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Home / Doubts and Solutions/JEE/Test Resources

Can you answer 24th question from rotation only give correct option?

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Asked by vikrammalikviku 2nd January 2018, 12:54 PM
Answered by Expert
Answer:
First we find the position of center of gravity for the semi-circular ring ABC.
Perimeter of semicircular ring is (πR+2R). Since mass is distributed uniformly, center of gravity will divide the whole lengeth equally and it is located such that
the curved length of the ring is R[(π/2)+1] as shown in figure. Hence the angle subtended by the chord PQ at center is approximately 90º ( arc / radius = angle)..
Hence PG = R/√2 and the force due to gravity (mg) acts at P to give a torque T = (mgR) / √2 .

begin mathsize 12px style T space equals space I space alpha w h e r e space I space i s space m o m e n t space o f space i n e r t i a space a n d space alpha space i s space a n g u l a r space a c c e l e r a t i o n m o m e n t space o f space i n e r t i a space o f space t h e space g i v e n space r i n g space a b o u t space t h e space g i v e n space a x i s space o f space r o t a t i o n space equals space fraction numerator 7 M R squared over denominator 12 end fraction T space equals space fraction numerator 7 M R squared over denominator 12 end fraction alpha space equals space m g fraction numerator R over denominator square root of 2 end fraction space H e n c e space alpha space equals space 12 over 7 cross times fraction numerator 9.8 over denominator 5 end fraction cross times fraction numerator 1 over denominator square root of 2 end fraction space equals space 2.38 space r a d divided by s squared end style

Answered by Expert 5th January 2018, 1:20 PM
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