A frictional track abcde ends in a circular loop of radius r, body slides down the track from point a which is at a height h of 5 cm. maximum value of r for the body to successfully complete the loop is:
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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 ??

7 years ago
Answers : (6)
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.
7 years ago
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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
7 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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Sagar Singh
B.Tech IIT Delhi
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7 years ago
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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.
#thanKyou..
I hope It's help..
@RamSharma
Here is your answer...
So, the answer is.. :
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Home»Forum»Mechanics»plz ans this
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 ??

7 years ago
Answers : (6)
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.
7 years ago
Disapproved
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
7 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
We are all IITians and here to help you in your IIT JEE preparation.
All the best.
If you like this answer please approve it....
win exciting gifts by answering the questions on Discussion Forum
Sagar Singh
B.Tech IIT Delhi
[email protected]
7 years ago
Approved
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.
#thanKyou..
I hope It's help..
@RamSharma
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