I want to solve the standard problem of a cylinder rolling down an inclined plane by only using conservation laws. In particular I want to use the equation:

$$E_{pot}=\frac{1}{2}mv^2+\frac{1}{2}I\omega ^2$$

where $m$ is the mass of the cylinder, $v$ the velocity of translation and $\omega$ the angular velocity. The moment of inertia $I$ of a cylinder rotating around the central axis is $\frac{1}{2}mr^2$ and I know the relation $v=\omega r$ $$\implies mgh=\frac{1}{2}mv^2+\frac{1}{2}(\frac{mr^2}{2}\frac{v^2}{r^2}) \\ \iff mgh=\frac{3}{4}mv^2 \\ \iff \boxed{v^2=\frac{4gh}{3}}$$

Is there any way I can get the acceleration $a$ from this result? Differentiating doesn't lead to anything here.

  • $\begingroup$ Well differentiation is how you get acceleration from velocity, you just have to know what $\dot{h}$ is and how it relates to other variables $\endgroup$ – Triatticus Dec 9 '18 at 17:32
  • $\begingroup$ The acceleration depends on the inclination angle, which you haven't specified. If you only know the height of the inclined plane and not its horizontal extent, you can't expect to find acceleration. $\endgroup$ – Ben51 Dec 9 '18 at 17:42
  • $\begingroup$ @ben51 The angle for which I want to find the acceleration is $\phi=30°$. Combining your comment and Triatticus comment, do I have to rewrite $h$ in terms of the angle? $\endgroup$ – Nullspace Dec 9 '18 at 17:50
  • $\begingroup$ @Triatticus If the plane has an angle $\phi$ then $h=(\sin{\phi}) \cdot L$. I am guessing $L$ (the distance traveled on the plane?) is changing with time right? $\endgroup$ – Nullspace Dec 9 '18 at 17:52
  • $\begingroup$ What you are obtaining is the velocity when the cylinder reaches the location of zero potential energy. The acceleration is simply $\g $. $\endgroup$ – nodarkside Dec 9 '18 at 18:05

If you take your answer and differentiate

$$2v v'={4g\over 3}h'$$

Of course $v'=a$ is the acceleration (I am using the apex to indicate differentiation) , but now you have the problem of.. what is $h'$?

If you call $h$ the height of the incline it does not make sense to differentiate (it would be $0$) it and the formula you found would just give you the velocity of the cylinder after leaving the incline. However if $h=h(t)$ is the height of the cylinder on the incline at a given time $t$ and then your formula for the velocity is valid at every timepoint and we can differentiate it. If you want to differentiate your energy, you need to write it in away which is valid at any instant $t$ rather than just at the end.

So now $h'$ is of course the velocity along the height of the incline, let me call that $v_z$ (to make it clear: in the direction of gravity, not of motion!).

So: $$2va={4g\over 3}v_z$$ Unfortunately this is only one component of the velocity. Luckily, because we know the geometry of the incline, and we know that, as the cylinder is rolling on the incline the velocity is directed parallel to the incline, by simple trigonometric projection, we also know that $v_z=v sin(\alpha)$ where $\alpha$ is the steepness of the incline.

Finally: $$2va={4g\over 3}v\sin(\alpha)$$ and by eliminating $v$: $$a={2\over 3}g\sin(\alpha)$$ which is the right result (double check with other methods for example here)

Another way to see this would have been to write the potential energy not in term of the height $h$ but of the position $x$ along the incline (so that $x'=v$) as $mgxsin(\alpha)+c$, where $c$ is a constant depending on where you put the $0$-value of the energy which disappears upon differentiation.

So you were close. When you do this kind of tricks (e.g. differentiating the energy) always make sure that your derivation holds at all times and that you are using the right coordinate systems or use vectors instead of scalars, you run the risk of loosing components otherwise, which are usually easier to handle with force diagrams (see video).


If we assume constant acceleration, then use the motion equation

$$s=s_0+\frac12 (v_0+v)t\,, $$

where presumably $s_0=0$ and $v_0=0$. The acceleration is then the simple fraction of velocity change over duration,

$$a=\frac{dv}{dt}=\frac{v-v_0} t\,.$$

The distance $s$ is the length of the slope. You'll have to find that with the help of the vertical height and the inclination angle. That should be a simple sine relation.


Once you have $v^2_{\rm final}= \frac{4gh}{3} $ you can use the constant acceleration kinematic equation $$v^2_{\rm final} =v^2_{\rm initial} + 2\, a \,s$$ with $v_{\rm initial} = 0$ and $h = s\,\sin \alpha$.

That kinematic equation is just the conservation of energy "in disguise" $$\frac 12 m v^2_{\rm final} -\frac 12mv^2_{\rm initial} =F_{\rm \parallel \, slope}\,s = m\, a \,s$$

  • $\begingroup$ That's also a cool way to do it and it follows straight from what I already calculated. Thanks! :) $\endgroup$ – Nullspace Dec 9 '18 at 19:52

This is a one-dimensional problem, so the simplest solution is as follows: Using

$$h = x\sin{\theta}$$

where $x$ is distance along the surface of the $\theta$-inclined plane, you have:

$$ mgx\sin{\theta} =\frac 3 4 m\dot x^2$$

so if you set:

$$ g' = \frac 2 3 g\sin{\theta} $$


$$ U = mg'x = \frac 1 2 m \dot x^2 $$

From here we know:

$$ -\frac{dU}{dx} = mg' = m{\ddot x}$$

so the acceleration is

$$ a = -g' \equiv -\frac 2 3 g\sin{\theta} $$


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