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