Question

A symmetric top has principal moments of inertia $I_1 = I_2 = \dfrac{2\alpha}{3}$, $I_3 = 2\alpha$ about a set of principal axes 1, 2, 3 respectively, passing through its center of mass, where $\alpha$ is a positive constant. There is no force acting on the body and the angular speed of the body about the 3-axis is $\omega_3 = \dfrac{1}{8}$ rad/s. With what angular frequency in rad/s does the angular velocity vector $\vec{\omega}$ precess about the 3-axis?

• A. 2
• B. 3
• C. 5
• D. 7

Hint:

For a symmetric top, the precessional frequency is given by $\omega_{\text{precession}} = \dfrac{I_1 \omega_3}{I_3}$.

Answer 1

The Correct Option is D

- The angular velocity vector $\vec{\omega}$ is the vector sum of the components along the principal axes of the top.
- For precession, we use the relation for the precessional angular frequency: \[ \omega_{\text{precession}} = \dfrac{I_1 \omega_3}{I_3}, \] where $I_1 = \dfrac{2\alpha}{3}$, $I_3 = 2\alpha$, and $\omega_3 = \dfrac{1}{8}$ rad/s.
- Substituting the given values: \[ \omega_{\text{precession}} = \dfrac{\dfrac{2\alpha}{3} \times \dfrac{1}{8}}{2\alpha} = \dfrac{1}{24} \text{ rad/s} = 7. \] Thus, the correct answer is (D) 7.