A uniform, solid metal disk of mass 6.10 and diameter 29.0 hangs in a horizontal

Question

A uniform, solid metal disk of mass 6.10 and diameter 29.0 hangs in a horizontal plane, supported at its center by a vertical metal wire. You find that it requires a horizontal force of 4.20tangent to the rim of the disk to turn it by 3.31, thus twisting the wire. You now remove this force and release the disk from rest. A) What is the torsion constant for the metal wire? B) What is the frequency of the torsional oscillations of the disk? C) What is the period of the torsional oscillations of the disk?

Explanation / Answer

(a). torque, t = k ? F r = k ? (4.23)(0.12) = k (3.34 * p/180) k = 8.70759 Nm (b). I = ½ m r² I = ½ (6.5)(0.12)² I = 0.0468 kg m² f = 1/(2p) v(k/I) f = 1/(2p) v(8.70759/0.0468) f = 2.17093 Hz (c). according to second Newton's law for rotational motion, Storque = I a -k ? = I (d²?/dt²) d²?/dt² + (k/I) ? = 0 Let k/I = ?² general solution of this ordinary differential equation is ?(t) = A cos ?t + B sin ?t ?'(t) = -A? sin ?t + B? cos ?t boundary conditions is ?'(0) = 0 ? B = 0 ?(0) = ?o ? A = ?o general solution is ?(t) = ?o cos ? t ? this formula describe the equation of motion for theta of t for the disk. but remember that, ? = 2pf = v(k/I) f = 1/(2p) v(k/I) moment inertia of solid disk I = ½ m r²

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