Question

A particle is attached to the lower end of a uniform rod which is hinged at its other end as shown in the figure. Them minimum speed is given to the particles so that the rod performs a circular motion in a vertical plane will be :
[ length of the rod is l, consider masses of both rod and particle to be the same]

• A. $\sqrt{5gl}$
• B. $\sqrt{4gl}$
• C. $\sqrt{4.5gl}$
• D. $noneofthese$

Answer 1

The correct option is C $\sqrt{4.5gl}$

$\omega =\frac{v}{l}$

Now, decrease in $K.E.=$ Increase in P.E

or, $\frac{1}{2}I{\omega }^2=2mgl+mgl$

$\therefore \frac{1}{2}(\frac{m{l}^2}{3}+m{l}^2){\left(\frac{v}{l}\right)}^2=3mgl$

$\therefore v=\sqrt{4.5gl}$

Hence, the answer is $\sqrt{4.5gl}.$