GATE 2026 PH – Question 55
A symmetric rigid body has moment of inertia I1, I2, I3 about its principal axes 1, 2, and 3, respectively, with I1 = I3 = I⊥ and I2 ≠I⊥. It is rotating in space with no torque on it so that its angular momentum L⃗ is constant. Let ω1, ω2, ω3 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?
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Correct answer: (B) $\omega_1^2+\omega_3^2$; (C) Angle between axis 2 and $\mathbf L$; (D) $\omega_2$
Explanation
For a torque-free symmetric rigid body, the angular momentum $\mathbf L$ is constant in space. In the body frame with principal moments $I_1=I_3=I_\perp$ and $I_2$, the Euler equations are
$$I_\perp\dot\omega_1=(I_\perp-I_2)\omega_2\omega_3,\quad I_2\dot\omega_2=(I_3-I_1)\omega_3\omega_1=0,\quad I_\perp\dot\omega_3=(I_2-I_\perp)\omega_1\omega_2 .$$
- **$\omega_2$ is constant** (the middle equation gives $\dot\omega_2=0$). ✓ (D)
- With $\omega_2$ constant, the pair $(\omega_1,\omega_3)$ rotates in its plane at the constant angular rate $\Omega_p=\dfrac{(I_2-I_\perp)}{I_\perp}\omega_2$. Check: $\dfrac{d}{dt}(\omega_1^2+\omega_3^2)=2\omega_1\dot\omega_1+2\omega_3\dot\omega_3=0$ since the two terms cancel. So **$\omega_1^2+\omega_3^2$ is constant**. ✓ (B)
- The component of $\mathbf L$ along axis 2 is $L_2=I_2\omega_2$ (constant) and $|\mathbf L|$ is constant, so the **angle between axis 2 and $\mathbf L$ is constant**. ✓ (C)
- $\omega_1+\omega_3$ is not constant because the vector $(\omega_1,\omega_3)$ rotates. ✗ (A)
Answer **B, C and D**.