Question

A disc of radius R and mass M is pivoted at the rim and is set for small oscillations. If a simple pendulum has to have the same period as that of the disc, the length of the simple pendulum should be
(A) $\dfrac{5}{4}R$
(B) \[\dfrac{2}{3}R\]
(C) \[\dfrac{3}{4}R\]
(D) \[\dfrac{3}{2}R\]

Answer 1

Hint: Moment of inertia is defined as the torque needed for a desired angular acceleration about a rotational axis. It depends upon the body mass distribution and the axis chosen. Here we will make use of the concept of moment of inertia.
The required formula is $I = {I_o}M{R^2}$

Complete step by step answer:
Now, let us start the question by calculating the moment of inertia of the disc about the axis of oscillation,
$I = {I_o} + M{R^2}$
We further know that ${I_o} = \dfrac{1}{2}M{R^2}$
So, we can write that,
$I = \dfrac{1}{2}M{R^2} + M{R^2}$
$I = \dfrac{3}{2}M{R^2}$
Time period of a pendulum,
$T = 2\pi \sqrt {\dfrac{I}{{Mgr}}} $
Now, by putting the value of I in T, we get,
$T = 2\pi \sqrt {\dfrac{{3M{R^2}}}{{2MgR}}} $
On simplifying the above equation, we get,
$T = 2\pi \sqrt {\dfrac{{3R}}{{2g}}} .........(1)$
Now, let us consider a simple pendulum having effective length l has the same time period,
$T = 2\pi \sqrt {\dfrac{l}{g}} ........(2)$
By comparing equation (1) and equation (2), we get,
$2\pi \sqrt {\dfrac{{3R}}{{2g}}} = 2\pi \sqrt {\dfrac{l}{g}} $
On further simplifying, we get,
$\sqrt {\dfrac{{3R}}{{2g}}} = \sqrt {\dfrac{l}{g}} $
$\dfrac{{3R}}{{2g}} = \dfrac{l}{g}$
By cancelling g by g, we get,
$l = \dfrac{{3R}}{2}$
Thus, the correct option is option (D), the length of the simple pendulum should be \[\dfrac{3}{2}R\]

Note:
The moment of inertia not only depends upon the mass of the body but also on the distribution of mass in relation to the axis of rotation.The moment of inertia usually depends upon the direction of the axis, and always upon the perpendicular distance from the axis to the centre of mass of the object.