Question

A uniform rod of length l is suspended by an end and is made to undergo small oscillations. Find the length of the simple pendulum having the time period equal to that of the road.

Answer 1

It is given that the length of the rod is l.

Let point A be the suspension point and point B be the centre of gravity.

Separation between the point of suspension and the centre of mass, l' = \[\frac{l}{2}\]

Also, h =\[\frac{l}{2}\]

Using parallel axis theorem, the moment of inertia about A is given as,

\[I =  I_{CG}  + m h^2 \] 

\[   = \frac{m l^2}{12} + \frac{m l^2}{4} = \frac{m l^2}{3}\] 

\[\text { the  time  period  }\left( T \right) \text { is  given  by, }\] 

\[T = 2\pi\sqrt{\frac{I}{mgl'}} = 2\pi\sqrt{\frac{I}{mg\frac{l}{2}}}\] 

\[   = 2\pi\sqrt{\frac{2m l^2}{3mgl}} = 2\pi\sqrt{\frac{2l}{3g}}\]

Let T' be the time period of simple pendulum of length x.

Time Period \[(T')\] is given by ,\[T' = 2\pi\sqrt{\left( \frac{x}{g} \right)}\] \[\text { As  the  time  period  of  the  simple  pendulum  is  equal  to  the  time  period  of  the  rod,} \] \[T' = T\] 

\[ \Rightarrow \frac{2l}{3g} = \frac{x}{g}\] 

\[ \Rightarrow x = \frac{2l}{3}\]