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We all know how easy it is for a bicycle to tip over when sitting on it at rest. But when riding the bicycle at a good pace, it is harder to tip it over because we must change the angular momentum vector of the spinning wheels.

View the video on gyroscope precession for a complete demonstration of precession of the bicycle wheel.

Also, when a spinning disk is put in a box such as a Blu-Ray player, try to move it. It is easy to translate the box in a given direction but difficult to rotate it about an axis perpendicular to the axis of the spinning disk, since we are putting a torque on the box that will cause the angular momentum vector of the spinning disk to precess.

We can calculate the precession rate of the top in [link] . From [link] , we see that the magnitude of the torque is

τ = r M g sin θ .

Thus,

d L = r M g sin θ d t .

The angle the top precesses through in time dt is

d ϕ = d L L sin θ = r M g sin θ L sin θ d t = r M g L d t .

The precession angular velocity is ω P = d ϕ d t and from this equation we see that

ω P = r M g L . or, since L = I ω ,
ω P = r M g I ω .

In this derivation, we assumed that ω P ω , 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.

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?

Strategy

We use [link] to find the precessional angular velocity of the gyroscope. This allows us to find the period of precession.

Solution

The moment of inertia of the disk is

I = 1 2 m r 2 = 1 2 ( 0.30 kg ) ( 0.05 m ) 2 = 3.75 × 10 −4 kg · m 2 .

The angular velocity of the disk is

20.0 rev / s = 20.0 ( 2 π ) rad / s = 125.66 rad / s .

We can now substitute in [link] . The precessional angular velocity is

ω P = r M g I ω = ( 0.05 m ) ( 0.3 kg ) ( 9.8 m / s 2 ) ( 3.75 × 10 −4 kg · m 2 ) ( 125.66 rad / s ) = 3.12 rad / s .

The precessional period of the gyroscope is

T P = 2 π 3.12 rad / s = 2.0 s .

Significance

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 [link] is a good approximation of the precessional angular velocity.

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

The Moon’s gravity is 1/6 that of Earth’s. By examining [link] , we see that the top’s precession frequency is linearly proportional to the acceleration of gravity. All other quantities, mass, moment of inertia, and spin rate are the same on the Moon. Thus, the precession frequency on the Moon is
ω P (Moon) = 1 6 ω P (Earth) = 1 6 ( 5.0 rad / s ) = 0.83 rad / s .

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Practice Key Terms 1

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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