Shown in the figure now the rotational inertia of the

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shown in the figure. Now, the rotational inertia of the system is the sum of the rotationalinertiaI1of the stick shown horizontal in the figure and the rotational inertiaI2of thestick shown vertical. Thus, we have
659=+=112+13=51212222IIIMLMLMLwhereL= 1.00 m andMis the mass of a meter stick (which cancels in the next step).Now, withm= 2M(the total mass), Eq. 15-29 yieldsTMLMghLg222565122whereh=L/4 was used. Thus,T= 1.83 s.45. From Eq. 15-28, we find the length of the pendulum when the period isT= 8.85 s:=4.22LgTThe new length isL´ =Ldwhered= 0.400 m. The new period is222224LLdTdTggggwhich yieldsT´ = 8.76 s.46. We requireo22LITgmghsimilar to the approach taken in part (b) of Sample Problem – “Physical pendulum, periodand length,” but treating in our case a more general possibility forI. Canceling 2,squaring both sides, and cancelinggleads directly to the result;Lo=I/mh.47. We use Eq. 15-29 and the parallel-axis theoremI=Icm+mh2whereh=d. For a soliddisk of massm, the rotational inertia about its center of mass isIcm=mR2/2. Therefore,2222222/ 22(2.35 cm)+2(1.50 cm)2220.367 s.22(980 cm/s)(1.50 cm)mRmdRdTmgdgd48. (a) For the “physical pendulum” we haveT= 2Imgh=2com2Imhmgh.
CHAPTER 15660If we substituterforhand use item (i) in Table 10-2, we have22212abTrrg.In the figure below, we plotTas a function ofr, fora= 0.35 m andb= 0.45 m.(b) The minimum ofTcan be located by setting its derivative to zero,/0dTdr. Thisyields2222(0.35 m)(0.45 m)0.16 m.1212abr(c) The direction from the center does not matter, so the locus of points is a circle aroundthe center, of radius [(a2+b2)/12]1/2.49. Replacingxandvin Eq. 15-3 and Eq. 15-6 withandd/dt, respectively, we identify4.44 rad/s as the angular frequencyThen we evaluate the expressions att= 0 anddivide the second by the first:at0/tddt=tan.(a) The value ofatt= 0 is 0.0400 rad, and the value ofd/dtthen is –0.200 rad/s, so weare able to solve for the phase constant:= tan1[0.200/(0.0400x4.44)] = 0.845 rad.(b) Onceis determined we can plug back in too=mcosto solve for the angularamplitude.We findm= 0.0602 rad.50. (a) The rotational inertia of a uniform rod with pivot point at its end isI=mL2/12 +mL2= 1/3ML2. Therefore, Eq. 15-29 leads to
661221323228MLgTTMg Lso thatL= 0.84 m.(b) By energy conservationbottom of swingend of swingmmEEKUwhereUMg(cos)1withbeing the distance from the axis of rotation to the centerof mass. If we use the small-angle approximation (cos1122within radians(Appendix E)), we obtain2210.5 kg9.8 m/s22mmLUwherem= 0.17 rad. Thus,Km=Um= 0.031 J. If we calculate (1cos)

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