Find the period of small oscillations of a uniform rod with length l, pivoted at one end.
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By the formula for a physical pendulum !
Time period = 2π*√I/mgd
where I is the inertia of the body from the axis at which it is pivoted
and d is the distance of the pivoted point from the centre of gravity which sometimes even coincides with the centre of mass of the body.
So,
Since the body is pivoted at one of it's ends
The inertia will become(using parallel axis theorem)
ml^2/12 + ml^2/4 = ml^2 /3
And d = l/2
So,
the time period = 2π√ml^2/3/mgl/2
= 2π √ 2l/3g
Time period = 2π*√I/mgd
where I is the inertia of the body from the axis at which it is pivoted
and d is the distance of the pivoted point from the centre of gravity which sometimes even coincides with the centre of mass of the body.
So,
Since the body is pivoted at one of it's ends
The inertia will become(using parallel axis theorem)
ml^2/12 + ml^2/4 = ml^2 /3
And d = l/2
So,
the time period = 2π√ml^2/3/mgl/2
= 2π √ 2l/3g
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