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Grade:

                        

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.

9 years ago

Answers : (1)

vikas askiitian expert
509 Points
							

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...

9 years ago
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