Page:Elementary Text-book of Physics (Anthony, 1897).djvu/57

where $I$ is the moment of inertia and $f$ a constant depending on the magnitude of the force.

If, now, another body, of which the moment of inertia can be calculated, be joined with the first, the time of oscillation changes to $t=\pi {\sqrt {\frac {I+I'}{f}}}$, where $I'$ is the moment of inertia of the body added. Combining the two equations, we obtain, as the value of the moment of inertia desired,

$I={\frac {I't^{2}}{t'^{2}-t^{2}}}\cdot$

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Fig. 1339. Rotation about a Fixed Point.— Suppose a body so conditioned that its only motion is a rotation about the fixed point $O$ (Fig. 13). Suppose the force $F$ applied at a point in the body, which moves under the action of the force through the infinitesimal distance $QR$. This motion is a rotation about the point $O$ through the angle $\phi ={\frac {QR}{OQ}}\cdot$ The work done by the force during this rotation is

$W=F\cdot {\frac {QR}{QS}}\cdot QR\cdot$

Since, in the limit, when $QR$ and $QS$ are infinitesimal, the triangles $QRS$ and $OPQ$ are similar, ${\frac {QR}{QS}}={\frac {OP}{OQ}}$, and hence

$W=F\,.\,OP\cdot {\frac {QR}{OQ}}=F\,.\,OP\,.\,\phi .$

The work thus done is equal to the kinetic energy gained by the rotating body, or to ${\tfrac {1}{2}}\omega ^{2}I$, where $I$ is the moment of inertia of the body and $\omega$ the angular velocity which it gains during the motion. Now $\omega ^{2}=2\alpha \phi$, where $\alpha$ is the angular acceleration (§ 20), and hence

$F\,.\,OP=I\alpha$

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The product $F\,.\,OP$, or the product of the force and the perpendicular let fall from the axis of rotation upon the line of direction