I saw the problem 1.213 of Irodov's Problems in General physics and modified it.

So the modified problem is a particle of mass $m$ is on the tip of cone of mass $M$ and volume $V$. What work needs to be performed by external agent in the process to get the particle from the tip to infinity against the gravitational force of the cone?

I've tried hard. But unsuccessful, any hint?

  • $\begingroup$ do you know how to perform double and triple integrals ? $\endgroup$ – Tofi Oct 22 '16 at 17:44
  • $\begingroup$ Please show your attempt. In order to advise you we need to know how far you got. $\endgroup$ – sammy gerbil Oct 22 '16 at 18:23

In order to know what work is needed, one way is to study the gravitationnal potential energy $E_p$ of $m$: in fact, if we consider that the $m$ is at rest in the beginning and in the end of the experience, then $W = \Delta E_p$. However, choosing the convention $E_p(\infty) = 0$, this relation becomes $W=-E_p$. So, first of all we will find a relation between the geometry of the cone and its volume; then, we will find the potential energy due to a infinitesimal slice of cone; finally, we will integrate this, which will gives us the total potential energy of $m$.

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The volume between $h$ and $h+dh$ is, forgetting the terms in $d^2h$ is $$dV = \pi r^2 dh$$ However, Thalès's theorem states that

$$ \frac{r}{R} = \frac{h}{H}$$

So $$V = \int_0^H dV = \int_0^H \frac{\pi R^2}{H^2}h^2dh = \frac{\pi}{3}R^2H$$

enter image description here

Now we will find the potential energy $dE_p$ related to the attraction of the slice delimited between $h$ and $h+dh$. First of all, we must know the potential energy $d^2E_p$ relate dto the attraction of a tiny ring starting at $\rho$ and ending at $\rho+d\rho$. This slice has a mass

$$ \delta m = \frac{2\pi\rho\, d\rho\, dh}{V}M$$

So we have $$d^2E_p = -\frac{Gm\,\delta m}{l}= -\frac{2\pi Gm\rho\, d\rho\,dh M}{Vl}$$

However, $$l^2 = \rho^2 + (h-H)^2$$

So $$l\,dl = \rho\,d\rho$$

Finally, $$-d^2E_p = \frac{2\pi GMm}{V}dh\,dl$$

We have $l \in [H-h, \sqrt{(H-h)^2+\rho^2}]$, so the slice $[h,h+dh]$ causes a potential energy $$dE_p = \int d^2E_p = -\frac{2\pi GMm}{V}(\sqrt{(H-h)^2+r^2}-(H-h))dh$$

Thalès's theorem states that $$\frac{r}{R} = \frac{H-h}{H}$$

So, finally, $$dE_p = -\frac{2\pi GMm}{V}(\sqrt{1+\frac{R^2}{H^2}}-1)(H-h)dh$$

We can now find what the total potential energy of $m$ on the tip of the cone is. $h$ varies in $[0,H]$, so we have

$$ E_p = \int dE_p = -\frac{\pi GMm H^2}{V}(\sqrt{1+\frac{R^2}{H^2}}-1)$$

However, $V = \frac{\pi}{3}R^2H$, so

$$ E_p = -\frac{3GMm}{R^2}(\sqrt{H^2+R^2}-H)$$

The minimum work needed to bring the mass $m$ from the tip of the cone to the infinity is thus $$W = \frac{3GMm}{R^2}(\sqrt{H^2+R^2}-H)$$

  • $\begingroup$ Isn't there a easy way? Well, I thought dividing the cone into infitesimally thin cylinders with dx breadth. And then I the whole mass of the cylinders would be centered at the COM and then I would calculate the PE for one cylinder and then integrate it to get the whole cone's PE. Isn't it a nice way? Give me your thoughts. $\endgroup$ – Mockingbird Oct 23 '16 at 12:08
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    $\begingroup$ @Mockingbird : Although a bit long-winded here, I think integrating over rings is the shortest derivation. A gravitating object can only be replaced by a point particle at the CoM when the object is spherically symmetric - which is not the case for a cylinder. (If it works for a cylinder, why not use the CoM of the whole cone instead?) See the Shell Theorem. $\endgroup$ – sammy gerbil Oct 23 '16 at 13:56
  • $\begingroup$ How is h/H=r/R? I don't get it. $\endgroup$ – Mockingbird Oct 26 '16 at 8:31
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    $\begingroup$ @Mockingbird : Unfortunately Spirine has changed the definitions of $h$ between the 1st and 2nd cone diagrams. This is confusing. In both cases the relation is the ratio of corresponding sides in "similar triangles," which is elementary geometry. In the 1st diagram, contrary to what that diagram indicates, $h$ is measured from the point of the cone. Then $r/R=h/H$. In the 2nd diagram, $h$ is measured from the base. Then $r/R=(H-h)/H$. $\endgroup$ – sammy gerbil Nov 1 '16 at 19:28

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