Video: Calculating the Precessional Period of the Gyroscope

A gyroscope spins with its tip on the ground which produces negligible frictional resistance. The gyroscope has a radius of 5.0 cm and a mass of 0.30 kg and spins at 20 rev/s. The center of mass of the gyroscope’s disc is at a 5.0 cm displacement from its tip along the rotational axis of the gyroscope. What is the precessional period of the gyroscope?

04:38

Video Transcript

A gyroscope spins with its tip on the ground, which produces negligible frictional resistance. The gyroscope has a radius of 5.0 centimeters and a mass of 0.30 kilograms and spins at 20 revolutions per second. The center of mass of the gyroscope’s disc is at a 5.0-centimeter displacement from its tip along the rotational axis of the gyroscope. What is the precessional period of the gyroscope?

We can call this precessional period capital 𝑇 sub 𝑝 and begin on our solution by drawing a diagram. The gyroscope in this example consists of a rotating disc with a mass of 0.30 kilograms and a radius, we’ve called π‘Ÿ, 5.0 centimeters. The center of the rotating disc is a distance, also π‘Ÿ, 5.0 centimeters above the ground. And the gyroscope rotates at an angular speed we’ve called capital Ξ© of 20 revolutions per second.

We know that, under these conditions, the axis that runs through the center of the spinning gyroscope and is perpendicular to it will begin to deviate from a vertical line. As the axis of the gyroscope moves off of that original vertical line, we know that this axis itself will slowly begin to rotate about the original vertical. That rotation will itself have an angular speed we can call πœ” sub 𝑝, the precession angular speed.

We can recall the mathematical relationship describing that angular precession speed, πœ” sub 𝑝. That angular rate is equal to the mass of our object times the acceleration due to gravity times the distance from the point of contact of the object with the ground to its center of mass all divided by its moment of inertia times its own angular speed, πœ”.

As we consider this relationship for our scenario, we’re given the mass, π‘š, of the gyroscope. And the acceleration due to gravity, 𝑔, we can treat as exactly 9.8 meters per second squared. The distance from the point of contact of our gyroscope with the ground to its center of mass is also given to us, 5.0 centimeters. So, all that remains is to solve for the object’s moment of inertia and its angular speed in radians per second.

Knowing that our gyroscope is a disc rotating about a line through its center, when we look up the moment of inertia for an object of that shape, rotating in that way, we see it’s equal to one-half the object’s mass times its radius squared. Plugging that in to our expression for πœ” sub 𝑝, we see that the mass value of our gyroscope cancels out, as does one factor of its radius π‘Ÿ. So, the angular precession speed is equal to two times the acceleration due to gravity over the radius of the disc times its angular speed in radians per second.

We’re not given the angular speed of the disc in those units, but we are given its angular speed in units of revolutions per second. Knowing that one revolution about a circle is equal to two πœ‹ radians, that means we can substitute capital Ξ© times two πœ‹ in for lowercase πœ”, the angular rotation rate of the disc in radians per second. This expression will let us solve for πœ” sub 𝑝, but what we wanna solve for is 𝑇 sub 𝑝, the period of the precession.

We know that, in general, period is equal to two πœ‹ divided by angular speed. If we apply this relationship to our scenario, we can say that 𝑇 sub 𝑝 is equal to two πœ‹ over πœ” sub 𝑝, or two πœ‹ over this expression we’ve arrived at. Simplifying this expression, we see it’s equal to π‘Ÿπœ” times four πœ‹ squared radians per revolution all divided by two times the acceleration due to gravity. A factor of two cancels from our numerator and denominator. And we’re now ready to plug in and solve for 𝑇 sub 𝑝.

When we do plug in, we’re careful to insert our radius π‘Ÿ in units of meters. Before we calculate this result, let’s take a look at the units in this expression to see that they work out. First, the units of meters in our numerator and denominator cancel one another out. And when we multiply our angular speed in revolutions per second by our angular conversion from radians to revolutions, the units of revolutions drop out. And with the factors of time in seconds involved, we can see that our final units will be radian seconds, or simply seconds. That agrees with what we’d expect for units for a period 𝑇. When we do enter these numbers on our calculator, we find a result, to two significant figures, of 2.0 seconds. That’s the precessional period of this rotating gyroscope.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.