1 (24.3.11) 2 10.2 The mechanical energy is then E=U+Km0=mg -cos)+,0, (24.3.12) with I=(1/3)md2.There are no non-conservative forces acting(by assumption),so the mechanical energy is constant,and therefore the time derivative of energy is zero, 0=de d E=mgsine ae do (24.3.13) d dt I:dt Recall that o.=de/dt and a.=do.dt=d2/dt2,so Eq.(24.3.13)becomes 0=0 m8号s血0+1， d (24.3.14) There are two solutions,=0,in which case the rod remains at the bottom of the swing, 0=m82sin8+1。 d28 dr (24.3.15) Using the small angle approximation,we obtain the simple harmonic oscillator equation (Eq.(24.3.6) de mg(d12）0=0. d2 (24.3.16) I Example 24.3 Torsional Oscillator A disk with moment of inertia about the center of mass Im rotates in a horizontal plane. It is suspended by a thin,massless rod.If the disk is rotated away from its equilibrium position by an angle 0,the rod exerts a restoring torque about the center of the disk with magnitude given by=b(Figure 24.6),where b is a positive constant.At t=0,the disk is released from rest at an angular displacement of 0.Find the subsequent time dependence of the angular displacement e(t). 24-724-7 Krot = 1 2 I p ω z 2 . (24.3.11) The mechanical energy is then E = U + Krot = mg d 2 (1− cosθ ) + 1 2 I p ω z 2 , (24.3.12) with 2 (1/ 3) PI = md . There are no non-conservative forces acting (by assumption), so the mechanical energy is constant, and therefore the time derivative of energy is zero, 0 = dE dt = mg d 2 sinθ dθ dt + I p ω z dω z dt . (24.3.13) Recall that ω z = dθ / dt and α z = dω z / dt = d2 θ / dt 2 , so Eq. (24.3.13) becomes 0 = ω z mg d 2 sinθ + I p d2 θ dt 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ . (24.3.14) There are two solutions, ω z = 0 , in which case the rod remains at the bottom of the swing, 0 = mg d 2 sinθ + I p d2 θ dt 2 . (24.3.15) Using the small angle approximation, we obtain the simple harmonic oscillator equation (Eq. (24.3.6)) d2 θ dt 2 + mg(d / 2) I p θ  0 . (24.3.16) Example 24.3 Torsional Oscillator A disk with moment of inertia about the center of mass Icm rotates in a horizontal plane. It is suspended by a thin, massless rod. If the disk is rotated away from its equilibrium position by an angle θ , the rod exerts a restoring torque about the center of the disk with magnitude given by τ cm = bθ (Figure 24.6), where b is a positive constant. At t = 0 , the disk is released from rest at an angular displacement of θ0 . Find the subsequent time dependence of the angular displacement θ (t)