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(A) The disc is suspended by two parallel ropes of length $l$. When the disc is displaced slightly in its own plane,it undergoes a motion similar to a

Vedclass · Accepted Answer

$2\pi \sqrt {\frac{l}{g}}$

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(A) The disc is suspended by two parallel ropes of length $l$. When the disc is displaced slightly in its own plane,it undergoes a motion similar to a physical pendulum.For a physical pendulum,the time period is given by $T = 2\pi \sqrt{\frac{I}{mgd}}$,where $I$ is the moment of inertia about the axis of rotation and $d$ is the distance from the pivot to the center of mass.In this case,the ropes are attached at a distance $R/2$ above the center of the disc. The effective length of the pendulum is $l_{eff} = l$. However,because the ropes are attached at the rim (or specific points),the disc swings such that the center of mass moves in an arc of radius $l$.Since the ropes remain parallel during the motion,the disc does not rotate about its center of mass; it moves in pure translation. The motion is equivalent to a simple pendulum of length $l$.Therefore,the time period is $T = 2\pi \sqrt{\frac{l}{g}}$.

$A$ disc of mass $m$ and radius $R$ is attached to the ceiling with the help of two ropes of length $l$. Find the time period of small oscillations of the disc in the plane of the disc.

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