Question

A thin uniform bar of length L and mass 8 m lies on a smooth horizontal table. Two point masses m and 2 m are moving in the same horizontal plane from opposite sides of the bar with speeds 2v and v respectively. The masses stick to the:

• A. $\frac{6v}{5l}$
• B. $\frac{3v}{5l}$
• C. $\frac{6v}{11l}$
• D. $\frac{6v}{l}$

Answer 1

The correct option is A $\frac{6v}{5l}$

Moment of inertia of uniform rod of length(l) -

$I=\frac{M{l}^2}{12}$

- wherein

About axis passing through its center & perpendicular to its length.

Law of conservation of angular moment -

$\overrightarrow{r}=\frac{d\overrightarrow{L}}{dt}$

- wherein

If net torque is zero

i.e $\frac{d\overrightarrow{L}}{dt}=0,$

$\overrightarrow{L}=$ constant

angular momentum is conserved only when external torque is zero .

Centre of mass from system from 0

$=>\frac{8m\times 0+m\left(\frac{L}{3}\right)-2m\left(\frac{L}{6}\right)}{8m+m+2m}=0$

So,centre of mass is at zero.

From conservation of ang.momentum; $L_i=L_f$

$L_i=m\left(2v\right)\times \left(\frac{L}{3}\right)+2mv\times \left(\frac{L}{6}\right)=mvL$

$L_f=\left[\right(8m)\times \frac{L^2}{12}+m{\left(\frac{L}{3}\right)}^2+2m{\left(\frac{L}{6}\right)}^2]\omega$

$=[\frac{2}{3}m{L}^2+\frac{m{L}^2}{9}+\frac{m{L}^2}{18}]\omega$

$=\left(\frac{12+2+1}{18}\right)m{L}^2\omega =\frac{5}{6}m{L}^2\omega$

$\frac{5}{6}m{L}^2\omega =mvl,$

$\therefore \omega =\frac{6v}{5L}$