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 = I_com+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...

Approved