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Grade: 11
        

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

7 years ago

Answers : (1)

TANAYRAJ SINGH CHOUHAN
65 Points
							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
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7 years ago
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