Solving Acceleration Problem: Decomposing into Tangential & Normal Components - Prof. Kurt, Assignments of Mathematics

Solutions to two problems involving the decomposition of acceleration into tangential and normal components in 2d and 3d space. The problems involve calculating the velocity, speed, tangent vector, normal vector, and curvature of a particle moving in a given trajectory, and finding the acceleration of a particle moving along a 3d curve.

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Pre 2010

Uploaded on 08/19/2009

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Acceleration Problem Examples
Here’s a couple typical problems involving the decomposition of acceleration into tangen-
tial and normal components.
1. Problem: (From class handout on Wednesday 4/1.) Suppose a particle moves in 2D
space with position r(t) =< R cos(qt/R), R sin(qt/R)>at time t, where qand Rare
some positive constants. Compute each of v(t),kv(t)k,T(t),a(t),N(t), and κ(t) (as
functions of t, though some may be constant). Write out aT, aN,aT,and aN. Then
compute aTand aNdirectly from the vector projection formulas and compare.
Solution: It’s easy to compute directly that
v=<sin(qt/R),cos(qt/R)>
kvk= 1
a=<cos(qt/R)/R, sin(qt/R)/R >
T=<sin(qt/R),cos(qt/R)>(same as v)
N=<cos(qt/R)/R, sin(qt/R)/R >
κ= 1/R.
The formula a=d(kvk)
dt T+κkvk2Nthen yields
a= 0T+1
RN=<cos(qt/R)/R, sin(qt/R)/R >
which is of course the same as above. Note that aT=d(kvk)
dt = 0 since the speed is
constant—there is no tangential acceleration here, and so aT=0(the zero vector).
Also, aN= 1/R; all of the acceleration is normal, due to changing direction, so that
a=aN.
You can also compute directly that
aT=a·v
v·vv=0
1v=0
aN=aaT=a=<cos(qt/R)/R, sin(qt/R)/R > .
2. Another Problem: A particle moves along a path parameterized by the 3D curve
x=u1, y =u2+ 2, z =u, moving in the direction of increasing x; note that
here uis just some parameter, not necessarily time. When the particle is at the point
(0,3,0) (corresponding to u= 1) it has a speed of 2 meters per second, and its speed
is increasing at a rate of 5 meters per second squared. Find the acceleration of the
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Acceleration Problem Examples

Here’s a couple typical problems involving the decomposition of acceleration into tangen- tial and normal components.

  1. Problem: (From class handout on Wednesday 4/1.) Suppose a particle moves in 2D space with position r(t) =< R cos(qt/R), R sin(qt/R) > at time t, where q and R are some positive constants. Compute each of v(t), ‖v(t)‖, T(t), a(t), N(t), and κ(t) (as functions of t, though some may be constant). Write out aT , aN , aT , and aN. Then compute aT and aN directly from the vector projection formulas and compare.

Solution: It’s easy to compute directly that

v = < − sin(qt/R), cos(qt/R) > ‖v‖ = 1 a = < − cos(qt/R)/R, − sin(qt/R)/R > T = < − sin(qt/R), cos(qt/R) > (same as v) N = < − cos(qt/R)/R, − sin(qt/R)/R > κ = 1 /R.

The formula a = d(‖ dtv ‖)T + κ‖v‖^2 N then yields

a = 0T +

R

N =< − cos(qt/R)/R, − sin(qt/R)/R >

which is of course the same as above. Note that aT = d(‖ dtv ‖) = 0 since the speed is constant—there is no tangential acceleration here, and so aT = 0 (the zero vector). Also, aN = 1/R; all of the acceleration is normal, due to changing direction, so that a = aN. You can also compute directly that

aT =

a · v v · v

v =

v = 0 aN = a − aT = a =< − cos(qt/R)/R, − sin(qt/R)/R >.

  1. Another Problem: A particle moves along a path parameterized by the 3D curve x = u − 1 , y = u^2 + 2, z = −u, moving in the direction of increasing x; note that here u is just some parameter, not necessarily time. When the particle is at the point (0, 3 , 0) (corresponding to u = 1) it has a speed of 2 meters per second, and its speed is increasing at a rate of 5 meters per second squared. Find the acceleration of the

1

particle.

Solution: Start with

a =

d(‖v‖) dt

T + κ‖v‖^2 N. (1)

We are told that d(‖ dtv ‖)= 5 and ‖v‖ = 2.

We can find κ from the parameterization, as

κ =

‖dT/du‖ ‖dr/du‖

where T = d dur /‖ (^) dudr ‖. Notice I’m using the u parameterization of the curve (rather than the position of the particle at time t, which I don’t know!) Also, I’m writing dr/du, not v, because if u isn’t time then dr/du isn’t velocity! (If we knew r(t), the position of the particle at any time, then we could that.) You can work out (feel free to use Maple) that

κ =

(1 + 2u^2 )^3 /^2

(not too ugly) and at u = 1 this gives curvature κ = 1/ 33 /^2 at the point of interest.

We already computed T = d dur /‖ (^) dudr ‖ above, which at u = 1 gives

T(1) =< 1 /

Note that we always have TWO choices for T (there are always two tangent vectors to the curve, and they point in opposite directions.) We want the one that points in the direction this particle is moving (increasing x coordinate) which is exactly the one I wrote out (it has positive x component.)

To compute N use the formula N = (dT/du)/‖dT/du‖, which gives

N =< −u/

1 + 2u^2 , 1 /

1 + 2u^2 , u/

1 + 2u^2 >.

At u = 1 this gives N(1) =< − 1 /

Finally, fill in d(‖ dtv ‖)= 5, ‖v‖ = 2, κ = 1/ 33 /^2 , and the vectors T(1), N(1) into equation (1) and find a ≈< 1. 60 , 4. 53 , − 1. 60 >.