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118. Problem 16.81P (HRW) A 2.5 kg disk, 42 cm in diameter, is supported by a massless rod, 76 cm long, which is pivoted at its end. We have to find (a) the period of oscillation when the massless torsion string is not connected. (b) When the torsion spring is connected so that, in equilibrium, the rod hangs vertically, we have to find the torsional constant of the spring so that the new period of oscillation is 0.50 s shorter than before.
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Solution: Click For PDF Version (a) In the first part of the problem we will consider oscillations of the pendulum when the torsion spring does not touch the rod connecting the disk with the pivot. In the approximation of negligible mass of the rod compared to the mass of the disk, the rotational inertia of the physical pendulum about the axis of oscillation passing through the pivot will be
where M is the mass and R is the radius of the disk, and d is the distance of the centre of the disk from the pivot. Data of the problem are
Period of oscillation of a physical pendulum is given by the mathematical expression
(b) We will next consider the physical pendulum when torsion spring is in contact
with the rod. Let the torsional constant of the spring be
It is an equation of SHM. Period
We thus have the relation
From this result we can compute the value of
We find
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Equation of motion of the physical pendulum in this situation will be
of
oscillations of the physical pendulum attached to a torsional spring will
therefore be
when
