Question 140: Centre of Mass and Rotational Motion

Question

A particle of mass $$\mathrm{m}$$ is projected with a velocity '$$\mathrm{u}$$' 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 :

Accepted Answer

$$\begin{aligned}
& \mathrm{L}=m u \cos 	heta H \
& =m u \cos 	heta 	imes \frac{u^2 \sin ^2 	heta}{2 g} \
& =\frac{m u^3}{2 g} 	imes \frac{\sqrt{3}}{2} 	imes\left(\frac{1}{2}ight)^2=\frac{\sqrt{3} m u^3}{16 g}
\end{aligned}$$

Answer

$$\frac{\mathrm{mu}^3}{\sqrt{2} \mathrm{~g}}$$

Answer

zero

Answer

$$\frac{\sqrt{3}}{2} \frac{\mathrm{mu}^2}{\mathrm{~g}}$$

Answer

$$\frac{\sqrt{3}}{16} \frac{\mathrm{mu}^3}{\mathrm{~g}}$$

Centre of Mass and Rotational Motion

A particle of mass $\mathrm{m} is projected with a velocity '\mathrm{u}' making an angle of 30^{\circ} with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height \mathrm{h}$ is :

Choose the correct answer: