In Figure, a 2.50 kgdisk of diameter D=42.0 cmis supported by a rod of length L=76.0 cmand negligible mass that is pivoted at its end.

1. With the massless torsion spring unconnected, what is the period of oscillation?
2. With the torsion spring connected, the rod is vertical at equilibrium. What is the torsion constant of the spring if the period of oscillation has been decreased by $\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{}\mathbf{s}$?

• The mass of the disk is, m=2.5 kg
• The length of the rod is, L=76.0 cm=0.76 m
• The diameter of disk is, D=42.0 cm=0.42 m

When a body is connected to a spring such that its vertical motion is allowed, then the body undergoes small oscillations due to the torque acting on it at one end that is pivoted. To calculate the period of oscillation, you can take the moment of inertia of the metal disk at an axis that is parallel to the central axis of the disk and is passing through the pivoted end. This causes a torsion balance of the disk giving an equilibrium concept. In the second case, you are given a change in the period of oscillation but as there is no change in setting up the change in period is due to the addition of the torsion constant due to the pivoted end.

Formulae:

The period of oscillation of a body undergoing SHM,

$\mathrm{T}=2\mathrm{\pi }\sqrt{\frac{{\mathrm{I}}_{\mathrm{p}}}{\mathrm{mgh}}}$ ….. (i)

where, I_p is the moment of inertia of the body about the axis, m is the mass of the system, g is the acceleration due to gravity, h is the height of the rise due to oscillation.

The moment of inertial of a disk about the central disk,

${\mathrm{I}}_{\mathrm{c}}=\frac{{\mathrm{mR}}^2}{2}$ ….. (ii)

Where, m is the mass of the disk, R is the radius of the disk.

The moment of inertia of a disk about a parallel axis according to parallel axis theorem,

${\mathrm{I}}_{\mathrm{p}}=\frac{{\mathrm{mR}}^2}{2}+{\mathrm{mh}}^2$ ….. (iii)

Where, m is the mass of the disk, R is the radius of the disk, h is the distance of the parallel axis from the central axis of the disk.

Here, the moment of inertia is acting about the pivot, thus, the distance of the parallel axis will be:

$\begin{array}{rcl}\mathrm{h}& =& \mathrm{L}+\mathrm{R}\\ & =& 0.76\mathrm{m}+\frac{0.42\mathrm{m}}{2}\\ & =& 0.97\mathrm{m}\end{array}$

Now, the equation for period of torsion pendulum at a distance h from the disk can be given using equation (iii) in equation (i) as follows:

$\mathrm{T}=2\mathrm{\pi }\sqrt{\frac{{\displaystyle \frac{{\mathrm{mR}}^2}{2}}+{\mathrm{mh}}^2}{\mathrm{mgh}}}$ ….. (iv)

Here, the radius is,

$\begin{array}{rcl}\mathrm{R}& =& \frac{\mathrm{D}}{2}\\ & =& \frac{0.42\mathrm{m}}{2}\\ & =& 0.21\mathrm{m}\end{array}$

The acceleration due to gravity is,

$\mathrm{g}=9.8\mathrm{m}/{\mathrm{s}}^2$

Now substitute known values into equation (iv).

$\begin{array}{rcl}\mathrm{T}& =& 2\times 3.14\sqrt{\frac{{\displaystyle \frac{\left(2.5\mathrm{kg}\right){\left(0.21\mathrm{m}\right)}^2}{2}+\left(2.5\mathrm{kg}\right){\left(0.97\mathrm{m}\right)}^2}}{\left(2.5\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(0.97\mathrm{m}\right)}}\\ & =& 6.28\sqrt{\frac{\left(0.055+2.352\right)\mathrm{kg}.{\mathrm{m}}^2}{23.765\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2}}\\ & =& 6.28\times 0.32\mathrm{s}\\ & =& 2.0\mathrm{s}\end{array}$

Therefore, the period of oscillation when the torsion spring is not connected is 2.0 s.

Now, when the torsion spring is connected, the torque acting at the center of the disk will become

$\mathrm{\zeta }=\mathrm{mgh}+\mathrm{\kappa }$

Where,$\mathrm{\kappa }$is the torsion constant of the spring.

So, the equation for new period will become using the above value in equation (i) as follows:

${\mathrm{T}}^{\text{'}}=2\mathrm{\pi }\sqrt{\frac{{\displaystyle {\mathrm{I}}_{\mathrm{p}}}}{\mathrm{mgh}+\mathrm{\kappa }}}$ ….. (v)

Here, T’ is the new period.

Therefore, the new period is,

$\begin{array}{rcl}{\mathrm{T}}^{\text{'}}& =& \mathrm{T}-0.5\mathrm{s}\\ & =& 2.0\mathrm{s}-0.5\mathrm{s}\\ & =& 1.5\mathrm{s}\end{array}$

Rearranging the equation (v) for the torsion constant, you get the required value of the constant using the given data and equation (iii) as follows:

$$\begin{gathered} \mathrm{\kappa }=\frac{4{\mathrm{\pi }}^2}{{\mathrm{T}}^{\text{'}2}}{\mathrm{I}}_{\mathrm{p}}-\mathrm{mgh} \\ \mathrm{\kappa }=\frac{4{\mathrm{\pi }}^2}{{\mathrm{T}}^{\text{'}2}}\left(\frac{{\mathrm{mR}}^2}{2}+{\mathrm{mh}}^2\right)-\mathrm{mgh} \end{gathered}$$

Substitute known values in the above equation.

$\begin{array}{rcl}\mathrm{\kappa }& =& \frac{4\times {\left(3.14\right)}^2}{{\left(1.5\mathrm{s}\right)}^2}\left(\frac{\left(2.5\mathrm{kg}\right){\left(0.21\mathrm{m}\right)}^2}{2}+\left(2.5\mathrm{kg}\right){\left(0.97\mathrm{m}\right)}^2\right)-\left(2.5\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(0.97\mathrm{m}\right)\\ & =& 17.528{\mathrm{s}}^{-2}\left(0.055\mathrm{kg}.{\mathrm{m}}^2+2.352\mathrm{kg}.{\mathrm{m}}^2\right)-23.\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2\\ & =& \left(42.21-23.7\right)\mathrm{N}.\mathrm{m}/\mathrm{rad}\\ & =& 18.5\mathrm{N}.\mathrm{m}/\mathrm{rad}\end{array}$

Hence, the torsion constant of the spring is 18.5 N.m/rad.