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So I've learned for any slope with a sliding object, say a box, the speed at the bottom only depends on the height it was dropped from because of conservation of energy, assuming absence of friction and all of that. Indeed, this makes a lot of sense to me if the slope is perfectly linear.

However, I'm not quite sure why this works if the slope has a different shape such that it is curved, say a quarter pipe. I understand that conservation of energy still applies here, but because the slope is curved the box's angle must also change during the slide, meaning that some of the potential energy should go into rotational kinetic energy. Because of this, the linear kinetic energy would have to decrease, and the box would have a smaller speed at the bottom. However, this seems to lead to a contradiction of the fact that the speed only depends on the height, since by my logic the shape of the ramp would also be a factor.

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  • $\begingroup$ You really want to look at this: en.wikipedia.org/wiki/Brachistochrone_curve $\endgroup$
    – Gert
    May 2, 2021 at 19:04
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    $\begingroup$ Are you thinking of a situation where the object continues to spin after it reaches the bottom (so that its final rotational kinetic energy is nonzero)? $\endgroup$ May 2, 2021 at 20:02

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Short answer: if the object is a point mass, then there will be no rotation. If the object is large, then it will rotate about its center of mass.

Long answer:

I will assume no friction to simplify the calculations.

Consider the following setup, where the blue square of mass $M$ slides down a circular path with center $O$:

enter image description here

Suppose the moment of inertia of the square about its center of mass $G$ is $I_G$. Then, by the parallel axis theorem, the moment of inertia about point $O$ is $I_O=I_G+MR^2$.

We can now calculate the kinetic energy of the box. Suppose the box rotates about point $O$ with angular speed $\omega$. We will calculate the kinetic energy about point $O$, which will be $$\mathrm{KE} = \dfrac12 I_O \omega^2$$

Substituting our relationship for $I_O$, we have, $$\mathrm{KE} = \dfrac12 \left(I_G + MR^2\right)\omega^2$$

which develops to $$\mathrm{KE}=\dfrac12 I_G \omega^2 + \dfrac12 MR^2 \omega^2$$

Since this is "prefect" circular motion, we know that $v=\omega R$ (where $v$ is the speed of the center of mass) therefore,

$$\mathrm{KE} = \underbrace{\dfrac12 I_G \omega ^2}_\mathrm{rotational \ KE} + \underbrace{\dfrac12 Mv^2}_{\mathrm{translational \ KE}}$$

The moment of inertia of an symmetrical object will usually be expressed of the form $I=CMd^2$ where $d$ is some geometric measurement (like a side length) and $C$ is a constant, usually $C\leq1$.

Therefore, you can now approximate, based on the size of the object $d$, how much rotational energy with respect to kinetic energy the object will have. Knowing that $I_G\approx CMd^2$ and that $\omega = v/R$, we have that,

$$ \mathrm{KE_{rot}} = \dfrac12 CMd^2 \left( \dfrac vR \right) ^2$$

Thus, the rotational KE is $$\mathrm{KE_{rot}} = \dfrac{CMd^2 v^2} {2R^2}.\tag{*}$$

A couple of noteworthy observations: the rotational KE is inversely proportional to $R^2$, and proportional $d^2$. So, if you're dealing with a point mass, you assume a dimensionless point and $d\to 0$ (point masses are assumed to have an $I_G=0$), so you get that that the rotational KE is also zero, and then you get $$\mathrm{KE_{point \ mass}} = \dfrac12 Mv^2$$

As pointed out in the comments by @Michael Seifert, using the $(*)$ equation, you can compare the order of magnitude of the object's rotational KE to its translational KE. In cases with small objects and small radii of curvature (the radii of curvature is $\kappa = 1/R$, so small radius of curvature describes flatter surface, whereas large radius of curvature describes a curved surface), you will notice that the rotational KE is very small compared to the translational KE, which, again, is why we can approximate the object as a point mass.

Let's put some numbers to it. Suppose $M=1.0 \ \rm kg$ with $d=10 \ \rm cm$, and for a cube rotating through its center, $C=1/6$. Say we're on a hill with $R=15 \ \rm m$. The block slides down a height of $h=3.0 \ \rm m$, so its gain in kinetic energy is $Mgh$. The resulting speed will be $v=7.7 \ \rm m/s$, and thus: the translational KE is $\rm KE_{trans} = 29 \ J$, whereas the rotational KE is $\rm KE_{rot}=2.1\times 10^{-6} \ J$, which is therefore negligible.

Next, I will offer a conceptual explanation as to why the object will rotate. Let's zoom in on the block, and imagine 3 points on the block as follows:

enter image description here

Each of these points is a longer distance away from the center $O$, therefore, by $v_i=\omega R_i$, the points on the block that are furthest from $O$ move faster, and thus have greater speed.

Now, if you suddenly imagine that no forces act on the block (i.e. it suddenly teleports into a vacuum), then, you realize that since some parts of the block move slower and some move faster, the block will be rotating. In other words, it would have gained rotational kinetic energy about its CoM.

However, if the dimensions of the block are assumed to be very small (such that $I_G\to 0$, then one can neglect this rotation). This is often assumed in entry-level physics courses.

With friction, this can get even more interesting -- if we had a ball, it would begin rotating about its center due to friction (as is the case with rolling on an incline).

This is also worth pointing out: make sure to not double count your kinetic energy.


Finally, back to your conservation of energy question: yes, for non point-like objects following a curved path (e.g. if their orientation changes) they would have gained some amount of rotational kinetic energy.

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    $\begingroup$ It might be worth pointing out explicitly that for an object of size $d$, $I_G \sim M d^2$. This means that the rotational kinetic energy is something like $M v^2 (d^2/R^2)$, which gives you a way to compare (order-of-magnitude-wise) how much rotational vs. how much translational kinetic energy a block sliding on a curve will have. $\endgroup$ May 3, 2021 at 18:04
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    $\begingroup$ @Michael Seifert thank you for the suggestion. I updated the post. $\endgroup$
    – user256872
    May 3, 2021 at 20:29

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