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A small block is projected with a speed v_( 0) on a horizontal track placed on a sufficientlyrough surface which turns into a semi circle ( vertical) of radius R. Find the min value of V_(o) so that the body will hit the point A after leaving the track at its highest point. The arrangement is shown in the figure, given that the straight part is rough & the curved path is smooth. The coefficient of friction is mu. |
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Answer» `(2R)=1/2g t^(2)`, whrer `t=` time of its fall. `rArr t=2 sqrt(R//g)` `:.` The distance `AB=2vsqrt(R//g)` `rArr d=2v_(x) sqrt(R//g)`...(a) Work energy theorem is applied to the motion of the body from A to B leads `Delta KE=W_(f)` `rArr 1/2 mv_(o)^(2)-1/2 mv_(1)^(2)=mu mgd` `rArr v_(o)=sqrt(v_(1)^(2)+2mu gd)` ...(b) Energyconservation between B & C yields `1/2 mv_(1)^(2)-1/2 mv^(2)=mg(2R)` `rArr v_(1)=sqrt(v^(2)+4gR)` ...(c) When the block escapes C, its minimum speed v can be given as `(mv^(2))/(R)=mg` ( `:.` the normal CONTACT force `=0`) `rArr v=sqrt(gR)`...(d) (NOTE that if the particle has speed LESS than `sqrt(gR)` then it will not reach to the highest point of the curvature. By using (c) and(d) we obtain `v_(1)=sqrt(5gR)` ...(e) using (a) and (d) we obtain `d=(sqrt(gR))2(sqrt(R/g))=2R`...(f) `v_(o)=sqrt(5gR+2 mu g (2R))=sqrt((5+4 mu )gR)`. 
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