Question

A disk of mass 3 kgand radius0.15 mhangs in the xy plane from a horizontal low-friction axle. The axle is $\text{0.09 m}$ from the center of the disk. What is the frequencyof small-angle oscillations of the disk? What is the period?

Short answer

The frequency of small angle oscillation of the disk is $1.07\text{Hz}$.

Step-by-step solution

Define the frequency of Oscillations.

The frequency of the oscillation is the number of oscillations in one second. If the particle completes one oscillation in $T$ seconds, then the number of oscillations in one second or frequency is given as,

$\mathit{f}\mathbf{=}\frac{\mathbf{1}}{\mathbf{T}}$

The frequency of oscillation is measured in Hertz.

About the period of oscillations:

The period of the oscillations of a disk in a vertical plane through the point $S$ is,

$T=2\pi \sqrt{\frac{I_s}{Mgh}}$ ….. (1)

Here, $I_s$ is moment of inertia of the disk relative to $M$is mass of the disk, $h$is distance from the centre of mass of disk to the suspension point $S$ and $g$ is acceleration due to gravity.

Figure shows that the oscillations of a disk in a vertical plane: 

The diagram shows the oscillations of a disk in a vertical plane supported by a horizontal axis through a point $S$.

Find the frequency of the disk: 

The moment of inertia of the disk relative to point$O$is,

$I_O=\frac{1}{2}M{R}^2$

Here, $M$ is mass of the disk and is radius of the disk.

Thus, the moment of inertia of the disk relative to is,

$\begin{array}{rcl}{I}_s& =& I_O+M{h}^2\\ & =& \frac{1}{2}M{R}^2+M{h}^2\\ & & \end{array}$

Substitute for into equation (1).

$\begin{array}{rcl}T& =& 2\pi \sqrt{\frac{\frac{1}{2}M{R}^2+M{h}^2}{Mgh}}\\ & =& 2\pi \sqrt{\frac{R^2+2h^2}{2gh}}\\ & & \end{array}$

Substitute $\text{0.15 m}$ for $R$, $0.09$ for$h$ , and $9.8\mathrm{m}/{\mathrm{s}}^2$ for $g$ in the above equation.

$T=2\times 3.14\sqrt{\frac{{\displaystyle {\left(0.15\text{m}\right)}^2+2\left(0.09\text{m}\right)}}{2(9.8\mathrm{m}/{\mathrm{s}}^2){\displaystyle \left(0.09\text{m}\right)}}}$

$\begin{array}{rcl}T& =& 6.28\sqrt{\frac{{\displaystyle \left(0.0225+0.0162\right){\mathrm{m}}^2}}{1.764{\mathrm{m}}^2/{\mathrm{s}}^2}}\\ & =& 6.28\left(0.148\text{s}\right)\\ & =& 0.93\text{s}\\ & & \end{array}$

Therefore, the period of the oscillations of a disk is $0.93\text{s}$.

The frequency of small angle oscillation of the disk is,

$f=\frac{1}{T}$

Substitute $0.93\mathrm{s}$for $\mathrm{T}$

$\begin{array}{rcl}f& =& \frac{1}{0.93\text{s}}\\ & =& 1.07\text{Hz}\\ & & \end{array}$

Hence, the frequency of small angle oscillation of the disk is $1.07\text{Hz}$.