Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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.

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.

Grade:

1 Answers

vikas askiitian expert
509 Points
10 years ago

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...

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free