PHYS 218sec. 517-520ReviewChap. 3Motion in 2 or 3 dimensionsPosition & velocity vectorsx1-dim2-dimxyOne variable is enough to specify the position.rGNow you need two variables, so use a vectorˆˆxy=+rijGIn 3-dim. the position vector isˆˆˆ.Since the particle is moving, the position changes with time,.Then you can also define the ˆˆˆ() () ()velocity vector and acceleration )v(ectxt yt ztxyz=++=++rijrijkkGG() () () ()ˆˆ ˆˆˆˆ() () () () () ()()()()()ˆˆ ˆˆˆˆ() () ()to() () ()r.xyzyxzxyzd t dx t dy t dz ttttvtvtvtdt dt dt dtdv tdv tdv tdttttatatatdt dt dt dt== + + =++== + + =++rvijkijkvaijkijkGGGGComponents of accelerationWe focus on 2-dim. motions.Any 2-dim. vector can be written as a sum of other two vectors.ˆˆThese two vectors could be and , which are always pointing the same dirction.You can also use other two vectors which may change dirijections with time butare perpendicular to each other.Decompose with respect to the direction of .avGG:the component of parallel to , i.e., the path:the component of perpendicular (normal) to ⊥⊥=+aaavaa vaa&&GGGGGGGGGParallel parta&GvGThis is similar to the case of 1-dim. motion.The magnitude of the velocity can change and it depends on the sign of the acceleration.But the direction of the velocity does not change as the acceleration is always along the velocity.During a time interval , the velocity changes as.ttδδ→ +vvaGGGPerpendicular partDuring a time interval , the velocity changes as, but ( ) is to perpendicu !lartt∆→ +∆∆= ∆vv v va vGGG G G G1vG∆vG2vG1212In a small time interval , is nearly perpendicular to .For a very short time interval, is also nearly perpendicular to .But and have the same magnitude.t∆∆∆vvvvvvGGGGGGThe magnitude of the velocity does not change.But the direction of the velocity changes.In general, has both components, parallel and perpendicular to the velocity .The particle's speed changes by and its direction changes by .: The directions of and are not constanaaNote⊥⊥avaa&&GGGGt, they change with time.When speed is constant along a curved path.Acceleration is normal to the path. Since the speed doesn’t change, there is no acceleration parallel to the velocity.PChange of velocity at point PvGaGWhen speed is increasing/decreasing along a curved path.PvGaGaG: its component parallel to is in the direction of the velocity is increasing.: its component parallel to is in the opposite direction of the velocity isbluer ed ⇒⇒vvvvaaGGGGGG decreasing.Projectile motionA projectile: any object that is given an initial velocity and then follows a path determined entirely by the gravitational acceleration. (We ignore air resistance.)xyv0GaGtrajectory2downward acceleration due to gravity0, ( 9.8 m/s )xyaagg==− =We choose the Cartesian (x-y) coordinate system to describe this motion,since the acceleration is always in the negative y-direction.You can easily find that, in general, the acceleration and velocity are not parallel norperpendicular to each other. So the acceleration changes the direction of the motion as well as its speed.00(, )xy0αProjectile motion (equations)x-directiony-direction00 000 0 0 00cos(cos )xxxxavv vxx vtx v tαα=∴ ===+ =+0002200 0 0 0sin11(sin )22yyyyagvv gtv gtyyvt gt y v t gtαα= −∴ = − = −=+ − =+ −0α0v0xv0 yv000000000000sin sincos cosyyxxvvvvvvvvαααα= ⇒ == ⇒ =These are equations of motion for a projectile.This completely describes the projectile motion and most problems can besolved using these equations.Of course, these equations are valid only when air resistance is neglected andthe gravitational acceleration is a constant.Projectile motion (application)xyv0GR0αOhMaximum height h110101112222000200101 10At the maximum height, ( ) 0 determines the time () 0The maximum height is the -position at sin11()2222yyyyyyyyyvt tvvt v gt tghy ttvvvvhyt vt gt v gggggα= ⇒== −⇒==⎛⎞⎟⎜⎟==− = − ==⎜⎟⎜⎟⎜⎝⎠at 0t =1at tt=2at tt=2001,2yyvvthgg==00 000 0cos , sinxyvv vvαα==Range RR is determined by the point where the projectile hits the ground again.222202220202If the projectile hits the ground at , we have ( ) 011() 0222yyytt ytyt v t gt t v gtvtg==⎛⎞⎟⎜⇒ == − = −⎟⎜⎟⎜⎝⎠∴ =20 is excluded as it corresponds to the initial point.t =22000000 020222 sin cos sin 2()xyxvvvvRxt vtgg gαα α∴ === = =()00So, for fixed , the maximal value of is obtained when 454vRπα ==DDifferent initial and final heightsh0vGαROxyExpress R with initial variables()111210112101210020101 0 0Assume that the projectile hits the ground at .Then is determined by the condition that ( ) .1()210212Thus, we have ( )yyyyxxyytttythyt h v t gtgt v t htvvghgvRxt vt v vg== −= − = −⇒−−=⇒ =++=== +()2gh+()2100 11If 2 , then 0 and is physically unacceptable.yytvvgh tg= − +<()22Solution of 0142ax bx cxbbaca++== − ± −Circular motionThis is another important example of a 2-dim. motion.The direction of the velocity is changing.This means that the acceleration must have a component which is perpendicular to the velocity even if the speed is constant.Cf. In the projectile motion, the direction of the acceleration is always fixed.Uniform circular motionA particle is moving on a circle with a constant speed.alwaysTherefore, the acceleration can be written as⊥⇒ ==a0aa&GGGGUniform Circular Motion (direction of the acceleration)vGvGaGaGt+ ∆vaGGt+ ∆vaGGThe acceleration is always pointing to the center of the circular path.counterclockwiseclockwiseIt is called a centripetal acceleration.Uniform Circular Motion (acceleration)φ∆1vG2vGRs∆RRs∆vvv∆φ∆φ∆These two triangles are similar!Explained in the classN200 0contantdefinition of ||||The magnitude of the acceleration||lim lim limtt tvvs vvsvR RvvsvsvatRtRtR∆→ ∆→ ∆→∆∆= ⇒∆= ∆∆∆∆== = =∆∆∆12||||vvv==GG2vaR=Uniform Circular Motion (acceleration)Period (T): the time for one revolution, i.e. the time for the object to complete one trip around the circle.In a time T, the particle travels a distance equal to the circumference, 2πR. Since this motion has a uniform speed, its average speed is the instantaneous speed. 22224RvTvRaRTππ=∴ ==Rcircumference = 2 RπNonuniform Circular MotionThis is a circular motion with varying speed.2vaR⊥=0a ≠&2in radial direction: , (in polar coordinate, this is the direction)||in tangent direction: ,
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