please show all work!!! A thin metal disk of mass m = 2.00 x 10-3 kg. and radius

Question

please show all work!!!

A thin metal disk of mass m = 2.00 x 10-3 kg. and radius R = 2.20 cm is attached at its center to a long fiber. (Figure 1) When the disk is turned from the relaxed state through a small angle theta. the torque exerted by the fiber on the disk is propotional to theta. tau = -k theta. The constant of proportionality k is called the "torsional 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. Express your answer in terms of some or all of the variables I and w. The disk, when twisted and released, oscillates with a period T of 1.00 s. Find the torsional constant k of the fiber. Give your numerical answer for the torsional constant to an accuracy of three significant figures.

Explanation / Answer

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 + )
max and 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 · 2·10³ kg · (0.022m)² = 0.484·10kgm²
The frequency is given by
= 2 / T
This leads to a torsional constant of:
k = I · (2 / T)² = 0.484·10kgm² · 4 · ² / 1s² = 19.11·10Nm

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.