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113. Problem 15.45 (RHK) A physical pendulum consists of a uniform solid disk of mass M = 563 g and radius R =14.4 cm supported in a vertical plane by a pivot located a distance d = 10.2 cm from the centre of the disk. The disk is displaced by a small angle and released. We have to find the period of the resulting simple harmonic motion.
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Solution: Click For PDF Version The centre of mass of the physical pendulum will be at the centre of the circular disk. Rotational inertia of a uniform disk of mass M and radius R about the axis passing through its CM and perpendicular to the plane of the disk is
By the parallel axis theorem the rotational inertia of the disk about the axis passing through the pivot will be
Let us consider small oscillations of the physical pendulum about its pivot.
Let at some instant the line joining the pivot to the CM be at angle
As we are considering small oscillations, for small angles we can approximate
Equation of motion of the physical pendulum in this approximation is
It is an equation of SHM with period T given by the relation
Substituting R =14.4 cm and d = 10.2 cm, we calculate the value of T,
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from the vertical. Torque exerted by the weight of the pendulum about the axis
passing through the CM will be
