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116. Problem 15.53 (RHK) A physical pendulum has two possible pivot points; one has a fixed position and the other is adjustable along the length of the pendulum. The period of the pendulum when suspended from the fixed pivot is T. The pendulum is then reversed and suspended from the adjustable pivot. The position of this pivot is moved until, by trial and error, the pendulum has the same period as before, namely T. We have to show that the free-fall acceleration is given by
in which L is the distance between the two pivot points. We note that g can be measured in this way without needing to know the rotational inertia of the pendulum or any of its other dimensions except L.
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Solution: Click For PDF Version A physical pendulum has two possible pivot points P1 and P 2 ; point P1
has a fixed position and P2 is adjustable along the length of the pendulum. Let
the period of oscillation of the pendulum when suspended from P1 be T.
Let d be the distance between P1 and the centre of mass (cm) of the
pendulum. Let
and its period of oscillation will be given by the relation
The period of oscillation when the physical pendulum oscillates about the pivot P2 will be
L is the distance between P1 and P2 . L is to be so adjusted such that
This condition gives the equation that fixes
Simplifying this algebraic equation, we find
Substituting this expression for
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be the
rotational inertia of the physical pendulum about the axis passing through the
cm and perpendicular to the plane of oscillation. The rotational inertia of the
pendulum about an axis perpendicular to the plane of oscillation and passing
through the pivot P1 will be
