Question

A uniform rod pivoted at upper end is released when it is making an angle of ${60}^o$. Find the radial force acting on a particle of mass dm at its tip when it makes an angle of ${37}^o$ with the vertical.

• A. $0.6$ dm g
• B. $0.8$ dm g
• C. $0.9$ dm g
• D. None

Answer 1

The correct option is C $0.9$ dm g

$\frac{1}{2}I{\omega }^2=mg\frac{1}{2}(cos37-cos60)$.
or
$\frac{m{l}^2}{3}{\omega }^2=mgl(.8-.5)$.
$\frac{M{\nu }^2}{l}=0.9mg$ or $\frac{{\nu }^2}{l}=0.9g$.
$F_{rad}=dm\frac{{\nu }^2}{l}=0.9\left(dm\right)g$