Question

A uniform disc of radius R is pivoted at a point O on the circumference. The time period of small oscillations about an axis passing through O and perpendicular to the plane of disc will be
A. $T = 2\pi \sqrt {\dfrac{{3R}}{{2g}}}$
B. $T = 2\sqrt {\dfrac{R}{g}}$
C. $T = 2\sqrt {\dfrac{{2R}}{g}}$
D. $T = 2\sqrt {\dfrac{{2R}}{{3g}}}$

Answer 1

The time period of a disc depends upon the moment of inertia of the disc about the axis of rotation. Since the axis of rotation is about a point on the circumference, we can use parallel axis theorem to find the value of moment of inertia about this axis.

Complete step by step answer:
It is given that a uniform disc has a radius $R$. It is pivoted at a point O on its circumference. We need to calculate the time period of small oscillations about an axis passing through O and perpendicular to the plane of the disc.
First let us calculate the moment of inertia of the disc.
It is given that the axis of rotation passes through a point on the circumference of the disc and perpendicular to the disc. We know that the moment of inertia about an Axis passing through the centre of the disc and perpendicular to the plane of the disc is given as
${I_{cm}} = \dfrac{{M{R^2}}}{2}$
Where M is the mass and R is the radius of the disc.
An axis passing through the circumference of the disc perpendicular to the plane will be parallel to the axis passing through the centre.
Hence, we can use the parallel axis theorem here.
According to the parallel axis theorem moment of inertia about an axis parallel to an axis passing through the centre of mass of the body is given by the sum of moment of inertia about the axis passing through centre of mass and the product of mass and square of distance between these two axes.
$\Rightarrow I = {I_{cm}} + M{R^2}$
Where, ${I_{cm}}$ is the moment of inertia about the axis passing through the centre, M is the mass and R is the distance between the two axes.
Now let us substitute the value of moment of inertia about the centre of mass, then we get
$\Rightarrow I = \dfrac{{M{R^2}}}{2} + M{R^2}$
$\Rightarrow I = \dfrac{3}{2}M{R^2}$
Now the time period can be calculated using the equation
$T = 2\pi \sqrt {\dfrac{I}{{Mgl}}}$
Where, $I$ is the moment of inertia, M is the mass, g is acceleration due to gravity, $l$ is the distance between centre of mass and the point of suspension.
Here the distance $l = R$
$\Rightarrow T = 2\pi \sqrt {\dfrac{I}{{MgR}}}$
On substituting a value of moment of inertia we get
$\Rightarrow T = 2\pi \sqrt {\dfrac{{\dfrac{3}{2}M{R^2}}}{{MgR}}}$
$\Rightarrow T = 2\pi \sqrt {\dfrac{{3R}}{{2g}}}$
This is the time period of the disc when it rotates through an axis passing through its circumference.

Therefore, the correct answer is option A.

Note:
Remember that the moment of inertia of a body changes with respect to the axis of rotation taken. The moment of inertia about an axis passing through the centre is given as ${I_{cm}} = \dfrac{{M{R^2}}}{2}$
Where M is the mass and R is the radius of the disc. Whereas the moment of inertia about an axis parallel to this axis on the circumference is $I = \dfrac{3}{2}M{R^2}$ .