Question

A non uniform rod $OA$ of liner mass density $\lambda ={\lambda }_0\times ({\lambda }_0=const.)$ is suspended from ceiling with hinge joint $O$ & light string as shown in figure. Find the angular acceleration of rod just after the string is cut

• A. $\frac{2g}{L}$
• B. $\frac{g}{L}$
• C. $\frac{4g}{3L}$
• D. $Noneofthese$

Answer 1

The correct option is C $\frac{4g}{3L}$

$X_{cm}=\frac{\int xdm}{\int dm}=\frac{{\int }_0^{L}x\times {\lambda }_{o}xdx}{{\int }_0^L{\lambda }_{o}xdx}$

$=\frac{{\lambda }_o{\int }_o^L{x}^{2}dx}{{\lambda }_o{\int }_o^{L}xdx}$

$X_{cm}=\frac{{\left(\frac{x^3}{3}\right)}_o^L}{{\left(\frac{x^2}{2}\right)}_o^L}=\frac{2}{3}\times \frac{L^3}{L^2}=\frac{2L}{3}$

Moment of inertia about O

$dI=dm\times x^2\Rightarrow I=\int dm{x}^2$

$\Rightarrow {\int }_o^L{\lambda }_{o}xdx\times x^2\Rightarrow {\lambda }_o\int x^{3}dx$

$\Rightarrow \frac{{\lambda }^o}{4}(x^4{)}_o^{2}I=\frac{{\lambda }_o}{4}\times L^4$

& mass can be given as:

$m={\int }_o^L{\lambda }_{o}xdx=\frac{{\lambda }_o}{2}{\left(x^2\right)}_0^L=\frac{{\lambda }_2}{2}\times L^2$

So, torque about the center can be given as,
$\tau =mg\times \frac{2L}{3}=I\alpha$

$\Rightarrow \frac{{\lambda }_0}{2}\times L^{2}g\times \frac{2L}{3}=\frac{{\lambda }_0}{4}\times L^4\alpha$

$\Rightarrow \frac{{\lambda }_{o}g{L}^3}{3}={\lambda }_o\alpha \frac{L^4}{4}$

$\Rightarrow \frac{{\lambda }_{o}g}{3}=\frac{{\lambda }_o\alpha L}{4}\Rightarrow \alpha =\frac{4g}{3L}$
Hence, the correct answer is option $\left(C\right)$