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Grade: 12th pass
        
Please sir help me with this question. I am posting this 10th time.It is from AIATS.Please sir solve it ,I tried it many days but I failed to solve it,so plssss. Plsssss solve it.I got anwer but in that equation was written as Fstring+Fhinge=Mg which was not possible bcz both the forces are acting in opposite direction. Please solve it,I am posting this question from many days.
one year ago

Answers : (1)

neeraj
20 Points
							
before snapping the string,the tension in the string are,
2T=mg=mg/2
torque on the rod is,
T=mg(1/2)=I alpha=mgl/2=1/3ml(square)(alpha)=mgl/2=alpha=3g/2l
so the acceleration with which the centre of mass falls down is ,
a=l(alpha)/2=3g/4
so,
ma=f(down)-f(up)=m 3g/4=mg-T’=T=mg-3mg/4
T’=mg/4
one year ago
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