# Intrinsic coordinates kinematics problem

#### Santilopez10

Homework Statement
A particle moves along a defined curve so that its acceleration tangential component: $a_t=-ks$, where k is a constant and s denotes the arc distance respect to a point Q.
a) Find an expression for the velocity as a function of s.
b) Supposing that at Q its velocity equals 3.6 m/s and at A (s=5.4 m) 1.8 m/s, find k and its curvature radius at A knowing that the acceleration has magnitude 3.0 m/s^2.
c) Find at which distance the particle inverts its movement.
Homework Equations
$\vec v= v e_t$
$\vec a= a_t e_t + a_n e_n = \frac{dv}{dt} e_t + \frac {v^2}{\rho} a_n$
So I know that $a_t = \frac{dv}{dt}=-ks$ and $\frac{dv}{dt}=v\frac{dv}{ds}$ then: $$v dv=-ks ds \rightarrow (v(s))^2=-ks^2+c$$ and using my initial conditions it follows that: $$(3.6)^2=c \approx 13$$ and $$(1.8)^2=13-5.4k \rightarrow k=1.8 \rightarrow (v(s))^2=13-1.8s$$
What bothers me is finding at which point it turns, I am having trouble even getting started, any help would be appreciated.

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#### Delta2

Homework Helper
Gold Member
I think it asks to find the s for which velocity changes sign that is from positive becomes negative. I guess the distance s for which $v(s)^2=-ks^2+c=0$.

btw I think you have a mistake in the calculation of k, shouldn't it be $(1.8)^2=13-(5.4)^2k$?

#### Santilopez10

I think it asks to find the s for which velocity changes sign that is from positive becomes negative. I guess the distance s for which $v(s)^2=-ks^2+c=0$.

btw I think you have a mistake in the calculation of k, shouldn't it be $(1.8)^2=13-(5.4)^2k$?
You are right about my mistake! Glad you could help me. One question though, does this represent a physical situation ? Considering that when the rhs is negative square root becomes complex

#### Delta2

Homework Helper
Gold Member
Well to be honest I found it hard to answer your question, but I THINK that the physical interpretation of the imaginary velocity we get for $s>\sqrt\frac{c}{k}$ is that the particle moves through the curve in the opposite direction.

EDIT: On second thought I believe imaginary velocity means that the particle can never be there and that its movement is confined for s between $-\sqrt\frac{c}{k}<s<+\sqrt\frac{c}{k}$

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#### Santilopez10

Well to be honest I found it hard to answer your question, but I THINK that the physical interpretation of the imaginary velocity we get for $s>\sqrt\frac{c}{k}$ is that the particle moves through the curve in the opposite direction.

EDIT: On second thought I believe imaginary velocity means that the particle can never be there and that its movement is confined for s between $-\sqrt\frac{c}{k}<s<+\sqrt\frac{c}{k}$
Maybe it could model a particle movement in an ellipse or circle.

#### Delta2

Homework Helper
Gold Member
Maybe it could model a particle movement in an ellipse or circle.
Maybe, we don't have any information about the normal force so we don't know what curve shape it is and whether it is a closed curve. However i believe that is not the point here. The point is that the particle is restricted within an interval. Like a particle that is restricted within a semi circle or a semi ellipse and moves back and forth.

To prove this, using the definition of acceleration $a=\frac{d^2s}{dt^2}=-ks$ and solving for s(t) we find that s(t) is sinusoidal. So indeed the particle moves back and forth (along a curve not necessarily in a straight line).

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"Intrinsic coordinates kinematics problem"

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