Video: Finding the Precessional Angular Speed of a Gyroscope

A gyroscope has a disk of mass 5.0 Γ— 10Β² g and radius 0.100 m. The disk spins at 40 rev/s. The center of mass of the disk is 0.10 m from a pivot. What is the precession angular velocity of the gyroscope?

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Video Transcript

A gyroscope has a disk of mass 5.0 times 10 to the two grams and radius 0.100 meters. The disk spins at 40 revolutions per second. The center of mass of the disk is 0.10 meters from a pivot. What is the precession angular velocity of the gyroscope?

In this problem, we’ll assume that no energy of the gyroscope is lost due to friction, and we’ll also assume that 𝑔, the acceleration due to gravity, is exactly 9.8 meters per second squared. Let’s start by highlighting some of the vital information the problem statement gives us. The gyroscope has a mass of 5.0 times 10 to the two grams and a radius of 0.100 meters. Its rotation rate is 40 revolutions per second. And we’re told that the disk’s center of mass is 0.10 meters from a pivot. We’re asked for the precession angular velocity of the gyroscope. We’ll call that πœ” sub 𝑝.

Let’s draw a quick sketch of the scenario, so we can familiarize ourselves with what’s going on. In our situation, we have a disk rotating at a speed of 40 revolutions per second, we call πœ”, and as it does that, it slowly precesses about an axis here, the vertical dotted line. And that precession rate, we’ve called the precession angular velocity, πœ” sub 𝑝. We’re told that the center of mass of the disk is a distance of 0.10 meters from the pivot. We also know the mass of the disk, π‘š, 5.0 times 10 to the two grams. And the radius of the disk, we’ll call π‘Ÿ. And that value is 0.100 meters.

For given conditions like this, there is a relationship that tells us what the precession angular velocity, πœ” sub 𝑝, is. πœ” sub 𝑝 is equal to the mass of the disk times 𝑔, the acceleration due to gravity, multiplied by the radius of the disk, π‘Ÿ, all divided by the moment of inertia of the disk times πœ”, its angular speed.

When we apply this relationship to our scenario, the first step we want to take is to figure out what is 𝐼, the moment of inertia of the disk. When a disk rotates about its center as in this case, then the moment of inertia of that disk is equal to one-half its mass times its radius squared. So we can enter this expression into our equation for the precession angular velocity. Since in our case the radius of the disk, referred to in the moment of inertia equation, and the precession radius, that is the distance from the center of mass of the disk to the axis of rotation for precession, are the same, we need not distinguish one from the other in our equation by using special notation.

And looking at this equation, we see that we can cancel out a few terms. First, the mass of the disk cancels out. It’s both in the numerator and denominator. When we simplify this equation for πœ” sub 𝑝, we are left with a result stating πœ” sub 𝑝 equals two times 𝑔, the acceleration due to gravity, divided by π‘Ÿπœ”. We can now plug in for 𝑔, π‘Ÿ and πœ” to solve for πœ” sub 𝑝. 𝑔 is 9.8 meters per second squared. π‘Ÿ is equal to 0.100 meters. And πœ” is equal to 40 revolutions per second.

Before we enter these values in our calculator, the one change we want to make is to convert πœ” from revolutions per second to radians per second. Since one revolution around a circle is defined as two πœ‹ radians, then if we multiply 40 revolutions per second by two πœ‹, then that value will be in units of radians per second as we want.

When we multiply these numbers together, we find a precession angular velocity, πœ” sub 𝑝, of 0.78 radians per second. This is how fast the disk precesses around the axis of rotation of the gyroscope.

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