The rotor (flywheel) of a toy gyroscope has mass 0.140 kg. Its moment of inertia about its axis is 1.20 * 10^-4 kg•m^2. The mass of the frame is 0.0250 kg. The gyroscope is supported on a single pivot (Fig. E10.51) with its center of mass a horizontal distance of 4.00 cm from the pivot. The gyroscope is precessing in a horizontal plane at the rate of one revolution in 2.20 s.
(a) Find the upward force exerted by the pivot.

Question

The rotor (flywheel) of a toy gyroscope has mass 0.140 kg. Its moment of inertia about its axis is 1.20 * 10^-4 kg•m^2. The mass of the frame is 0.0250 kg. The gyroscope is supported on a single pivot (Fig. E10.51) with its center of mass a horizontal distance of 4.00 cm from the pivot. The gyroscope is precessing in a horizontal plane at the rate of one revolution in 2.20 s.
(a) Find the upward force exerted by the pivot.

Diagram illustrating the gyroscope's rotation axes and components for angular momentum conservation.

(a) Find the upward force exerted by the pivot.

Accepted Answer

Hey, everyone in this problem. During an experiment, the wheel of a laboratory gyroscope processes in a horizontal plane at 30 rpm. The wheel with a mass of 200 g that is fixed to a frame is free to spin about the shaft and the shaft has a mass of 25 g. The moment of inertia of the wheel about the shaft is 10 to the negative four kg meter squared. The gyroscope is supported by a pivot near the right hand end of the shaft located five cm away from its center of gravity. Shown in the figure we were asked to determine the force exerted by the pivot on the shaft. Now, the diagram were given, he shows this gyroscope With the shaft five cm rotating about the wheel. Alright. So let's think about this. We are asked to determine a force. Let's go ahead and draw out the forces in our diagram. Alright. So our pivot we're gonna force from our pivot acting upwards. We're gonna call that F P, we're gonna have the weight of the wheel. I'm gonna call it W W acting downwards. We're going to also have the weight of the shaft, we'll call that W s acting downwards as well. So we have three forces. They're all acting in the Y direction we're gonna take up to be the positive Y direction. Now, we know we have no net external forces, we have an equilibrium situation. And so the sum of the forces and in this case, we're looking in the Y direction, it's going to be equal to zero. That's our equilibrium condition. What forces do we have while acting in the positive Y direction, we have the force F P okay. The force exerted by the pivot on the shaft acting in the negative Y direction. We have the weight W of the wheel and we also have the weight sws of the shaft. So we have FP minus W W minus W S is equal to zero. Okay. So we want to find the force F P, we can isolate and move those weights to the right hand side. We get that FP is equal to the weight W of the wheel plus the weight W s of the shaft. Okay. Recall that the weight is going to be equal to the mass times G, the gravitational acceleration. So we have the mass of the wheel times G plus the mass of the shaft times G. We can factor out that gravitational acceleration. So we have G times M W plus M S. Now we know the value of G that's a gravitational acceleration that constant that we know. And we're also given M W and M s. Let's go ahead and convert those were given those values and grams, let's convert them into kilograms are standard unit. Okay. We're gonna do that separately. So we don't um get our equation super messy. And so the mass w of the wheel we're told is g. We're gonna take that 200 g and multiply by one kg per 1000 g. The unit of Graham cancels, we essentially take 200 divided by 1000 And we get 0.2 kg in the mass of the shaft M S we're told is 25 kg and we're gonna do the same thing. Whoops, not 25 kg 25 g. We're going to do the same thing multiply by one kg per 1000 g. The unit of Graham cancels were essentially dividing by 1000 and we get a value of 0.25 kilograms. So going back to our equation for the force F P, We have the gravitational acceleration 9. m/s squared. The mass of the wheel, 0.2 kg plus the mass of the shaft. 0.025 kg. Okay. And if we work this out on our calculator, we are going to get a force of 2.205. The units we have meters per second squared times kilograms, we get kilogram meter per second squared, which is equivalent to a newton. And so that is our force in newtons like we want. And if we look at the answer choices, We found that the force exerted by the pivot on the shaft is approximately 2.2 newtons that's going to correspond with answer choice. C Thanks everyone for watching. I hope this video helped see you in the next one.

Verified Solution

Key Concepts

Moment of Inertia

Precession

Torque and Equilibrium