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Q. FIND THE DISTANCE FROM THE TOPOF AUNIFORM BAR OF LENGTH 24cm WHICH IS TO BE MOUNTED ON A WALL ABOUT AN AXIS PERPENDICULAR TO ITS LENGTH,SO THAT TIME PERIOD OF OSCILLATION IS MINIMUM.
T = 2pi[I/mgx]1/2.......................1
x is perendicular distance from com to point axis of rotation...
I = Icom+mx2 = ml2/12 + mx2
T = 2pi[(ml2/12 + mx2)/mgx]1/2 ..............................2
now , from eq 2 it is clear that time period is function of x ...
T proportional to root(fx) ,
here f(x) = [ml2/12+mx2/mgx]
here we can use concept of minima maxima , we have to differentiate f(x) wrt x & then put it to 0...
d/dx(fx) = -l2/12x2 + 1 = 0
x = l/2root3 ..............3
therefore x = l/2root3 = 24/2root3 = 4root3cm
this is the distance measured from center of mass...
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