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

The object is projected in the \( xy \)-plane at an angle of \( 45^\circ \) with the x-axis. The velocity at any point can be written as:

\[ \vec{v} = v_0 (\cos 45^\circ \hat{i} + \sin 45^\circ \hat{j}). \]

Step 1: Velocity components at maximum height

At the maximum height, the vertical component of the velocity becomes zero, i.e., \( v_y = 0 \), and only the horizontal component remains: \[ v_x = v_0 \cos 45^\circ = \frac{v_0}{\sqrt{2}}. \]

Step 2: Position at maximum height

The object’s position at maximum height is given by: \[ y = \frac{v_0^2 \sin 45^\circ \cos 45^\circ}{g} = \frac{v_0^2 \sin 90^\circ}{2g} = \frac{v_0^2}{2g}. \] The horizontal position at maximum height is: \[ x = v_0 \cos 45^\circ \times \frac{v_0}{g} = \frac{v_0^2}{g\sqrt{2}}. \]

Step 3: Angular momentum at maximum height

At the maximum height, the angular momentum \( L \) with respect to the origin is given by: \[ \vec{L} = \vec{r} \times m\vec{v}. \] The position vector \( \vec{r} = x \hat{i} + y \hat{j} \) and the velocity \( \vec{v} = v_x \hat{i} \) at maximum height. The magnitude of the angular momentum is: \[ L = m \left( x v_y - y v_x \right), \] where \( v_y = 0 \) and \( v_x = \frac{v_0}{\sqrt{2}} \). Thus, the angular momentum simplifies to: \[ L = m \left( \frac{v_0^2}{g\sqrt{2}} \times \frac{v_0}{\sqrt{2}} \right) = \frac{mv_0^3}{4\sqrt{2}g}. \]

Step 4: Direction of angular momentum

Since the object was projected at an angle with the x-axis, the direction of angular momentum is along the negative z-axis, due to the direction of the rotational motion.

Final Answer:

The magnitude and direction of the angular momentum are:

\[ \boxed{\frac{mv_0^3}{4\sqrt{2}g}} \quad \text{along the negative z-axis}. \]