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This question refers to the solution of problem 12 here. It involves a spherical shell of mass $M$ filled with frictionless fluid of mass $M$ rolling down an inclined plane.

(This is problem 12 of the 2013 F=Ma exam, a high school competitive physics exam).

I understand the solution, but I don't understand how the given rotational inertia is obtained. We have a spherical shell of mass $M$ filled with frictionless fluid of mass $M$. The rotational inertia is apparently $$I_{contact} = \frac{2}{3}MR^2 + MR^2 + MR^2$$

Why? In particular, why is the $MR^2$ term added twice?

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Minor comment to the post (v2): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. –  Qmechanic Jul 27 at 15:15

2 Answers 2

There are two MR^2 terms added. Once for the shell (the parallel axis theorem requires this additional term, since the torques are being written about an axis passing through the point of contact), and once for the fluid (since it is frictionless, the fluid does not rotate, behaves as point mass, thus only MR^2 suffices).

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This seems to be an application of the parallel axis theorem. From Wikipedia's http://en.wikipedia.org/wiki/Parallel_axis_theorem:

Suppose a body of mass $m$ is made to rotate about an axis $z$ passing through the body's center of mass. The body has a moment of inertia $I_\textrm{cm}$ with respect to this axis. The parallel axis theorem states that if the body is made to rotate instead about a new axis $z′$ which is parallel to the first axis and displaced from it by a distance $d$, then the moment of inertia $I$ with respect to the new axis is related to $I_\textrm{cm}$ by

$I = I_\mathrm{cm} + md^2.$

Explicitly, $d$ is the perpendicular distance between the axes $z$ and $z′$.

Since, in the case at hand, one has computed the torque about an axis through the point of contact and not the center of mass, the moment of inertia about the center of mass $\big(I_\textrm{cm} = \frac{2}{3}MR^2\big)$ must be corrected by the factor $md^2 = MR^2$ to obtain the moment of inertia about the point of contact and keep things consistent.

The other $MR^2$ term corresponds to the moment of inertia of the frictionless fluid inside the shell, which has the same mass as the spherical shell and does not rotate at all.

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