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a frictionless track ABCDE ends in a circular loop of radius "R". A body slides down the track from point "A" which is at height "h" = 5 cm. Maximum value of "R", for which the body successfully completes the loop is ???


What if the track is not frictionless ??



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6 years ago

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Answers : (4)

										

Just at the topmost point of loop take centrifugal force eqal to mg.take out  v^2 term from energy eqn.At 1st only p.e then at the topmost point of loop p.e.+ k.e.

6 years ago
										

the initial energy of the particle is mgh. It will complete the loop if it reaches the highest point on the loop which has a  height of 2R. therefore by equating mgh =mg2R we get h=2R. R=2.5cm. 


if there is frictional force the radius required will decrease even more depending on the coeff. of friction

6 years ago
										

Dear mohit,


At the top you have to take centrifugal force to be equal to mg and apply energy conservation law, You will get Maximum value of R=2 cm


 






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6 years ago
										Hi Mohit!                For a particle to undergo circular motion,vel at higqest pt=rg^1/2 nd at lowest pt=5rg^1/2  ; proof is simple,at top most point,by balancing the force,we get weight mg= centrifugal forbe mv^2 /r ,now we get v= rg^1/2. At the bottom,K.E = K.E at top + P.E lost ,that is 0.5mv^2 = 0.5 mu^2 + mg(2r) {since,the particle has dropped by a diameter height of 2r},where u= vel at the top most pt nd u= rg^1/2.substituting this,we get v= 5rg^1/2 .NOW BETTER REMEMBER THIS AS FORMULA .now K.E at topmost point= P.E lost ,which means 0.5mv^2= mgh. Substituting v= rg^1/2 nd value of h=5 cm,we get,r = 2h = 10cm.   HOPE U R CLEAR WITH THE SOLUTIONS.IF NOT,FEEL FREE 2 CALL ME ON 07209736303. ALL THE BEST, PLZZ APPROVE MY ANSWER IF U LIKE IT 
										
6 years ago

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