Question

A ring of redius r is suspended from a point on its circumference. Determine its angular frequency of small oscillations.

answer fast

Answer 1

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★ It acts as physical pendulum, the time period of which is

T = 2π√I/mgl

Here,

I = moment of inertia of ring about point of suspension

= mr^2 + mr^2

=2mr^2

and

l = distance of point of suspension from center of gravity
=r

now,

T = 2π√2mr^2/mgr

=2π√2r/g

omega= w = 2π/T (angular frequency )

putting the value of T

we get,

w = √g/2r

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Answer 2

Answer:

Explanation:

It is a physical pendulum the time period of which is  

 $T=2π\sqrt{1/mgl}$

Here, I = moment of inertia of the ring about point of suspension = mr2 + mr2 = 2mr2  

and l = distance of point of suspension from centre of gravity

$T=2π\sqrt{2mr^2/mgr} =2π\sqrt{2r/g}$

$∴ angular frequency  ω   \frac{2π}{T}                                         ω    \sqrt[]{g/2r}$