Normal and Tangential Components

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Normal and Tangential

Components

In mathematics, given a vector at a point on the

curve, that vector can be decomposed uniquely asa sum of two vectors, one tangent to the curve,called the tangential componentof the vector,and another one perpendicular to the curve, calledthe normal componentof the vector. Similarly avector at a point on a surface can be broken downthe same way.

More generally, given a submanifoldNof amanifold M, and a vector in the tangent space toMat a point N, it can be decomposed into the component tangent to Nand the

component normal to N.

Surface

More formally, let Sbe the surface, andx be the point on the surface. Let V be avector atx. Then one can write uniquely Vas a sum

V=V V!

where the "rst vector in the sum is the tangential component and the second one isthe normal component. It follows immediately that these two vectors areperpendicular to each other.

To calculate the tangential and normal components, consider a unit normal to thesurface , that is, a unit vector # perpendicular to S andx.Then,

V$ %V& # ' #

(nd thus

V $V) V

where * & + denotes the dot product. (nother formula for tangential component is

V $ )# %# V',

where * * denotes the cross product.


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-ote that these formulas do not depend on the particular unit normal # used %thereeist two units normal to any surface at a given point, pointing in the oppositedirections, so one of the units normal is negative of the other one'.

Submanifold

More generally, given a submanifoldNof a manifold Mand a pointp / N, we get ashort eact sequence involving the tangent spaces0

The quotient space is a generali1ed space of normal vectors.

If Mis a 2iemannian manifold, the above sequence splits, and the tangent space of

Matp decomposes as a direct sum of the component tangent to Nand thecomponent normal to N.

Computations:

Suppose Nis given by non)degenerate equations.

If Nis given eplicitly, via parametric equations %such as a parametric curve', thenthe derivative gives a spanning set for the tangent bundle %it3s basis if and only ifthe parametri1ation is an immersion'.

If Nis given implicitly %as in the above description of the surface, more generally ashypersurface' as a level set or intersection of level surfaces for gi, then thegradients of gispan the normal space.

In both cases, we can again compute using the dot product4 the cross product is

special to 5 dimensions though.

References

6 2o7ansky, 8ladimir %9:;:'.


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6 @en7amin Erowell %FAA5' -ewtonian physics. %online version G9H' IS@-A:;ABD;A9.


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Projectile motionProjectile motionis a form of motion in which an ob7ect or particle %calleda pro7ectile' is thrown near the earths surface, and it moves along a curved pathunder the action of gravity only.The only force of signi"cance that acts on theob7ect is gravity, which acts downward to cause a downward acceleration. There areno hori1ontal forces needed to maintain the hori1ontal motion J consistent with theconcept of inertia.

The initial velocity

If the pro7ectile is launched with an initialvelocity , then it can be written as

.

The components and can be found if theangle, is known0

,

.

If the pro7ectiles range, launch angle, and dropheight are known, launch velocity can be foundusing -ewtons formula

.

The launch angle is usually epressed by thesymbol theta, but often the symbol alpha is used.

Kinematic quantities of pro7ectile motionIn pro7ectile motion, the hori1ontal motion and the vertical motion are independentof each other4 that is, neither motion aects the other.
http://en.wikipedia.org/wiki/Velocityhttp://en.wikipedia.org/wiki/Velocityhttp://en.wikipedia.org/wiki/Velocityhttp://en.wikipedia.org/wiki/Velocity


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Acceleration

Since there is only acceleration in the vertical direction, the velocity in the

hori1ontal direction is constant, being equal to . The vertical motion of the

pro7ectile is the motion of a particle during its free fall. ere the acceleration isconstant, being equal to .The components of the acceleration are0

,

.

Velocity

The hori1ontal component of the velocity of the ob7ect remains unchangedthroughout the motion. The vertical component of the velocity increases linearly,

because the acceleration due to gravity is constant. The accelerations in the anddirections can be integrated to solve for the components of velocity at any time asfollows0

,

.

The magnitude of the velocity %under the ?ythagorean theorem '0

Displacement

At any time , the projectile's horizontal and

vertical displacement:

,

.

The magnitude of the displacement:

.

The maimum height of projectile
http://en.wikipedia.org/wiki/Displacement_(vector)http://en.wikipedia.org/wiki/Displacement_(vector)


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The highest height which the ob7ect will reachis known as the peak of the ob7ects motion.

The increase of the height will last, until

, that is,

.

Time to reach the maimum height0

.

Nrom the vertical displacement of the maimum height of pro7ectile0

.

2elation during hori1ontal range and maimum height

The relation between the range on the hori1ontal plane and the maimum

height reached at is0

Application of the !or" energy theorem

(ccording to the work)energy theoremthe vertical component of velocity are listedbelow0

.

2eferences

@udOPgoston0 Ksrleti!"i#a$.,@udapest, TankQnyvkiadO, 9:CD. IS@- :D5 9;

C;;F :
http://en.wikipedia.org/wiki/Work_(physics)#Work_and_kinetic_energyhttp://en.wikipedia.org/wiki/Work_(physics)#Work_and_kinetic_energy


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If7. RtonyiSndor0 %i"i#a &.,@udapest, -em1etiTankQnyvkiadO, FAA:. IS@-

:;C):D5)9:)DACF)F

ack Nrigyes0 Ngy'egy(f)ggvnyt*+l*"ato#, -ss"ef)ggse#sadato#,

@udapest, -em1etiTankQnyvkiadO, FAAB. IS@- :D5)9:)5AD)

Sample problem:

Example 1 If the acceleration of an ob7ect is given by

. Nind the ob7ects velocity and position functions given

that the initial velocity is and the initial

position is .SolutionUe3ll "rst get the velocity. To do this all %well almost all' we need to do is integratethe acceleration.

To completely get the velocity we will need to determine the *constant+ ofintegration. Ue can use the initial velocity to get this.


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The velocity of the ob7ect is then,

Ue will "nd the position function by integrating the velocity function.

Vsing the initial position gives us,

So, the position function is,

Example 2 Nor the ob7ect in the previous eample determine the tangential andnormal components of the acceleration.Solution

There really isn3t much to do here other than plug into the formulas. To do this we3llneed to notice that,


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Let3s "rst compute the dot product and cross product that we3ll need for theformulas.

-et, we also need a couple of magnitudes.


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The tangential component of the acceleration is then,

The normal component of the acceleration is,


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PR#$%&': (Normal and tangential components)

%9' Nind velocity, speed and acceleration of a particle whose tra7ectory is given byr%t' $ %5 cos%Ft', t, 5 sin%Ft''.

%F'Suppose a particle moves with a constant speed. Show that the acceleration isnormal to the tra7ectory.


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%5' Nind the tangential and normal components of the acceleration if r%t' $ %t, cosF%t',sinF %t''.

%B'( water)skier is pulled forward by a rope at u $ ;.9 m sW9 , following a curvedpath which can be approimated by a circular arc of radius r $ FA m. Nind thevelocity and acceleration components if X $ BY

%' Starting from rest, a motorboat travels around a circular path of Z $ A m at aspeed that increases with time, v $ %A.F tF' m[s. Nind0 The magnitudes of the boat3svelocity and acceleration at the instant t $ 5 s. ?lan0 The boat starts from rest %v $A when t $ A'. 9' Ealculate the velocity at t $ 5s using v%t'. F' Ealculate thetangential and normal components of acceleration and then the magnitude of theacceleration vector.


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Solution:

%9' Let us "rst "nd the derivatives. The velocity is

v%t' $ r A %t' $ %WD sin%Ft', , D cos%Ft''.

The acceleration is

a%t' $ v A %t' $ %W9F cos%Ft', A, W9F sin%Ft''.

Speed is the magnitude of the velocity vector, so

v $ \v%t'\ $ q 5D sinF %Ft' F 5D cosF %Ft' $ ] 5D F $ ] D9.

-ote, that the speed is independent of time, so by the net problem the

acceleration is normal to the tra7ectory.%F' Since the speed is constant, its square is constant as well0

v ^ v $ \v%t'\ F $ const.

>ierentiating this equation with respect to t, we get

Fa ^ v $ A,

i.e. acceleration is orthogonal to velocity.

%5' The derivatives are

v%t' $ %9, WF cos%t' sin%t', F sin%t' cos%t'' $ %9, W sin%Ft',sin%Ft''

and a%t' $ %A, WF cos%Ft', F cos%Ft''.

The tangential component of the acceleration is


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To "nd the normal component of the acceleration, compute the cross product "rst0

The normal component of the acceleration can be found using the formula

%B'


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. Solution0 9' The velocity vector is v $ v ut , where the magnitude is given by v $%A.FtF' m[s. (t t $ 5s0

v $ A.FtF $ A.F%5'F $ 9.C m[s

F' The acceleration vector is a $ atut anun $ vut %vF[Z'un.

Tangential component0 at $ v $ d%.FtF'[dt $ A.Bt m[sF

(t t $ 5s0 at $ A.Bt $ A.B%5' $ 9.F m[sF .

-ormal component0 an $ vF[Z $ %A.FtF'F[%Z' m[sF

(t t $ 5s0 an $ G%A.F'%5F'HF[%A' $ A.ADBC m[sF

The magnitude of the acceleration is

a $ G%at'F %an'FHA. $ G%9.F'F %A.ADBC'FHA. $ 9.FA m[sF


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Problem: (projectile motion)

9. ( ball is thrown o a balcony and has an initial velocity of 9C m[s hori1ontally.>etermine the position of the ball at t 9.A s, F.A s, 5.A s, and B.A s.

F. ( child travels down a water slide, leaving it with a velocity of B.F m[shori1ontally, as in Nigure ;%a'. The child then eperiences pro7ectile motion,landing in a swimming pool 5.F m below the slide. Nor how long is the childairborneY


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5. ( helicopter, travelling hori1ontally, is CF m above the ground. The pilotprepares to release a relief package intended to land on the ground :D mhori1ontally ahead. (ir resistance is negligible. The pilot does not throw the

package, but lets it drop. Uhat is the initial velocity of the package relative tothe groundY

B. ( golfer strikes a golf ball on level ground. The ball leaves the ground with aninitial velocity of BF m[s G5F_ above the hori1ontalH. The initial conditions areshown in Nigure 9F. If air resistance is negligible, determine the ball3shori1ontal range %assuming that it lands at the same level from which itstarted'

. ( ball is tossed from an upper)story window of a building. The ball is given aninitial velocity of Cm[s at an angle of FAA below the hori1ontal, where itstrikes the ground 5 seconds later.a. ow far hori1ontally from the base of the building does the ball strike thegroundYb. (t what height was the ball thrownY

Solution:

9. Let the direction be to the right and the y direction be downward %which isconvenient since there is no upward motion' %see Nigure D%a''. ori1ontally%constant vi '08i$ 9C m[st $9.A s$ Y$ vit


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$%9C m[s'%9.A s' $9C m

F. ori1ontally %constant vi '0vi $B.F m[s$ Yt$ Y8ertically %constant ay '0viy $A y$5.F may $g $:.C m[sF vfy $Y

T$YThe hori1ontal motion has two unknowns and only one equation vit. Uecan analy1e the vertical motion to determine t 0

5. ori1ontally %constant vi '0 $:D mt$ Yvi$ Y8ertically %constant ay '0viy $A m[s y$ CF may $g$ :.C m[sF t $Y


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B. Ue begin by "nding the hori1ontal and vertical components of the initialvelocity.8i$vicos v viy $visin v $ %BF m[s'%cos 5F_' $%BF m[s'%sin 5F_'vi $5D m[s viy $FF m[s

ori1ontally %constant vi '0vi $5D m[s$ Y

t$ Y8ertically %constant ay '0(y$ )g$) :.C m[sF y $Aviy $FF m[s t $Yvfy $)FF m[sSince the hori1ontal motion has two unknowns and only one equation, we canuse the vertical motion to solve for t0

Therefore, the ball was hit at t A and the ball lands at FF m[s J B.: m[sF t A.Solving for t, we "nd that t B. s, which we can use to "nd the hori1ontalrange. $vit$ %5D m[s'%B. s' $9.D 9AF m

The hori1ontal range is 9.D 9AF m.

. %a' $ vit $ %C.AA cos FA.A_'%5.AA'$ FF.D m

%b' Taking y positive downwards,y $ vyit 9 F gtF $ C.AA%cos FA.A_'5.AA 9 F %:.CA'%5.AA' F

$ F.5 m

Normal and Tangential Components

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