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# Argghh! Dynamics Question? Hints given - Just don't know how to do the question?

The acceleration history of a moving object, that starts from rest at time zero, is recorded against time as follows:

ax = t / 2

ay = (1 - 0.2 t)

where ax and ay are the x- and y- components of the acceleration in m/s^2 and t is the time in s.

Find the instantaneous radius of curvature (m) when the time is T.

T[s] = 7.2;

Here are some notes I have been given but don't know how to interprest them in the question:

http://s589.photobucket.com/albums/ss337/johnnigan...

Also here is an image of the question asked:

http://s589.photobucket.com/albums/ss337/johnnigan...

Any help is appreciated as I have no idea how to solve this.

### 2 Answers

- EMLv 71 decade agoFavorite Answer
The steps that were given provide a sound approach, albeit a slightly longer method than I would take. Those steps can be reduced to:

ρ = |v|^2 / |ā x êv| = |v|^3 / [(ax)(vy) - (ay)(vx)]

With my simplification, you'll only need to find vx and vy by integrating ax and ay with respect to t and evaluating from 0 < t < 7.2s, the magnitude of the velocity = sqrt[(vx)^2 + (vy)^2], and the evaluations of ax and ay at t = 7.2s.

Following the steps you were given will get you there, too; you'll just have to do additional calculations.

- JacyLv 71 decade ago
ρ = |V|²/a(normal)

ā = ia(x) + ja(y) = it/2 + j(1- 0.2t) = i(7.2/2) + j[1 – 0.2(7.2)]

ā = i3.6 – j0.44 m/s²

ύ = ∫ â dt = iv(x) + jv(y) = it²/4 + j(t – 0.1t²)

ύ = i(7.2²/4) + j(7.2 – 0.1(7.2)²)

ύ = i(12.96) + j(2.02), |ύ| = √(12.96)² + (2.02)² = 13.116 m/s

|a(tang)| = ā•ê,

where, ê = ύ/|ύ| = [i(12.96) + j(2.02)]/ √(12.96)² + (2.02)²

ê = i0.988 + j0.15

|a(tang.)| = [i3.6 – j0.44]•[i0.988 + j0.15]

|a(tang.)| = 3.6(0.988) – 0.44(0.15) = 3.491 m/s²

|ā| = √(ax)²+(ay)² = √3.6²+0.44² = 3.627 m/s²

|a(norm.)| = √{|ā|² - a(tang.)²}

|a(norm.)| = √{3.627² - 3.491²} = √0.9665 = 0.983 m/s²

ρ = |V|²/a(normal) = 13.116²/0.983 = 175 m