A thin metal disk of mass m=2.00 x 10^-3 kg and radius R=2.20cm is attached at its center to a long fiber. When the disk is turned from the relaxed state through a small angle theta, the torque exerted by the fiber on the disk is proportional to theta. Find an expression for the torsional constant k in terms of the moment of inertia I of the disk and the angular frequency w of small, free oscillations. The disk, when twisted and released, oscillates with a period T of 1.00 s. Find the torsional constant k of the fiber.

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inyangotobong inyangotobong · Answer 1 · 2019-12-08T13:28:09+01:00

Answer:

k = 1.91 × 10^-5 N m rad^-1

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ogorwyne ogorwyne · Answer 2 · 2022-04-08T15:42:43+02:00

The torsional contant of the fiber is 1.9 * 10^-5 Nm/rad

What is Torsion constant?

This is a geometric property that represents the ability of a body to resist twist

Given

m = 2 * 10^-3 = 0.002 kg

R = 2.20cm = 0.022 m

T = 1.0 s

moment of inerta, I = 0.5 * m * r ^2

I = 0.5 * 0.002 * 0.022^2

I = 0.000000484 kgm^2 = 4.84*10^-7 kgm^2

Period of torsionn pendulum, T = 2pi sqrt ( I / K)

K = 4 pi^2 * I / T^2

K = 4 (3.142)^2 * 4.84*10^-7 / ^2

K = 1.9 * 10^-5 Nm/rad

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