A symmetric rigid body has moment of inertia $I_1, I_2, I_3$ about its principal axes 1, 2, and 3, respectively, with $I_1 = I_3 = I_\perp$ and $I_2 \neq I_\perp$. It is rotating in space with no torque on it so that its angular momentum $\vec{L}$ is constant. Let $\omega_1, \omega_2, \omega_3$ be the components of its angular velocity along the principal axes 1, 2, and 3, respectively. Which of the following quantities is/are constant during the motion of this rigid body?

Question

On a horizontal plane, a projectile of mass $m$ is launched from the ground with speed $v_0$ at an angle $\theta_0$ with the horizontal. In addition to the gravitational force ($mg$), it also experiences a drag force $\vec{F}_{drag} = -\gamma \vec{v}$, where $\vec{v}$ is its velocity and $\gamma$ is a constant. It hits the ground at a distance $R$ from the point of launch with its velocity making an angle $\theta$ with the horizontal, as shown schematically in the figure below.

Then which of the following options is correct?

Answer 1

Step 1: List the two conserved quantities available without touching Euler's equations directly.
For a torque-free rigid body, the space-frame angular momentum vector $\vec{L}$ is completely fixed in magnitude and direction, and the rotational kinetic energy $T$ is conserved since no torque does any work. In body-axis components, using $I_1=I_3=I_\perp$, these read
\[ L^2 = I_\perp^2(\omega_1^2+\omega_3^2) + I_2^2\,\omega_2^2 \]
\[ 2T = I_\perp(\omega_1^2+\omega_3^2) + I_2\,\omega_2^2 \]
with $L^2$ and $T$ both constant in time.

Step 2: Solve this pair of equations for the two combinations that appear.
Treat $X = \omega_1^2+\omega_3^2$ and $Y=\omega_2^2$ as unknowns in the linear system $I_\perp^2 X + I_2^2 Y = L^2$ and $I_\perp X + I_2 Y = 2T$. Since $I_\perp \neq I_2$, this system has a unique solution for $X$ and $Y$ in terms of the constants $L^2, T, I_\perp, I_2$. So both $X=\omega_1^2+\omega_3^2$ and $Y=\omega_2^2$ come out as fixed numbers, not functions of time.

Step 3: Read off statements (B) and (D).
$X = \omega_1^2+\omega_3^2$ being constant directly confirms statement (B) is TRUE. $Y=\omega_2^2$ being constant means $\omega_2$ keeps a fixed magnitude throughout the motion; since $\omega_2$ varies continuously in time from one starting value, it cannot jump in sign, so $\omega_2$ itself stays constant, confirming statement (D) is TRUE.

Step 4: Check statement (A) using the same constants.
Knowing only that $\omega_1^2+\omega_3^2$ is fixed does not fix the individual values of $\omega_1$ and $\omega_3$, they can still trade off with each other over time, tracing a circle of fixed radius in the $\omega_1$-$\omega_3$ plane. Their plain sum $\omega_1+\omega_3$ is not one of the conserved combinations $X$ or $Y$, and it changes as the body turns, so statement (A) is FALSE.

Step 5: Check statement (C) using $L^2$ and the body-frame components.
The angle $\theta$ between the symmetry axis (axis 2) and $\vec{L}$ satisfies $L_2 = L\cos\theta$, where $L_2 = I_2\omega_2$ is the body-frame component of $\vec{L}$ along axis 2. Since $I_2$ is fixed, $\omega_2$ is constant (Step 3), and $L=|\vec{L}|$ is constant, the ratio $\cos\theta = I_2\omega_2/L$ is built from constants alone. So $\theta$ does not change with time, confirming statement (C) is TRUE.

Final Answer:
Using only the conservation of $\vec{L}$ and kinetic energy, without solving Euler's equations directly, again shows that $\omega_1^2+\omega_3^2$, $\omega_2$, and the axis-2 to $\vec{L}$ angle stay fixed, while $\omega_1+\omega_3$ does not. \[ \boxed{\text{B, C, D}} \]

Show Hint

The Correct Option is B, C, D

Solution and Explanation

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