How can I find the instantaneous acceleration and velocity of an object sliding down a curve?

I know the following information:

  • The curve follows the shape of $e^{-ax}$
  • The object will always encounter non-negligible forces of weight, normal force, and friction
  • a(t) = v'(t) = x"(t)
  • objects are all alloys of copper, nickel, and zinc
  • the curve is made of sheet aluminum

I only have some knowledge of calculus and physics and I am not sure how to go about solving this problem. If any additional info is needed let me know, I wasn't sure exactly what would be needed.

Note This question involves calculating instantaneous velocity of an object on a curve in 2D at various intervals of time. I attempted using the x and y components to find the average velocity during my first attempt, and also attempted to solve for acceleration using a summation of forces and then integrating a(t) to get v(t) however I was wondering if a) my methodology was correct and b) if there was a more mathematically elegant solution


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  • $\begingroup$ Use Hamiltonian or Lagrangian mechanics. That makes the problem simple. Of course, you need to go study how to do that. Get a copy of John Taylor's Classical Mechnics. $\endgroup$ – Bill N Dec 28 '16 at 17:40
  • 2
    $\begingroup$ @BillN The Lagrangian method is not so simple when friction is involved. $\endgroup$ – garyp Dec 28 '16 at 18:18
  • $\begingroup$ You will need to specify the coefficient of friction, and the initial conditions. Then you can use F=MA , together with the fact that the downward force is the force of gravity minus the component of friction in the downward direction. The lateral force is due only to friction (lateral component). Note that the direction of friction will be along the curve and will depend on x. Note also that friction magnitude will vary with the component of gravitational force normal to the surface. $\endgroup$ – Lewis Miller Dec 28 '16 at 18:19
  • $\begingroup$ @LewisMiller ok, I thought the materials would be enough to find the coefficient online but i will add it to the given info. Correct me if im wrong, but you're saying i can calculate this simply with the summation of forces on the x and y? If so how can I turn that into a single function as either x(t), v(t), or a(t)? $\endgroup$ – GracefulLemming Dec 28 '16 at 18:28
  • $\begingroup$ @garyp True. One usually encounters Lagrangian on the way to Hamiltonian, so I had this internal compulsion to mention it. :) $\endgroup$ – Bill N Dec 28 '16 at 18:32

$$\Delta E=W_{nc}$$

At any point on the curve, considering an infinitesimal displacement $dx$ on the x axis (right hand Cartesian x-y coordinate system), from calculus, we have:

$$dy=f'(x)dx\ ,\ ds=\sqrt{1+f'(x)^2}dx$$

Where $ds$ is the infinitesimal displacement of the curve.

The work done by friction (The only non-conservative force here) during the displacement $ds$ is:

$$W_{nc}=-\mu ds$$

Where $\mu$ is the coefficient of friction between the ball and the surface of the curve.

The change of Mechanical energy is, then:

$$\Delta E=\frac12 m((v+dv)^2-v^2)-mgdh$$

Where $m$ is the mass of the ball and $dh$ is the infinitesimal (positive) change in height. If we assume:

$$\forall x;f'(x)\lt0$$

Which assures that the ball won't ever jump off the curve, we can say:


Then we get:

$$\Delta E=\frac12 md(v^2)+mgf'(x)dx=-\mu \sqrt{1+f'(x)^2}dx=W_{nc}$$

$$d(v^2)=-2\left(\frac{\mu}{m} \sqrt{1+f'(x)^2}+gf'(x)\right)dx$$

$$v(x)=\sqrt{-2\int_{x \text{ of start point}}^{x \text{ of end point}}\left(\frac{\mu}{m} \sqrt{1+f'(x)^2}+gf'(x)\right)dx}$$

The above is an explicit expression of $v(x)$.

In your case, $f(x)=e^{-ax}$, so plug in parameters. And give the above expression to Computer (for example Mathematica) to solve it for you numerically.


  • $\begingroup$ This is exactly what i was looking for, an elegant solution that has easy to understand math and physics behind it. Ill post the v(t) graph soon $\endgroup$ – GracefulLemming Dec 29 '16 at 6:21

You definitely do not want to solve this problem using Lagrangian or Hamiltonian mechanics. Given that you know the coefficient of static friction you can solve for the accelerations in the x and y directions etc.

What is important here is the geometry of the problem. We are interested in the direction tangent to the curve $y = e^{-ax}$, and the direction perpendicular to it. The direction tangent to the curve at point $x$ has the slope $f'(x)$, where $f'(x)$ is the derivative of $f(x)$ with respect to $x$. The direction normal to the curve at point $x$ has the slope $-1/f'(x)$. Now we proceed with Newton's 2nd Law:

$$ F_x = N\sin \theta - f_k \cos \theta, $$ $$ F_y = -mg+N\cos \theta + f_k \sin \theta, $$ where $f_k$ is the force of kinetic friction (given by $N\mu_k$), and $N$ is the normal force. We can solve for the normal force right away summing the forces along the direction perpendicular to the curve to find that $N = mg\cos \theta$. Now all that remains is to find what $\theta$ is. I defined $\theta$ above to be the angle subtending the force of gravity and the direction anti-parallel to the Normal force's direction. Once you do a bit more geometry you will be able to find expressions for $\sin \theta$ and $\cos \theta $ in terms of x. Now substituting these expressions into your 2nd law equations you essentially get: $$ m\ddot{x} = F(x), $$ $$ m \ddot{y} = G(x). $$ The key to solving the first equation is to recognize that $\ddot{x} = \dot{x}(d\dot{x}/dx)$. Using that fact you can solve for the velocity in the x direction as a function of x by just separating the differential equation and integrating. For the 2nd equation you must invert x to y and use the same trick--that is find $H(y)$ such that $m\ddot{y} = G(x) = H(y)$. We then write the double time derivative in y analogously to what we wrote for x and proceed as usual.

Hope I pointed you in a useful direction!


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