A thin metal disk of mass m=2.00 x 10^-3 kg and radius R=2.20cm is attached at i

Question

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.

Explanation / Answer

a) The equation of motion for the disk is given by I·d²?/dt² = - k·? d²?/dt² + k/I·? = 0 d²?/dt² + ?²·? = 0 with ?² = k/I This simple second order differential equation has the general solution: ? = ?max · sin(?·tˆ + f0) ?max and f0 are constants determined by the boundary conditions. ? is the angular frequency of the oscillator. The torsional constant in terms of moment of inertia and frequency is: k = I · ?² (b) The moment of inertia for a rotating disk is: I = ½ · m · R² = ½ · 2·10?³ kg · (0.022m)² = 0.484·10?6kgm² The frequency is given by ? = 2p / T This leads to a torsional constant of: k = I · (2p / T)² = 0.484·10?6kgm² · 4 · p² / 1s² = 19.11·10?6Nm

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