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A bobof mass m is suspeded by a light stringof length L . It is imparted a horizontal velocity v_(0)at the lowest point A such that it completes a semicircular trajectroy in the vertical plane with the string becoming slack only on reaching the topmost point C . Thisis shown in figure .Obtain a expression for (i) v_(0) (ii) the ratio of the kinetic energies (K_(B)/(K_(C ))) at B and C . Comment on the nature of the trajectroy of the bob after it reaches the point C . |
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Answer» Solution :Two forces acting on bob. (i)Gravity and (ii) the tension T in the string .the work is not done by tensionin the string.because dispalcement of bob is always normal to the string . The potential energy of the bob is thus associated with the gravitational force only and mechanical energy of the system is conserved . (i) LET potential energy of the system is conserved . L=0 ` :. mgL =0` Total mechanical energy point A, `E =1/2 mv_(0)^(2) [ :. "Newton.s Second law"] ` ` E =1/2 mv_(0^(2) ""....(1)` The velocity of bob at point A , is `v_(0)` Bob moving circular . ` :. ` Resultant force is string = centripetal force `T_(A)-mg =(mv_(0)^(2))/L [ :. "Newton.s Second law"] ` Where `T_(A)` is the tensionin the string at A. At the highest point C, the string = centripetal force `T_(A) -mg =(mv_(0)^(2))/L [ :. "Newton.s Second Law"]` Where `T_(A)` is the tension in the tring at A. At the highest [point C, the string slackens , as the tension in the string `T_(C)` become zero. Thus , at C `E=1/2 mv_(C )^(2) +2mgL ""....(2)` Where `T_(A)` is the tension in the string t A. At the highest point C, the string slackness, as the tension in the string `T_(C)`becomes zero. Thus at C ` E=1/2 mv_(C)^(2)+2mgL ""...(2)` `(mv_(C )^(2))/L = mg"" {:. "Newton.s second law "]` Where `v_(c)` is the speed at C ` :. v_(C)^(2) =mgL ""....(3)` And `v_(C)^(2) -gL ""...(4)` ` :. ` From EQUATION (2). , ` E = 1/2 m cc"gL"+2ML` ` :. E=(5mgL)/2 ""...(5)` `1/2 mv_(0)^(2) =(5mgL)/2 ""` [ From equation (1)] ` :. v_(0)=SQRT(5gL) ""...(6)` (ii) From equation (4), `v_(C )=sqrt(gL) ""...(7)` and at the Btotal mechanical energy . `E =1/2 mv_(B)^(2) +mgL "" [ :. h = L]` From equation (6) and (1), `1/2 mv_(0)^(2) =1/2 mv_(B)^(2) +mgL` `1/2 m cc 5gL =1/2 mv_(B)^(2)+mgL` ` [ :. "From equation (7)"]` ` :. 1/2mv_(B)^(2)=5/2 mgL-mgL` `1/2 mv_(B)^(2)=3/2 mgL` ` :. v_(B) =sqrt(3gL)` (iii)The ratio of the kinetic energies at B and C is : The ratio of the kinetic energies at B and C is : `(K_(B))/(K_(C))=(1/2mv_(B)^(2))/(1/2mv_(C)^(2))=(v_(b)^(2))/(v_(C )^(2))` ` :. (K_(B))/(K_(C ))=(3gL)/("gL")=3:1` At point C , the string becomes slack and the to the left . If the string is cut at this instant , the bob will execute a projectile to MOTION with horizontal and to the left . If the string is cit at this instant , the bob will execute a projectile to motion with horizontal projection akin to motion with horizontal projection akin to a rock kicked horizontally fromthe edge of a sliff. Otherwise the bob will continue on its circular pathand complete the revolution . |
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