Question

A massless rod $S$ having length $2 l$ 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. $\left|\frac{d L}{d t}\right|$ equals

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

The radius of the circle followed by the masses is $r=1 \sin \alpha$ As, angular momentum, $L = r \times P$ $= r \times m v$ $\Rightarrow | L |=I \sin \theta(m \omega I \sin \theta)$ On differentiating, we get $ \frac{d| L |}{d t} =m \omega l^{2} 2 \sin \theta \cdot \cos \theta \frac{d \theta}{d t} $ $\Rightarrow \left|\frac{d L}{d t}\right| =2 m l^{2} \omega^{2} \sin \theta \cdot \cos \theta $ $=m l^{2} \omega^{2} \sin 2 \theta$