A uniform thin rod of length 0.807 m is hung from a horizontal nail passing thro

Question

A uniform thin rod of length 0.807 m is hung from a horizontal nail passing through a small hole in the rod located 0.031 m from the rod's end. When the rod is set swinging about the nail at small amplitude, what is the period of oscillation? Ok so I got this:

I= m(L^2/12 + (L/2 - .031)^2)

I=m(.054271 + .138756)

I= .193027m

Then I plugged that into

T=2pi sqrt(I/mgd) The m's canciled out so..

T=2pi sqrt (.193027/(9.8*.138756)
T=2pi(.376764)
T=2.367

This is incorrect. My friend tried to solve it for me and got 1.59. This is also incorrect. Where am I going wrong??

Explanation / Answer

I what you did is correct If there is is mistake then it should be in taking the length of nail from end

u have to take the length of nail such that

Moment of inertia of the rod about the nail
I' = I + Md² = the d is Distance to center of gravity:

I am gving u example As I am confused about which end u took the length

A uniform rod length L is nailed to a post such that 2/3 of its length is below the nail.
Find the period of small oscillation of the rod

ans

Distance to center of gravity:
d = 2/3 L - 1/2 L = 1/6 L

Moment of inertia of the rod about center of gravity
I = 1/12 ML²

Moment of inertia of the rod about the nail
I' = I + Md² = M(1/12 L² + 1/36 L²) = 1/9 ML²

Torque:
Q() = Mg d =1/6 Mg L

Angular velocity of oscillations:
² = 1/I' Q/ = 9/ML² 1/6Mg L = 3/2 g/L

Answer:
Period T = 2/ = 2 (2/3) (L/g)

hope it should solve the problem

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