Question

A disc of radius r and mass m is pivoted at its rim an asset to small oscillation of a simple pendulum has time period then effective length of pendulum is

Answer 1

Effective length of pendulum is, $\bf l = \frac{3}{2}R$

Explanation:

Moment of inertia of the disc about the axis of oscillation is,

$I = I_o + MR^2$

$I = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2} MR^2$

Time period of the pendulum,

$T = 2\pi \sqrt{ \frac{I}{Mgr}} = 2\pi \sqrt{ \frac{3MR^2}{2MgR}}$

$T = 2\pi \sqrt{\frac{3R}{2g}$..............(1)

Consider a simple pendulum having effective length l has the same time period,

$T = 2\pi \sqrt{ \frac{l}{g}$.................(2)

Equating eq. 1 and 2

$2\pi \sqrt{\frac{3R}{2g}} = 2\pi \sqrt{ \frac{l}{g}$

$l = \frac{3}{2}R$

Thus, $\bf {l = \frac{3}{2}R}$ is the length of pendulum.

Answer 2

Given that,

Radius of disc= r

Mass = m

We need to calculate the moment of inertia at rim

Using parallel theorem

$I_{rim}=I_{c}+mr^2$

Put the value into the formula

$I_{rim}=\dfrac{mR^2}{2}+mR^2$

$I_{rim}=\dfrac{3mR^2}{2}$

We need to calculate the time period

Using formula of time period

$T=2\pi\sqrt{\dfrac{I}{mgd}}$

Here, d = distance between point of suspension and center of gravity

$d = R$

Put the value into the formula

$T=2\pi\sqrt{\dfrac{3mR^2}{2mgR}}$

$T=2\pi\sqrt{\dfrac{3R}{2g}}$......(I)

We need to calculate the effective length of the pendulum

Using formula of time period

$T=2\pi\sqrt{\dfrac{l}{g}}$

On compare of time period from equation

$T =2\pi\sqrt{\dfrac{3R}{2g}}$

The effective length is

$l=\dfrac{3R}{2}$

Hence, The effective length of the pendulum is $\dfrac{3R}{2}$.