A thin, round disk made of acrylic plastic (density is 1.1 g/cm3) is 20 cm in di

Question

A thin, round disk made of acrylic plastic (density is 1.1 g/cm3) is 20 cm in diameter and 2 cm thick (see figure below). A very small hole is drilled through the disk at a point 8 cm from the center. The disk is hung from the hole on a nail and set into simple harmonic motion with a maximum angular displacement (measured from vertical) of 7°. Calculate the period of the motion i 8 cm 7°1 A hole is drilled through a thin, round disk of mass M and radius R = 0.10 m. The hole is located h = 0.08 m from the center of mass of the disk. The disk is hung on a nail and set into simple harmonic motion. The period of a physical pendulum is related to the moment of inertia of the pendulum. Since the center of mass of the disk is located a distance h from the pivot point, we will need to use the parallel-axis theorem. As a reminder, the moment of inertia of a thin disk is MR2

Explanation / Answer

Given  

a physical pendulum with mass M , moment of inertia is I = 0.5*M*r^2 and d is the distance from center of mass to the point of pivot

we know that the time period of a physical pendulum is

T = 2pi sqrt(I/M*g*d)

T = 2pi sqrt(0.5*M*r^2 /M*g*d)

T = 2pi sqrt(0.5*r^2 /g*d)

T = 2pi sqrt(0.5*0.1^2/9.8*0.08)s

T = 0.5018 s

so the time perios of the given physical pendulum is T = 0.5018 s

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