JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A particle of mass \(m\) projected with a velocity ' \(u\) ' making an angle of \(30^{\circ}\) with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height \(\mathrm{h}\) is _______.
- A \(\frac{\sqrt{3}}{16} \frac{\mathrm{mu}^3}{\mathrm{~g}}\)
- B \(\frac{\sqrt{3}}{2} \frac{m u^2}{g}\)
- C \(\frac{m u^3}{\sqrt{2} g}\)
- D zero
Answer & Solution
Correct Answer
(A) \(\frac{\sqrt{3}}{16} \frac{\mathrm{mu}^3}{\mathrm{~g}}\)
Step-by-step Solution
Detailed explanation
\( \mathrm{L}=m u \cos \theta H \) \( =m u \cos \theta \times \frac{u^2 \sin ^2 \theta}{2 g} \) \( =\frac{m u^3}{2 g} \times \frac{\sqrt{3}}{2} \times\left(\frac{1}{2}\right)^2=\frac{\sqrt{3} m u^3}{16 g}\)
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