A physical pendulum consists of a meter stick that is pivoted at a small hole drilled through the stick a distance$\mathit{d}$ from the$\mathbf{50}\mathbf{cm}$mark. The period of oscillation is$\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{s}$. Find$\mathit{d}$.

1. Distance of pendulum from a small hole,$L=1\mathrm{m}$.
2. Time period of oscillation,$T=2.5\sec$.

We can write the period in terms of rotational inertia and torque. We can find the rotational inertial of a pendulum using the parallel axis theorem. By substituting the equation for rotational inertia and torque in the formula for the time, we can find the time.

Formula:

The total moment of inertia using the parallel axis theorem, $I=I_{cm}+m{h}^2$………(i) The moment of inertia of the center of mass of the pendulum,$I_{cm}=\frac{m{L}^2}{12}$……...(ii)

The time period of a pendulum, $T=2\pi \sqrt{IIt}$……….(iii)

Torque of a body in motion,$t=d\times mg$………(iv)

where, is the weight of the body.

From equations (i) & (ii), we get the moment of inertia as:

$I=\frac{m{L}^2}{12}+m{h}^2$

We haveL as thelength of the scale and h = d. So, we can write equation (iii) as:

$$\begin{gathered} T=2\tau \tau \sqrt{\frac{{\displaystyle \frac{m{L}^2}{12}}+m{d}^2}{mgh}} \\ {T}^2=4\tau {\tau }^2\left(\frac{{\displaystyle \frac{m{L}^2}{12}+m{d}^2}}{mgh}\right) \\ =4\tau {\tau }^2\left(\frac{{\displaystyle \frac{L^2}{12}+d^2}}{gd}\right) \end{gathered}$$

Substitute the values.

$$\begin{gathered} {\left(2.5\right)}^2\times \left(9.8\right)d=4{\left(3.14\right)}^2\left(\frac{1}{12}+d^2\right) \\ 61.25d=39.43\left(0.083+d^2\right) \\ 61.25d=3.27+39.43d^2 \\ 0=39.43d^2-61.25d+3.27 \end{gathered}$$

Now it becomes quadratic in d, so now we can find its roots. After solving it, we get,$d=0.056\mathrm{m}$

Hence, the distance is found to be 0.056 m.