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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 ??

Mohit Mittal , 15 Years ago

Grade 12

8 Answers

TANMOY KUNDU

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.

Last Activity: 15 Years ago

pratik nayak

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

Last Activity: 15 Years ago

SAGAR SINGH - IIT DELHI

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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Last Activity: 15 Years ago

Sukhendra Reddy Rompally B.Tech Mining Machinery Engg, ISM Dhanbad

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

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Last Activity: 15 Years ago