# finding velocity along a curve with kinematic equations using time

(i'm "not" looking for coding help. i need help setting up the math.)

i'm writing a program for a physics class to find the velocity of an object across a random curve. where the only force acting on it is gravity (g).

the goal is to compare the found results from the kinematic equations. with the value found from conservation of energy.

i'm currently finding the instantaneous velocities at each point along the curve and summing the values. so basically this V^2 = 2*A*S. where A = sin(theta)*g , theta = arctan(dy/dx) , and S = the arc length. i sum the V^2s for each interval and take its root. this is giving the right answer.

how ever my professor wants me to use an equation that uses time as well. and this is the problem i don't have a clue how to do this. as i'm not sure how to determine the change in acceleration with respect to time across a fixed path (example a curve that looks like y=x^2).

i could be misunderstanding what he is requesting. however even if i am, if this is possible i would like to know how it's done.

i think this deals with jerk but i don't know how i would find a value to fit a fixed curve. i've looked for jerk examples and found some that have been helpful but none that deal with following a path along some surface.

as a side note i'm already defining my curves parametrically as it allowed me to get around the problem of "dx = 0". and just in case that turns out to be the way to solve this.

i thought i could numerically integrate (r = x +v*t+.5*a*t^2) by finding the instantaneous acceleration for each time interval and using it to find V and then the change in position. however i'm not always getting answers that make sense no mater the size of delta t.

here's a picture

thank you all for your help.

chris

• Are you thinking of doing this by numerical integration? For an arbitrary curve it seems like the only possible way - at which point time will show up in your equations if you integrate for a small fixed time step. Could that be what you are after? – Floris Nov 30 '15 at 1:30
• that is what i've been thinking i need to do. it's what i have been doing with the first method already. however i've not been able to get anything that makes sense. – chris Nov 30 '15 at 1:39

So if the path is described parametrically with $\vec{r}(q) = (x(q),y(q))$ you need to define the tangent and normal vectors in order to describe the motion $\ddot{q}$ and the reaction force $F$.

The kinematic velocity vector is $$\vec{v}(q,\dot{q}) = (\dot{x},\dot{y}) = ( \frac{{\rm d}x}{{\rm d}q} \dot{q}, \frac{{\rm d}y}{{\rm d}q} \dot{q} ) = (x' \dot{q}, y' \dot{q})$$

where the $\square{ }'$ notation is derivative with the parameter $q$ and $\dot{\square}$ derivative with time. So speed is $v(q,\dot{q}) = \dot{q}\sqrt{x'^2+y'^2}$, or the inverse when speed is known, the parameter varies by $\dot{q} = \frac{v}{ \sqrt{x'^2+y'^2}}$

The tangent vector is $$\vec{e}(q) = ( \frac{x'}{\sqrt{x'^2+y'^2}}, \frac{y'}{\sqrt{x'^2+y'^2}})$$

and the normal vector $$\vec{n}(q) = ( -\frac{y'}{\sqrt{x'^2+y'^2}}, \frac{x'}{\sqrt{x'^2+y'^2}})$$

Finally the kinematic acceleration vector is

$$\vec{a}(q,\dot{q},\ddot{q}) = ( x' \ddot{q} + x'' \dot{q}^2, y' \ddot{q} + y'' \dot{q}^2 )$$

The acceleration vector can be decomposed into tangential acceleration $\vec{a}_T = \dot{v} \vec{e}$ and lateral acceleration $\vec{a}_L = \frac{v^2}{\rho} \vec{n}$ where $\rho$ is the instantaneous radius of curvature. From the above is found that

$$\vec{n} \cdot \vec{a} = \frac{v^2}{\rho} = \dot{q}^2 \frac{x'^2+y'^2}{\rho}$$ $$\rho(q) = \frac{ (x'^2+y'^2)^{\frac{3}{2}}}{y'' x'-y' x''}$$

Now you can state the equations of motion

$$F \vec{n} + m \vec{g} = m \vec{a}$$

which results in

\boxed{ \begin{aligned} F &= m \frac{ \dot{q}^2 (y'' x'-y' x'')+g x'}{\sqrt{x'^2+y'^2}} \\ \ddot{q} &= - \frac{ \dot{q}^2 (x' x'' + y' y'')+g y'}{x'^2+y'^2} \end{aligned} }

## Example

A parabolic path with $\vec{r} = (q,q^2)$ has $\vec{v} = (\dot{q},2q \dot{q})$ and $\vec{a} = (\ddot{q}, 2 q \ddot{q} + 2 \dot{q}^2)$. This is due to the partial derivatives \begin{aligned} x & = q & y & = q^2 \\ x' &= 1 & y' &= 2 q \\ x'' &= 0 & y'' &= 2 \end{aligned}

The formulas for reaction force and motion above are \begin{align} F&= m \frac{g + 2 \dot{q}^2}{\sqrt{1+4 q^2}} \\ \ddot{q} &= -q \frac{2(g+2 \dot{q}^2)}{1+4 q^2} \\ v &= \dot{q} \sqrt{1+4 q^2} \end{align}

The small motion approximation ($q \ll 1$, $\dot{q} \rightarrow 0$) results in

$$F = m g$$ $$\ddot{q} = -2 g q$$

which is simple harmonic motion with period $\tau = \frac{\sqrt{2} \pi}{\sqrt{g}}$.

• If you only care about motion along the curve can't you significantly simplify this? Just put everything in terms of the parameter along the curve and know the acceleration along the curve as the component of gravity resolved along the curve. While there may be acceleration along the other axis you know you stay on the curve. Or am I missing something? – Floris Nov 30 '15 at 3:06
• This is as simple as it gets with the general curve scenario. See physics.stackexchange.com/a/83592/392 for an example. For a specific shape curve, the equations simplify. – ja72 Nov 30 '15 at 4:20
• Thats what the $\ddot{q}=\ldots$ equation is. All the derivatives $x'$ $y'$ $x''$ and $y''$ are only a function of the parameter $q$. – ja72 Nov 30 '15 at 4:21
• Floris if i understand your comment correctly, that is what i'm currently doing. basically finding delta S along the curve and combining it with the component of gravity at each point to get the velocity. the trick is trying to find velocity based on delta T. – chris Nov 30 '15 at 21:27
• ja72 i'll try to work through your example and post the result to see if i found the correct answer. hopefully i'll be close to correct – chris Nov 30 '15 at 21:30