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