Question

A uniform rod of mass $m$ and length $L$ is rotated about an axis passing through the point $P$ as shown in figure. The magnitude of angular momentum of the rod, about the rotational axis $y{y}^{\prime }$ passing through the point $P$ is

• A. $\frac{M{L}^2}{18}$
• B. $\frac{M(x+y{)}^2}{9}$
• C. $\frac{M(L^2+x^2)}{9}$
• D. $\frac{M(L^2+2y^2)}{18}$

Answer 1

The correct option is B $\frac{M(x+y{)}^2}{9}$

We have $\frac{x}{y}=\frac{2}{1}$ or $\frac{x+y}{y}=\frac{3}{1}$
or $y=\frac{x+y}{3}=\frac{L}{3}$
$\Rightarrow x=2y=\frac{2L}{3}$
Now, $I_P=I_{com}+M(d{)}^2$
$=\frac{M(L{)}^2}{12}+M{(\frac{L}{2}-\frac{L}{3})}^2$
$=\frac{M{L}^2}{12}+\frac{M{L}^2}{36}$
$=\frac{4M{L}^2}{36}=\frac{M{L}^2}{9}=\frac{M(x+y{)}^2}{9}$.