Question

A ring is Suspended from a point s on its rim as shown in the figure displacement from equilibrium it oscillating with time period of 2 second the radius of ring is

Answer 1

The radius of ring is 0.5 m

Explanation:

When ring is slightly displaced from its equilibrium state

then we will have

$\tau = I\alpha$

so we have

$mgR sin\theta = (mR^2 + mR^2)\alpha$

$mgR sin\theta= 2mR^2\alpha$

now for small angle of displacement

$mgR\theta = 2mR^2\alpha$

$\alpha = \frac{g}{2R} \theta$

so we have

$T = 2\pi\sqrt{\frac{2R}{g}}$

$2 = 2\pi\sqrt{\frac{2R}{9.8}}$

$R = 0.5 m$

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Topic : Time period of oscillation

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Answer 2

The radius of the ring is 0.0620 m.

Explanation:

Given that,

Time period = 2 sec

We need to calculate the angular velocity

Using formula of angular velocity

$\tau=I\times\alpha$

$\alpha=\dfrac{\tau}{I}$

Where, I = moment of inertia of the ring about the rim

Put the value into the formula

$\alpha=\dfrac{mg\dfrac{R}{2}}{2mR^2}$

$\alpha=\dfrac{g}{4R}$

We need to calculate the radius of the ring

Using formula of time period

$T = \dfrac{2\pi}{\omega}$

$T=\dfrac{2\pi}{\alpha\times t}$

Here, $t =\dfrac{n}{T}$

Put the value into the formula

$2=\dfrac{2\pi}{\sqrt{\dfrac{g}{4R}}\times\dfrac{1}{2}}$

$R=\dfrac{g}{\pi^2\times 16}$

$R=0.0620\ m$

Hence, The radius of the ring is 0.0620 m.

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Topic : Moment of inertia

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