(b) We use τ= Iα, where τis the torque and Iis the rotational inertia. The contribution of the rod to Iis MA212/(Table 10-2(e)), where Mis its mass and Ais its length. The contribution of each ball is mA/,22bgwhere mis the mass of a ball. The total rotational inertia is IMm=+=+AA221224640120121062..kgmkgmbgbgbgwhich yields I= 1.53 kg⋅m2. The torque, therefore, is τ=⋅−=−⋅1537 66117.kg mrad / sN m.chch(c) Since the system comes to rest the mechanical energy that is converted to thermal energy is simply the initial kinetic energy KIi==⋅=×1212239459 100224ωkg mrad / sJ.2chchπ(d) We apply Eq. 10-13: θω=+−0221223932012766320ttπbgbgchbgchbgrad / ssrad / ss2.which yields 3920 rad or (dividing by 2π) 624 rev for the value of angular displacement
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.