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A pendulumbob is givenan intialvelocity v at thisbottommostpointand it isfoundthatboblosescirculartrackat a certainpoint and hitsthe pointof suspension. If l is length of the threadthen findv . |
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Answer» Solution :Let VBE the velocityof bobat thepointwhereitloses circulartrackand assumethatstringmakes an angle`theta `with thevertical atthis pointas shown infigure. Let O be thepointwherebobleavesthepointO ison thecircularpathhencewecan applycirculardynamicsfor this point . `T + mg cos theta = ("mu"^(2))/l ` Bob loses circulartrackat this pointhencewe cansustituteT = 0 in aboveequation. ` u = sqrt(gl cos theta ) "" (i)` Let O THEORIGIN and axes are as shownin FIGURE. Bobmovesunder gravity like a projectileafter point O till ithits thepointof suspension `S_(x) = u_(x)t +1/2 a_(x)t^(2) rArr l sin theta= ( u cos theta ) t + 1/2(0)t^(2)` `rArr "" t = (l san theta )/(u cos theta )"" ...(ii)` ` S_(y) - u_(y)t + 1/2 a_(y)T^(2) = l cos theta = (-u sin theta ) t + 1/2 "gt"^(2)` Substituting valueof t from equation (ii) we get ` l COSTHETA = (-u sin theta) ((lsin theta )/(u cos theta )) + 1/2 g ((l sin theta )/(u cos theta ))^(2)` `rArr"" costheta= (- sin theta) ((sin theta )/(cos theta )) + 1/2 gl ((sin theta)/(u costheta))^(2)` `rArr " " cos theta+(sin^(2)theta)/(costheta) =1/2gl((sin^(2)theta)/(u^(2)cos^(2)theta))` ` rArr "" 1/(cos theta)=1/2 gl((sin^(2)theta)/(u^(2)cos^(2)theta))` `rArr "" 1/2 gl((sin^(2)theta)/(u^(2)cos theta))=1 ` `rArr 1/2((glsin^(2)theta)/(gl cos theta xx cos theta))=1 ` `rArr"" tan^(2) theta = 2 rArr tan theta= sqrt(2)"" ....(i)` We can now apply conservationof mechnical ENERGY betweenthe bottommost pointand thepoint where it losesthe circular track . Lossof kinetic energy = gainin gravitational potentialenergy . 
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