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I am experimenting in lab with an iron sphere rolling down a smooth aluminum ramp.

The final velocity is smaller than the one predicted assuming energy conservation (see calculations below), which implies some energy is lost.

At the top of the ramp (height $H$) the sphere is at rest and energy $E_{tot}=E_{pot}=mgH$. At the end of the ramp the sphere has height 0 and so all potential energy, in absence of lossed, should be converted in kinetic energy, which has two components (rotation and translation). Hence I get $mgH=\frac{7}{10}mv^2\implies v=\sqrt{\frac{10}{7}gH}$.

Also, using the Euler-Lagrange equations I obtain that the balls should have constant acceleration: $$a=\frac{5}{7} g\sin{\alpha}$$ and the time it should take to get to the bottom og the ramp is: $$t = \sqrt{\frac{14}{5} \frac{l}{g\sin{\alpha}}}$$ where $\alpha$ is the angle the incline makes with the horizonal and $l$ is the length it travels down the incline.

What are the mechanical causes of the energy loss in this experiment? I am interested in a qualitative conceptual answer, not in formulas. Searching through various posts I have found so far the following:

  • air friction (not so important here I think, as velocity is not very high and iron has a large density, so a small surface)
  • rolling friction: the ball and the plane are not perfectly rigid, they deform a little so that the ball is always climbing over a small hump, causing a small resultant force which opposes motion (suppose this is quite small for an iron ball and aluminum ramp)
  • slippage friction: there is no single point of contact but a small but finite area of contact and the velocity of all those points of contact is not equal to zero. So in realtiy there is no such thing as rolling without slipping.

What else?

  • perhaps the surfaces are not perfect so it makes micro bounces, instead of a perfect rolling?
  • perhaps the ball sticks to the surface at microscopic level?
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    $\begingroup$ Are you taking into account the kinetic energy due to rolling? It might help to show how you are using energy conservation just to make sure there are no mistakes there first. $\endgroup$ – Aaron Stevens Mar 13 at 14:05
  • $\begingroup$ Along with friction due to uneven surfaces(as you mentioned).The center of mass of the ball can also cause change in your predictions because it would change the value of the initial and final potential energies.(If you are using a small ball however, this effect isn't potent) $\endgroup$ – Vaishakh Sreekanth Menon Mar 13 at 14:49
  • $\begingroup$ Updated the question adding clarifications on my calculations, which assume energy conservation. $\endgroup$ – Fabio Mar 14 at 2:29
  • $\begingroup$ I have $(1/2) (2/5) + (1/2)=7/10$. This is because $E_{kin\_rot} = (1/2) I \omega^2=(1/2) [(2/5) M r^2] (v/r)^2=(1/5)M v^2$ $\endgroup$ – Fabio Mar 14 at 3:05
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – Aaron Stevens Mar 14 at 3:11
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If you can hear the ball rolling down the ramp, then it means that some of the ball's energy is being lost by the radiation of sound waves off the ramp. If the ball is iron and the ramp is wood, this is almost inevitable.

Furthermore, for the ramp to excite vibrations in the air, it means that the ramp is deflecting under the influence of the rolling ball, and that means that energy is being dissipated in the wood itself.

Both of these energy loss mechanisms are difficult to predict and model.

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When the sphere rolls down the ramp it acquires both rotational $\frac {Iω^2}{2}$ and translational $\frac {mv^2}{2}$ kinetic energy, where $I$ is the mass moment of inertia, $ω$ is the angular velocity at the end of rolling, $m$ is the mass and $v$ is the linear velocity of the center of mass at the end of rolling.

Neglecting energy losses due to friction on the ramp and air drag, the sum of these kinetic energies should equal the change in potential energy, $mg\Delta h$ where $\Delta h$ is the change in height of the center of mass of the ball.

Hope this helps.

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