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The arrangement consists of
two identical uniform solid cylinders each of
mass 5kg on which two light threads are wound
symmetrically. Find the tensions of each
thread in the process of motion. The friction
in the axle of the upper cylinder is assumed to
be absent.
 
1)4.9N 2)9.8N 3) 88N 4) 5.8N

Asked by chandanbr6004 25th November 2017, 2:22 PM
Answered by Expert
Answer:
The arrangement consists of
 
Solution:
 
begin mathsize 12px style Torque space equation space for space the space the space top space sphere comma space let space it space be space called space as space sphere space straight A.
2 straight T subscript 1 straight r equals mr squared over 2 straight alpha subscript 1
2 straight T subscript 1 equals mrα subscript 1 over 2 space... left parenthesis 1 right parenthesis

For space the space lower space sphere space straight B comma space the space torque space equation space is comma
2 straight T subscript 1 straight r equals mr squared over 2 straight alpha subscript 2 space... left parenthesis 2 right parenthesis
From space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis space we space get comma space straight alpha subscript 1 equals straight alpha subscript 2

Force space on space sphere space straight B comma
mg minus 2 straight T subscript 1 equals ma subscript straight b space... left parenthesis 3 right parenthesis
Acceleration space of space point space on space sphere space straight A space through space the space thread space joining space both space straight A space and space straight B comma
straight r space straight alpha subscript 1 equals Acceleration space of space sphere space straight A equals straight a subscript straight b minus straight r space straight alpha subscript 2
Therefore comma space acceleration space of space straight B space is space
straight a subscript straight b equals space straight r space straight alpha subscript 1 plus straight r space straight alpha subscript 2 equals space 2 straight r space straight alpha subscript 1
putting space above space equation space in space equation space left parenthesis 3 right parenthesis
mg minus 2 straight T subscript 1 equals straight m.2 space straight r space straight alpha subscript 1
mg minus 2 straight T subscript 1 equals 2 straight m space straight r space straight alpha subscript 1 space end subscript... left parenthesis 4 right parenthesis
from space equation space left parenthesis 1 right parenthesis
2 straight T subscript 1 equals mrα subscript 1 over 2
mrα subscript 1 equals 4 straight T subscript 1 space space... left parenthesis 5 right parenthesis
putting space left parenthesis 5 right parenthesis space in space left parenthesis 4 right parenthesis
mg minus 2 straight T subscript 1 equals 2 left parenthesis 4 straight T subscript 1 right parenthesis
mg equals 8 straight T subscript 1 plus 2 straight T subscript 1
mg equals 10 straight T subscript 1
straight T subscript 1 equals mg over 10 equals space fraction numerator 5 space straight x space 9.8 over denominator 10 end fraction equals 49 over 10 equals 4.9 space straight N



end style
1)4.9N is the right answer.
Answered by Expert 26th November 2017, 12:20 AM
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