A grandfather clock has a pendulum that consists of a thin brass disk of radius r = 15.00 cm and mass 1.000 kg that is attached to a long thin rod of negligible mass. The pendulum swings freely about an axis perpendicular to the rod and through the end of the rod opposite the disk, as shown in Fig. 15-5 If the pendulum is to have a period of 2.000 s for small oscillations at a place where $\mathbf{g}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{800}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$,what must be the rod length L to the nearest tenth of a millimeter?

• Radius of Disc, r = 15.00 cm 0r 0.15 m.
• Mass of Disc, m = 1.000 kg.
• Period of small oscillations, T = 2.000 s.
• Acceleration due to gravity, $g=9.80\mathrm{m}/{\mathrm{s}}^2$.

The time taken by the pendulum to complete one oscillation is known as the time period of the pendulum. As we are given the period for the system, we can take the length of the pendulum as (Length of rod + Radius of the disc). From this and the formula for the period for the pendulum, we can find the length of the rod.

Formula:

The time period of a physical pendulum, $T=2\tau \tau \sqrt{\frac{l_o}{mgh}}$ (i)

Here, m is mass, $l_0$ is rotational inertia, g is gravitational acceleration, h is the length of the pendulum.

Referring the figure, we can write the length of the pendulum as, h = (Length of rod) + (Radius of Disc) as:

h = L + R

Using the parallel axis theorem,

$l_0=l_c+M{h}^2$

For a disk,

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

So, we can write

$l_0=\frac{1}{2}M{R}^2+M{(L+R)}^2$

Putting this equation in the equation (i) for time period, we have

$$\begin{gathered} T=2\tau \tau \sqrt{\frac{{\displaystyle \frac{1}{2}}M{R}^2+M{(L+R)}^2}{Mg{(L+R)}^2}} \\ =2\tau \tau \sqrt{\frac{{\displaystyle \frac{1}{2}{R}^2+{(L+R)}^2}}{g(L+R)}} \end{gathered}$$

Substituting the values of T = 2 s, R = 0.15 m , we can get the length of the rod.

$$\begin{gathered} 2s=2\tau \tau \sqrt{\frac{{\displaystyle \frac{1}{2}}{(0.15\mathrm{m})}^2+{(\mathrm{L}+0.15\mathrm{m})}^2}{9.8\mathrm{m}/{\mathrm{s}}^2(\mathrm{L}+0.15\mathrm{m})}} \\ {\left(\frac{1s}{\tau \tau }\right)}^2=\frac{{\displaystyle \frac{1}{2}{(0.15\mathrm{m})}^2+{(\mathrm{L}+0.15\mathrm{m})}^2}}{9.8\mathrm{m}/{\mathrm{s}}^2(\mathrm{L}+0.15\mathrm{m})} \\ {(L+0.15\mathrm{m})}^2-(0.99\mathrm{m})(\mathrm{L}+0.15\mathrm{m})+0.011{\mathrm{m}}^2=0 \\ (L+0.15\mathrm{m})=\frac{(0.99\mathrm{m})\pm \sqrt{{(-0.99\mathrm{m})}^2-4\times 1\times (0.011{\mathrm{m}}^2)}}{2\times 1} \end{gathered}$$

On further solving,

$$\begin{gathered} (L+0.15\mathrm{m})=0.978\mathrm{m} \\ L=(0.98-0.15)\mathrm{m} \\ L=0.83\mathrm{m} \end{gathered}$$

Hence, the value of the length of the rod is 0.83 m.