a rod is supported horizontally by means of two strings of equal length. of one of the string is cut . Then tension in other string at the same instant will..

Question

Achal Sood · Answer

Initially, tension in both string is M/2. Let there be a point b on the end of the rod from which the string has been cut.Acc of b is l*alpha. Let the end attached with string be a. Torque about a=moment of inertia of a *alpha.Now Mg*l/2=Ml 2 alpha/3. alpha=3g/l. a c =l/2alpha= 3g/4. Mg-T 1 =Ma c . Mg-T 1 =M*3g/4. T 1 =Mg/4.

ankit singh · Answer

e q u a t i o n o f f o r c e g i v e s m g − T = m a t o r q u e a b o u t t h e p o i n t o n t h e r o d w h e r e t e n s i o n a c t s m g 2 l ​ = I α = 3 m l 2 ​ α a n d a = α 2 l ​ S o l v i n g t h e s e w e g e t T = m g − 4 3 ​ m g = 4 1 ​ m g i n i t i a l l y T w a s 2 1 ​ m g , s o t h e T e n s i o n i s d e c r e a s e d .

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a rod is supported horizontally by means of two strings of equal length. of one of the string is cut . Then tension in other string at the same instant will..

a rod is supported horizontally by means of two strings of equal length. of one of the string is cut . Then tension in other string at the same instant will..

2 Answers

equationofforcegivesmg−T=matorqueaboutthepointontherodwheretensionactsmg2l=Iα=3ml2αanda=α2lSolvingthesewegetT=mg−43mg=41mginitiallyTwas21mg,sotheTensionisdecreased.

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a rod is supported horizontally by means of two strings of equal length. of one of the string is cut . Then tension in other string at the same instant will..

a rod is supported horizontally by means of two strings of equal length. of one of the string is cut . Then tension in other string at the same instant will..

2 Answers

equationofforcegivesmg−T=matorqueaboutthepointontherodwheretensionactsmg2l​=Iα=3ml2​αanda=α2l​SolvingthesewegetT=mg−43​mg=41​mginitiallyTwas21​mg,sotheTensionisdecreased.

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equationofforcegivesmg−T=matorqueaboutthepointontherodwheretensionactsmg2l=Iα=3ml2αanda=α2lSolvingthesewegetT=mg−43mg=41mginitiallyTwas21mg,sotheTensionisdecreased.