Question

Find $KE$ of the rotating composite rod comprising two rods of mass $m$ and $2m$ and each of length $l=1m$. Assume $\omega =10rad{s}^{-1}$ and $m=1kg$.

Answer 1

According to the question

$l=1m$ $w=10rad/s$ $m=1kg$

Moment of interia of rod of mass m anbout hinge

$l_1=I_{cm}+m{l}^2$

$I_1=\frac{m{l}^2}{12}+m{\left(\frac{l}{2}\right)}^2=\frac{m{l}^2}{12}+\frac{m{l}^2}{4}=\frac{m{l}^2}{3}$

$I=\frac{1}{3}$

moment of inertia of rod of ,mass $2m$ about hings

$I_2=I_{cm}+\left(2m\right)d^2$

$I_2=\frac{\left(2m\right)l^2}{12}+\left(2m\right){\left(\frac{3l}{2}\right)}^2=\frac{m{l}^2}{2}=\frac{Jm{l}^2}{2}=\frac{28m{l}^2}{6}=\frac{14m{l}^2}{3}$

$I_2=\frac{14}{3}$

$K=\frac{1}{2}\times \frac{1}{3}\times (10{)}^2+\frac{1}{2}\times \frac{14}{3}\times (10{)}^2=\frac{{10}^2}{2}(\frac{1}{3}\times \frac{14}{3})=50\times \frac{15}{3}$

$KE=250J$

Thus the total rimetic energy of the system is $250J$