A flat uniform circular disk has a mass of 3.00kgand a radius of 70.0cm. It is suspended in a horizontal plane by a vertical wire attached to its center. If the disk is rotated 2.50 radabout the wire, a torque of 0.600 N.mis required to maintain that orientation.

1. Calculate the rotational inertia of the disk about the wire.
2. Calculate the torsion constant.
3. Calculate the angular frequency of this torsion pendulum when it is set oscillating.

• Mass of the disk, m=3.00 kg.
• Radius of the disk, r=70.0cm or 0.7m.
• Rotation about the wire,$\theta =2.50\mathrm{rad}$.
• Value of torque, role="math" localid="1657272842504" $\tau =0.0600\mathrm{Nm}$.

The torque acting on the object is equal to the moment of force. The toque is also equal to the product of the moment of inertia and angular acceleration of the object. It can be written as the time rate of change of angular momentum of the object.

The moment of inertia of an object is equal to the sum of the product of mass and the perpendicular distance of all the points from the axis of rotation.

The angular frequencyis related to the period and frequency of the motion by,

$\mathit{\omega }\mathbf{=}\frac{\mathbf{2}\mathbf{\tau \tau }}{\mathbf{T}}\mathbf{=}\mathbf{2}\mathit{\tau }\mathit{\tau }\mathit{f}\mathbf{=}\sqrt{\frac{\mathbf{k}}{\mathbf{m}}}$

Here T is the time period, f is the frequency, k is force constant, and mis mass.

Using the expression for moment of inertia, torque, and angular frequency we can find the required values from the concept of simple harmonic oscillations of a body.

Formula:

The moment of inertia of a pendulum, $l=\frac{1}{2}m{r}^2$ (i)

Where is mmass, ris the radius.

The torque of an oscillation, $\tau =-k\theta$ (ii)

Where is k torsion constant, $\theta$is angular displacement.

The angular frequency of a body, localid="1657273170598" $\omega =\sqrt{\left(k/l\right)}$ (iii)

Using equation (i), find the rotational inertia of the disk by substituting the values of mass and radius in equation (i).

$$\begin{gathered} l=\frac{1}{2}\times 3\mathrm{kg}\times {\left(0.7\mathrm{m}\right)}^2 \\ =\frac{1}{2}\times 3\times 0.49 \\ =0.735\mathrm{kg}.{\mathrm{m}}^2 \end{gathered}$$

Hence, the value of rotational inertia is 0.735 kg. ${\mathrm{m}}^2$.

Using equation (ii), the torsion constant can be found by substituting the value of torque and angular displacement.

$$\begin{gathered} k=\frac{\tau }{\theta } \\ =\frac{0.0600\mathrm{N}.\mathrm{m}}{2.50\mathrm{rad}} \\ =0.024\mathrm{N}.\mathrm{m}/\mathrm{rad} \end{gathered}$$

Hence, the value of the torsion constant is.0.024 N.m/rad.

Using equation (iii), the value of angular frequency of the torsion pendulum is given as:

$$\begin{gathered} \omega =\sqrt{\frac{0.024}{0.735}} \\ =0.181\mathrm{rad}/\mathrm{s} \end{gathered}$$

Hence, the value of angular frequency is 0.181 rad/s.