/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 53 The rotor (flywheel) of a toy gy... [FREE SOLUTION] | 91Ó°ÊÓ

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The rotor (flywheel) of a toy gyroscope has mass 0.140 kg. Its moment of inertia about its axis is \(1.20 \times 10^{-4} \mathrm{kg} \cdot \mathrm{m}^{2} .\) The mass of the frame is 0.0250 \(\mathrm{kg}\) . The gyroscope is supported on a single pivot (Fig. E10.53) with its center of mass a horizontal distance of 4.00 \(\mathrm{cm}\) from the pivot. The gyroscope is precessing in a horizontal plane at the rate of one revolution in 2.20 \(\mathrm{s}\) . (a) Find the upward force exerted by the pivot. (b) Find the angular speed with which the rotor is spinning about its axis, expressed in rev/min. (c) Copy the diagram and draw vectors to show the angular momentum of the rotor and the torque acting on it.

Short Answer

Expert verified
(a) The upward force is 1.62 N. (b) The rotor spins at 7058 rev/min. (c) Draw the vectors as described.

Step by step solution

01

Understand the Problem

The problem involves a toy gyroscope with specific mass properties and motions around a pivot. We need to find the force exerted by the pivot, the spin speed of the rotor, and draw vectors representing the system.
02

Calculate the Upward Force by the Pivot

The total mass of the system (rotor + frame) is 0.140 kg + 0.0250 kg = 0.165 kg. The gravitational force is the total weight, given by \( F = mg \), where \( m = 0.165 \) kg and \( g = 9.81 \text{ m/s}^2 \). Thus, the upward force by the pivot is \( F = 0.165 \times 9.81 = 1.61965 \text{ N} \).
03

Calculate the Precession Rate and Relation to Angular Velocity

The precession rate \( \omega_p \) is one revolution every 2.20 s, or \( \omega_p = \frac{2\pi}{2.20} \text{ rad/s} \). We know from gyroscopic motion that torque induced by gravity \( \tau = rmg \) is equal to \( I \omega \omega_p \). Here, \( r = 0.04 \) m, \( I = 1.20 \times 10^{-4} \text{ kg} \cdot \text{m}^2 \), and \( \omega \) is the angular speed of the rotor.
04

Calculate the Spin Angular Velocity of the Rotor

Solving for \( \omega \), \( rmg = I \omega \omega_p \), we have \( 0.04 \times 1.61965 = 1.20 \times 10^{-4} \times \omega \times \frac{2\pi}{2.20} \). This simplifies to \( \omega = \frac{0.064786}{8.76923 \times 10^{-5}} \). Solve the expression to find \( \omega \approx 738.46 \text{ rad/s} \).
05

Convert Angular Velocity to Revolutions per Minute

Convert the angular velocity from rad/s to rev/min: \( \omega = 738.46 \times \frac{1}{2\pi} \text{ rev/s} \). To get rev/min, multiply by \( 60 \), resulting in \( \omega \approx 7057.7 \text{ rev/min} \).
06

Draw Angular Momentum and Torque Vectors

Copy the diagram of the gyroscope system and note that the angular momentum vector of the rotor points along its axis of rotation due to the spinning motion. The torque vector caused by gravity points upwards when considering the lever arm outwards from the pivot.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Moment of Inertia
When discussing gyroscopes, understanding the moment of inertia is crucial. It is a measure of how difficult it is to change the rotational motion of an object. Imagine spinning a wheel on its axle; the heavier it is, or the more mass it has further from the center, the harder it is to spin.
In mathematical terms, the moment of inertia (I) depends on both the object's mass distribution and the distance of this mass from the axis of rotation. For the gyroscope rotor in this example, it is given as \(1.20 \times 10^{-4} \mathrm{kg} \cdot \mathrm{m}^2\). This means, with this specific distribution of mass, that's the resistance the gyroscope will face against changing its spinning state.
The larger the moment of inertia, the more torque is needed to change its rotational speed. Conversely, less torque is required if the moment of inertia is smaller. This foundational concept helps us predict and calculate how gyroscopes behave when forces act upon them.
Precessional Motion
Precessional motion is one of the fascinating behaviors of gyroscopes. It's the movement of the axis of a spinning object, like a gyroscope, around another axis due to an applied torque, typically from gravity.
In this problem, precession is observed as the gyroscope spins and the entire system rotates in a horizontal plane. The precession rate described is one revolution every 2.20 seconds. This is a critical insight into how gyroscopes shift their orientation when balanced delicately on a pivot.
Understanding precessional motion helps us recognize that the application of gravitational torque doesn’t cause it to topple over immediately. Instead, it causes it to trace a circular path, and this unique motion is integral in technologies like navigational systems and stabilizers.
Angular Momentum
A key player in gyroscope physics is angular momentum, which hints at the rotational equivalent of linear momentum. For any rotating body, angular momentum (\(L\)) is given by \(L = I\omega\), where \(I\) is the moment of inertia and \(\omega\) is the angular velocity. It expresses the extent to which an object will maintain its rotational motion unless acted upon by an external force.
For gyroscopes, angular momentum is the steady, directional quantity that aids in retaining the orientation of the spinning rotor. As the gyroscope spins rapidly, it builds up significant angular momentum. This momentum stabilizes the gyroscope against changes in direction, hence providing its surprising ability to "resist" toppling.
The angular momentum vector points along the spin axis of the rotor, which is a crucial aspect to visualize because it helps predict the direction the gyroscope will precess.
Torque Calculation
Torque is the rotational counterpart of force that causes objects to rotate. It is calculated with the formula \(\tau = r \times F\), where \(r\) is the distance from the pivot point and \(F\) is the force applied perpendicular to the moment arm.
In the case of the gyroscope, gravitational forces create a torque that influences its precessional motion. In essence, the torque here results from the weight of the system exerting force, \(F = mg\), at a distance \(r\) from the pivot.
Utilizing this torque helps in determining parameters like angular velocity which informs about the spinning dynamics of the gyroscope. By quantifying torque, we can relate it directly to the gyroscope's stability and performance, illustrating how forces interact in rotational systems.

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Most popular questions from this chapter

A cord is wrapped around the rim of a solid uniform wheel 0.250 \(\mathrm{m}\) in radius and of mass 9.20 \(\mathrm{kg} .\) A steady horizontal pull of 40.0 \(\mathrm{N}\) to the right is exerted on the cord, pulling it off tangentially from the wheel. The wheel is mounted on frictionless bearings on a horizontal axle through its center. (a) Compute the angular acceleration of the wheel and the acceleration of the part of the cord that has already been pulled off the wheel. (b) Find the magnitude and direction of the force that the axle exerts on the wheel. (c) Which of the answers in parts (a) and (b) would change if the pull were upward instead of horizontal?

A \(50.0\)-kg grindstone is a solid disk 0.520 \(\mathrm{m}\) in diameter. You press an ax down on the rim with a normal force of 160 \(\mathrm{N}\) (Fig. P10.57). The coefficient of kinetion between the blade and the stone is 0.60 , and there is a constant friction torque of 6.50 \(\mathrm{N} \cdot \mathrm{m}\) between the axle of the stone and its bearings. (a) How much force must be applied tangentially at the end of a crank handle 0.500 \(\mathrm{m}\) long to bring the stone from rest to 120 \(\mathrm{rev} / \mathrm{min}\) in 9.00 \(\mathrm{s} ?\) (b) After the grindstone attains an angular speed of 120 \(\mathrm{rev} / \mathrm{min}\) , what tangential force at the end of the handle is needed to maintain a constant angular speed of 120 \(\mathrm{rev} / \mathrm{min}\) ? (c) How much time does it take the grindstone to come from 120 \(\mathrm{rev} / \mathrm{min}\) to rest if it is acted on by the axle friction alone?

A hollow, spherical shell with mass 2.00 \(\mathrm{kg}\) rolls without slipping down a \(38.0^{\circ}\) slope. (a) Find the acceleration, the friction force, and the minimum coefficient of friction needed to prevent slipping. (b) How would your answers to part (a) change if the mass were doubled to 4.00 \(\mathrm{kg}\) ?

BIO Gymnastics. We can roughly model a gymnastic tumbler as a uniform solid cylinder of mass 75 \(\mathrm{kg}\) and diameter 1.0 \(\mathrm{m}\) . If this tumbler rolls forward at 0.50 rev/s, (a) how much total kinetic energy does he have, and (b) what percent of his total kinetic energy is rotational?

A uniform ball of radius \(R\) rolls without slipping between two rails such that the horizontal distance is \(d\) between the two contact points of the rails to the ball. (a) In a sketch, show that at any instant \(v_{\mathrm{cm}}=\omega \sqrt{R^{2}-d^{2} / 4} .\) Discuss this expression in the limits \(d=0\) and \(d=2 R\) . (b) For a uniform ball starting from rest and descending a vertical distance \(h\) while rolling without slipping down a ramp, \(v_{\mathrm{cm}}=\sqrt{10 g h / 7}\) . Replacing the ramp with the two rails, show that $$ v_{\mathrm{cm}}=\sqrt{\frac{10 g h}{5+2 /\left(1-d^{2} / 4 R^{2}\right)}} $$ In each case, the work done by friction has been ignored. (c) Which speed in part (b) is smaller? Why? Answer in terms of how the loss of potential energy is shared between the gain in translational and rotational kinetic energies. (d) For which value of the ratio \(d / R\) do the two expressions for the speed in part (b) differ by 5.0\(\% ?\) By 0.50\(\% ?\)

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