# what are the sources of energy loss for a ball rolling down slope?

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?
• 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. Mar 13 '19 at 14:05
• 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) Mar 13 '19 at 14:49
• I have $(1/2) (2/5) + (1/2)=7/10$. This is because $E_{kin\_rot} = (1/2) I \omega^2=(1/2) [(2/5) M r^2] (v/r)^2=(1/5)M v^2$ Mar 14 '19 at 3:05
• Mar 14 '19 at 3:11

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.