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i have a problem in Q no.22 pg no.115 of circular motion from hc verma..can anybody plz give me the soln...:)

i have a problem in Q no.22 pg no.115 of circular motion from hc verma..can anybody plz give me the soln...:)

Grade:11

1 Answers

TANAYRAJ SINGH CHOUHAN
65 Points
8 years ago
Radius of the curves = 100m Weight = 100kg Velocity = 18km/hr = 5m/sec a) at B mg –mv2/R = N N = (100 × 10) –100 x 25/100 = 1000 – 25 = 975N At d, N = mg +mv2/R = 1000 + 25 = 1025 N b) At B & D the cycle has no tendency to slide. So at B & D, frictional force is zero. At ‘C’, mg sin ? = F F = 1000 ×1/2^1/2 = 707N c) (i) Before ‘C’ mg cos ? – N =mv2/R ? N = mg cos ? –mv2/R = 707 – 25 = 683N (ii) N – mg cos ? =mv2/R ? N =mv2/R+ mg cos ? = 25 + 707 = 732N d) To find out the minimum desired coeff. of friction, we have to consider a point just before C. (where N is minimum) Now, u N = mg sin ? u × 682 = 707 So, u = 1.037 PLEASE LIKE MY ANSWER IF YOU ARE SATISFIED BY IT AND BEST OF LUCK!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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