d kinematics x z y An object travels through 3d-space in time. r(t) G X r( t) Z Y t ) XYZ ˆˆˆ r (t)i r (t)j r(t)k =++ G It has position vector, The objects position is projected onto each coordinate axis resulting in three coupled 1d equations. These each give the position along the 1d coordinate as a function of time.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Consider the r x (t) component. We plot it as a function of time and look at the object’s position at two times. r x t r xi t i The average speed between those times is, Xf Xi X X(Ave.) fi rr r v tt t Δ == Δ If we take the limit as t 0 Δ= − → r xf t f t Δ v X (t) XX t0 X rd r Lim v (t) td t Δ→ Δ Δ We get the instantaneous speed , v X (t) , in the x direction at the time t. v Ave
Background image of page 2
We can similarly plot the instantaneous speed component v X as a function of time and similarly define the average x acceleration between those times as, fi tt t 0 Δ =− Again taking the limit as Xf Xi X X(Ave.) vv v a t Δ == Δ XX t0 X vd v Lim a (t) td t Δ→ Δ Δ We get the instantaneous acceleration at the time t. With similar definitions for v Y , a Y , v Z and a Z .
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Using the fundamental theorem of integral calculus, we derived for the r X component the expressions (for constant acceleration only ): 2 1 XX O X O X 2 r( t ) r v t at =+ + O X v( t ) v = + & And upon eliminating time from these: ( ) 22 O X X X O vv 2 a r r = +− With similar equations for y & z: 2 1 YY O Y O Y 2 t + O Y t ( ) O Y Y Y O 2 a r r 2 1 ZZ O Z O Z 2 t = ++ O X t = + ( ) O Z Z Z O 2 a r r = By orienting our coordinate system properly we can minimize labor making some of these vanish.
Background image of page 4
The requirement for constant acceleration is a rather severe constraint but consider a plot of the x component of the v-t plot under some constant acceleration and corresponding a-t plot: time v dv a dt = O v 0 Dropping the subscript & rearranging, Integrating both sides, dv adt = o vt t v0 0 dv adt a dt == ∫∫ time a 0 t o 0 vv a d ta t t Area under curve ( a ) from t = 0 to t
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
But suppose now that the acceleration is not constant time v dv(t) a(t) dt = O v 0 Rearranging, Integrating both sides, dv a(t)dt = o vt v0 dv = t o 0 vv a ( t ) d t −= Now if a(t) is an integrable function (or can be approximated as such) then we do the integral to still get a closed form solution. E.g. suppose , qt = ( q a constant) Then, t 2 o 0 q v v qtdt t 2 = 2 o q t 2
Background image of page 6
2 O q v(t) v t 2 =+ Giving Now since, X X dr v dt = dr vdt = Integrating, O rt r0 dr vdt = t 2 OO 0 q rr ( v t ) d t 2 −= + tt 2 00 q v d t t d t 2 + 3 XX O X O q r( t ) r v t t 6 + So for the time dependent position we get: where we dropped the x subscript but remember that we are along x (only).
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 8
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 29

9-15-08LectPHY2048 - Summary of 3d kinematics An object...

This preview shows document pages 1 - 8. Sign up to view the full document.

View Full Document Right Arrow Icon bookmark
Ask a homework question - tutors are online