Physics, asked by anshikapandey8317, 10 months ago

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

Answers

Answered by suskumari135
29

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.

Answered by CarliReifsteck
9

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}.

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