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11.22: Precession of a Gyroscope

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  • A drawing of a gyroscope, consisting of a disk that can spin on an shaft, perpendicular to the plane of the disk and through its center. Two rings surround the gyroscope. One is attached to the shaft above and below the disk, and the other is attached to the first ring and is in the plane of the disk so that this second ring is concentric with the disk.
    Figure 11.21, we see that the magnitude of the torque is

    $$\tau = rMg \sin \theta \ldotp$$


    $$dL = rMg \sin \theta dt \ldotp$$

    The angle the top precesses through in time dt is

    $$d \phi = \frac{dL}{L \sin \theta} = \frac{rMg \sin \theta}{L \sin \theta} dt = \frac{rMg}{L} dt \ldotp$$

    The precession angular velocity is \(\omega_{P} = \frac{d \phi}{dt}\) and from this equation we see that

    $$\omega_{P} = \frac{rMg}{L} \ldotp$$

    or, since L = I\(\omega\),

    $$\omega_{P} = \frac{rMg}{I \omega} \ldotp \tag{11.12}$$

    In this derivation, we assumed that \(\omega_{P}\) << \(\omega\), that is, that the precession angular velocity is much less than the angular velocity of the gyroscope disk. The precession angular velocity adds a small component to the angular momentum along the z-axis. This is seen in a slight bob up and down as the gyroscope precesses, referred to as nutation.

    Earth itself acts like a gigantic gyroscope. Its angular momentum is along its axis and currently points at Polaris, the North Star. But Earth is slowly precessing (once in about 26,000 years) due to the torque of the Sun and the Moon on its nonspherical shape.

    Example \(\PageIndex{1}\): Period of Precession

    A gyroscope spins with its tip on the ground and is spinning with negligible frictional resistance. The disk of the gyroscope has mass 0.3 kg and is spinning at 20 rev/s. Its center of mass is 5.0 cm from the pivot and the radius of the disk is 5.0 cm. What is the precessional period of the gyroscope?


    We use Equation 11.12 to find the precessional angular velocity of the gyroscope. This allows us to find the period of precession.


    The moment of inertia of the disk is

    $$I = \frac{1}{2} mr^{2} = \frac{1}{2} (0.30\; kg)(0.05\; m)^{2} = 3.75 \times 10^{-4}\; kg\; \cdotp m^{2} \ldotp$$

    The angular velocity of the disk is

    $$20.0\; rev/s = (20.0)(2 \pi)\; rad/s = 125.66\; rad/s \ldotp$$

    We can now substitute in Equation 11.12. The precessional angular velocity is

    $$\omega_{P} = \frac{rMg}{I \omega} = \frac{(0.05\; m)(0.3\; kg)(9.8\; m/s^{2})}{(3.75 \times 10^{-4}\; kg\; \cdotp m^{2})(125.66\; rad/s)} = 3.12\; rad/s \ldotp$$

    The precessional period of the gyroscope is

    $$T_{P} = \frac{2 \pi}{3.12\; rad/s} = 2.0\; s \ldotp$$


    The precessional angular frequency of the gyroscope, 3.12 rad/s, or about 0.5 rev/s, is much less than the angular velocity 20 rev/s of the gyroscope disk. Therefore, we don’t expect a large component of the angular momentum to arise due to precession, and Equation 11.12 is a good approximation of the precessional angular velocity.

    Exercises \(\PageIndex{1}\)

    A top has a precession frequency of 5.0 rad/s on Earth. What is its precession frequency on the Moon?


    • Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).