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Fraction Change in Angular Momentum of Cone Full of sand compared with half full.

by zerakith
Tags: cone, emptying, inertia, sand
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Apr24-11, 09:15 AM
P: 7
1. A light, hollow cone is filled with sand set spinning about a vertical axis through its apex on a frictionless bearing. Sand is allowed to drain slowly through a hole in the apex. Calculate the fractional change in angular velocity when the sand level has fallen to half its original value. You may neglect the contribution of the hollow cone to the moment of inertia.

2. Moment of Inertia of Solid Cone: [tex]I=\frac{\rho\pi R^4h}{10} [/tex]
Conservation of Angular Momentum: [tex]I_0\omega_0=I_1\omega_1 [/tex]
Dimensions of the cone: Length: h. Radius of circle at end of cone: R

3. So its clear to me that the way to proceed is to consider the conservation of angular momentum. At the start the cone is full of sand and thus the system is just a solid cone so:
[tex]I_0=\frac{\rho\pi R^4h}{10} [/tex]
For [tex]I_1 [/tex] i need to calculate the moment of inertia, the sticking point for me is the shape the sand will take within the cone. Intuitivly i think that the sand will form a conical ring around the cone (i.e so there is a smaller cone of empty space inside the cone and the rest is filled with sand). I'm not really happy with jumping to that conclusion however if I do that and work it through I get:
I've no real way to determine whether this is correct. Am I right in the assumption about the shape of the sand (and ideally why?), and does the fractional change I have ended up with make sense?
There is no rush to this, it's not examined and term is over, it is however, bugging me.

Thanks in advance

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May26-11, 10:22 AM
P: 7
I have now figure this out, the cone was vertical with the apex pointing downwards, hence it was simply matter of finding the new height of the cone, given that there was half the mass.

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