$I$A uniform steel bar swings from a pivot at one end with a period of$1.2s$ How long is the bar?

A pendulum is an extended object that, for tiny angular displacements, may be modeled to maneuver in simple periodic motionwitha couple of pivots that don't undergo the middle of mass.

$T=2\pi \sqrt{\frac{I}{mgd}}$

where $I$is the moment of inertia of the thing about an axis through the pivot and $d$is that the distance from the pivot to the middle of a mass of the article.

• We have an even steel bar swinging from a pivot at one end.
• The period of the pendulum is:
• $T=1.2\mathrm{s}$

We are asked to see the length of the bar.

The moment of inertia ofthe same bar rotating about an axis through one end is found from Table $12.2$

$I=\frac{1}{3}m{L}^2$

Where$L$is the length of the bar.

The period of the pendulum is found in Equation $\left(1\right):$

$T=2\pi \sqrt{\frac{I}{mgd}}$

Since the bar is uniform,the middle of massis found inthe middle and $d=L/\mathbf{2}:$

$T=2\pi \sqrt{\frac{I}{mgL/2}}$

$=2\pi \sqrt{\frac{2I}{mgL}}$

Substitute for $I$Equation $\left(2\right):$

$\Gamma =2\pi \sqrt{\frac{2\left(\frac{1}{3}ma{L}^2\right)}{man\pi }}$

$=2\pi \sqrt{\frac{2L}{3g}}$

Solve for $L$:

$L=\frac{3g{T}^2}{8{\pi }^2}$

Substitute of numerical value.

$L=\frac{3\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)(1.2\mathrm{s}{)}^2}{8{\pi }^2}$

$=0.54m$