Question

An object of mass \( m \) is projected from the origin in a vertical \( xy \)-plane at an angle \( 45^\circ \) with the x-axis with an initial velocity \( v_0 \). The magnitude and direction of the angular momentum of the object with respect to the origin, when it reaches the maximum height, will be:

• A. \( \frac{mv_0^3}{2\sqrt{2}g} \) along negative z-axis
• B. \( \frac{mv_0^3}{2\sqrt{2}g} \) along positive z-axis
• C. \( \frac{mv_0^3}{4\sqrt{2}g} \) along positive z-axis
• D. \( \frac{mv_0^3}{4\sqrt{2}g} \) along negative z-axis

Hint:

The angular momentum for projectile motion can be found by using the position and velocity vectors at the point of interest. For maximum height, only the horizontal velocity component contributes to angular momentum.

Answer 1

The Correct Option is D

This problem requires calculating the angular momentum of an object at its peak altitude. We will proceed with a step-by-step solution:

1. An object projected at a \(45^\circ\) angle with an initial velocity \(v_0\) has the following initial velocity components:

\(v_{0x} = v_0 \cos 45^\circ = \frac{v_0}{\sqrt{2}}\) and \(v_{0y} = v_0 \sin 45^\circ = \frac{v_0}{\sqrt{2}}\)

1. At the maximum height, the object's vertical velocity component is zero: \(v_y = 0\).
2. The horizontal velocity component remains constant throughout the object's trajectory: \(v_x = \frac{v_0}{\sqrt{2}}\)
3. Since the vertical velocity is zero at the maximum height, only the horizontal velocity influences the angular momentum at this point.
4. The time taken to reach the maximum height is calculated as:

\(t = \frac{v_{0y}}{g} = \frac{v_0/\sqrt{2}}{g} = \frac{v_0}{\sqrt{2}g}\)

1. Consequently, the horizontal distance (x-coordinate) covered at the maximum height is:

\(x = v_{0x} \cdot t = \frac{v_0}{\sqrt{2}} \cdot \frac{v_0}{\sqrt{2}g} = \frac{v_0^2}{2g}\)

1. The object's position coordinates at the maximum height are \((x, y) = \left(\frac{v_0^2}{2g}, \frac{v_0^2}{4g}\right)\), where \(y = \frac{v_{0y}^2}{2g} = \frac{v_0^2}{4g}\).
2. Angular momentum \(L\) relative to the origin is determined by the cross product of the position vector and momentum:

\(L = \mathbf{r} \times \mathbf{p} = m(x \, v_y - y \, v_x)\)

1. Substituting \(v_y = 0\) simplifies the equation to:

\(L = -m \left(\frac{v_0^2}{4g}\right) \left(\frac{v_0}{\sqrt{2}}\right) = -\frac{mv_0^3}{4\sqrt{2}g}\)

1. The negative sign indicates that the angular momentum is directed along the negative z-axis.

Therefore, the magnitude of the object's angular momentum with respect to the origin at its maximum height is \( \frac{mv_0^3}{4\sqrt{2}g} \), and its direction is along the negative z-axis.