Question 408: Centre of Mass and Rotational Motion

Question

A body of mass '$$m$$' is projected with a speed '$$u$$' making an angle of $$45^{\circ}$$ with the ground. The angular momentum of the body about the point of projection, at the highest point is expressed as $$\frac{\sqrt{2} m u^3}{X g}$$. The value of '$$X$$' is _________.

Accepted Answer

$$\begin{aligned}
& \mathrm{L}=\mathrm{mu} \cos 	heta \frac{\mathrm{u}^2 \sin ^2 	heta}{2 \mathrm{~g}} \
& =\mathrm{mu}^3 \frac{1}{4 \sqrt{2} \mathrm{~g}} \Rightarrow \mathrm{x}=8
\end{aligned}$$

Centre of Mass and Rotational Motion

A body of mass '$m' is projected with a speed 'u' making an angle of 45^{\circ} with the ground. The angular momentum of the body about the point of projection, at the highest point is expressed as \frac{\sqrt{2} m u^3}{X g}. The value of 'X$' is _________.

Integer / Numeric Answer Type