Question

A particle of mass m is projected from ground with speed u at an angle of 30° with the horizontal. Find its angular momentum about the point of projection when it reaches its maximum height.

• A. \(\frac{mu^3}{16g}\)
• B. \({\frac{\sqrt3mu^3}{16g}}\)
• C. \(\frac{mu^3}{3g}\)
• D. Zero

Answer 1

The Correct Option is B

To determine the angular momentum of the particle at its maximum height relative to its projection point, follow these steps:

1. Calculate the initial velocity components for a projection angle of \( \theta = 30^\circ \):
   • Horizontal velocity: \(u_x = u \cos 30^\circ = \frac{\sqrt{3}u}{2}\)
   • Vertical velocity: \(u_y = u \sin 30^\circ = \frac{u}{2}\)
2. At maximum height, the vertical component of velocity is zero.
3. Determine the time to reach maximum height \( t_m \) using \(v_y = u_y - gt\), setting \(v_y = 0\). This yields \(0 = \frac{u}{2} - gt_m \Rightarrow t_m = \frac{u}{2g}\).
4. At this point, the vertical velocity is zero, and the horizontal velocity remains \(u_x = \frac{\sqrt{3}u}{2}\).
5. Calculate the maximum height \( H \) using \(H = u_y t_m - \frac{1}{2} g t_m^2\). Substituting values gives \(H = \frac{u}{2} \cdot \frac{u}{2g} - \frac{1}{2} g \left(\frac{u}{2g}\right)^2 = \frac{u^2}{8g}\).
6. The angular momentum \( L \) about the projection point is given by \( L = m(v_x \cdot h - v_y \cdot x) \). Since \(v_y = 0\) at maximum height, this simplifies to \( L = m(v_x \cdot h) \).
7. Substitute the values for horizontal velocity and height at maximum height:
   • \(v_x = \frac{\sqrt{3}u}{2}\)
   • \(h = \frac{u^2}{8g}\)

Therefore, the angular momentum is \( L = m \left(\frac{\sqrt{3}u}{2}\right) \left(\frac{u^2}{8g}\right) = {\frac{\sqrt3mu^3}{16g}} \).