Question

A massless rod S having length $2l$ has equal point masses attached to its two ends as shown in figure. The rod is rotating about an axis passing through its centre and making angle $\alpha$ with the axis. The magnitude of change of momentum of rod i.e. $\left|\begin{array}{c}\frac{dL}{dt}\end{array}\right|$ equals

• A. $2m{l}^3{\omega }^2\sin \theta \cdot \cos \theta$
• B. $m{l}^2{\omega }^2\cdot \sin 2\theta$
• C. $m{l}^2\sin 2\theta$
• D. $m^{1/2}{l}^{1/2}\omega \sin \theta \cdot \cos \theta$

Answer 1

The correct option is B $m{l}^2{\omega }^2\cdot \sin 2\theta$

We know, that rate of change of angular momentum is equal to the tirque acting on the system.

$\Rightarrow |\frac{dL}{dt}|=\tau$

From the figure, Force on the mass m is given as:

$F=mr{\omega }^2$ (Due to centipetal force)

$r=lSin\alpha$

$F=m\times lSin\alpha \times {\omega }^2$

$\Rightarrow \tau =r_{perpendicukar}\times F$

$r_{perpendicukar}=lCos\alpha$

$\Rightarrow \tau =lCos\alpha \times mlSin\alpha {\omega }^2$

$\Rightarrow \tau =m{l}^2{\omega }^{2}Sin\alpha Cos\alpha$

Total Torque due to both the masses is given by

${\tau }_{net}={\tau }_1+{\tau }_2=m{l}^2{\omega }^{2}Sin\alpha Cos\alpha +m{l}^2{\omega }^{2}Sin\alpha Cos\alpha$

$\Rightarrow {\tau }_{net}=2m{l}^2{\omega }^{2}Sin\alpha Cos\alpha =m{l}^2{\omega }^{2}Sin2\alpha$

$\therefore |\frac{dL}{dt}|={\tau }_{net}=m{l}^2{\omega }^{2}Sin2\alpha$

Hence, the correct answer is OPTION B.