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Harlamova29_29 [7]
3 years ago
11

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 turn

ed 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.

Engineering
1 answer:
topjm [15]3 years ago
6 0

Answer:

k = 1.91 × 10^-5 N m rad^-1

Workings and Solution to this question can be viewed in the screenshot below:

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b) creep rate is    = 1.751 \times 10^{-5} \% per hr

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we know Arrhenius expression is given as

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activation energy for creep is \frac{\epsilon_{800}}{\epsilon_{700}} = = \frac{C\times e^{\frac{-Q}{R(800+273)}}}{C\times e^{\frac{-Q}{R(700+273)}}}

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