Question

Two identical rods A and B are connected by a wire of length l in the middle as shown in figure. The wire is bent into an arc of a circle of radius R and released. If Y is Young's modulus of wire. Find time period of small oscillations. MOI of each rod is $I_b$ :

• A. $\sqrt{\frac{2l}{Y{r}^4}}$
• B. $\sqrt{\frac{2l{I}_b}{Y{r}^4}}$
• C. $\sqrt{\frac{2I_b}{Y{r}^4}}$
• D. $\sqrt{\frac{2l{I}_b}{Y\pi r^4}}$

Answer 1

The correct option is D $\sqrt{\frac{2l{I}_b}{Y\pi r^4}}$

$\frac{Yi}{R}=-I_{bar}\frac{d^2\theta }{d{t}^2}$
$\frac{l}{R}=2\theta ,i=\frac{\pi r^4}{4}$ and is called geometric moment of inertia.
$\frac{Y\pi r^4}{4l}\left(2\theta \right)=-I_{bar}\frac{d^2\theta }{d{t}^2}$
or $\frac{d^2\theta }{d{t}^2}=-\frac{Y\pi r^4}{2l{I}_b}\theta$
comparing with $a={\omega }^2\theta$
$\omega =\sqrt{\frac{Y\pi r^4}{2l{I}_b}}$
$w=2\pi f=\frac{2\pi }{T}$
so $T=\sqrt{\frac{2l{I}_b}{Y\pi r^4}}$