 # 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 ?? 11 years ago

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.

11 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

11 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

All the best.

Sagar Singh

B.Tech IIT Delhi

sagarsingh24.iitd@gmail.com Sukhendra Reddy Rompally B.Tech Mining Machinery Engg, ISM Dhanbad
93 Points
11 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 Kushagra Madhukar
one year ago
Dear student,

At A,the energy of the body is = PE ,due to height H
At B = K.E
At B,
V = 2gH
Also
V= 5gR
Equating
5gR = 2gH
5R = 2H
R = 2/5H
H = 5
R = 2 cm

Thanks and regards,
Kushagra