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A gyroscope is precessing about a vertical axis. Describe what happens to the precession angular speed if the following changes in the variables are made, with all other variables remaining the same: (a) the angular speed of the spinning flywheel is doubled; (b) the total weight is doubled; (c) the moment of inertia about the axis of the spinning flywheel is doubled; \((\mathrm{d})\) the distance from the pivot to the center of gravity is doubled. (e) What happens if all four of the variables in parts (a) through (d) are doubled?

Short Answer

Expert verified
Halve, double, halve, double, and no change, respectively.

Step by step solution

01

Understanding Precession

The precession of a gyroscope is caused by the torque due to gravity acting on the rotating wheel. This torque causes an angular change in the gyroscope's orientation. The precession angular speed \( \omega_p \) is given by the formula \( \omega_p = \frac{rMg}{L_s} \), where \( r \) is the distance from the pivot to the center of gravity, \( M \) is the mass of the gyroscope, \( g \) is the acceleration due to gravity, and \( L_s \) is the angular momentum of the spinning flywheel.
02

Effect of Doubling Angular Speed

When the angular speed of the spinning flywheel \( \omega_s \) is doubled, the angular momentum \( L_s = I_s \cdot \omega_s \) (where \( I_s \) is the moment of inertia) also doubles. Since \( \omega_p = \frac{rMg}{L_s} \), doubling \( L_s \) results in halving the precession angular speed \( \omega_p \).
03

Effect of Doubling Total Weight

Doubling the total weight \( M \) doubles the torque \( \tau = rMg \) acting on the gyroscope. Since \( \omega_p = \frac{rMg}{L_s} \), doubling \( M \) directly doubles \( \omega_p \).
04

Effect of Doubling Moment of Inertia

When the moment of inertia \( I_s \) is doubled, the angular momentum \( L_s = I_s \cdot \omega_s \) also doubles (assuming \( \omega_s \) remains constant). Therefore, \( \omega_p = \frac{rMg}{L_s} \) is halved because \( L_s \) is in the denominator.
05

Effect of Doubling Distance from Pivot

Doubling the distance from the pivot \( r \) doubles the torque \( \tau = rMg \) as well, which doubles \( \omega_p \) since \( \omega_p = \frac{rMg}{L_s} \) is directly proportional to \( r \).
06

Combined Effect of Doubling All Variables

If all variables (angular speed, total weight, moment of inertia, and distance from pivot) are doubled, the doubling of \( I_s \) and the doubling of \( \omega_s \) will cause the angular momentum \( L_s \) to quadruple. Thus, \( \omega_p = \frac{4rMg}{4L_s} \) which effectively results in no change to the precession angular speed \( \omega_p \); it remains the same.

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Key Concepts

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

Angular Speed
Angular speed is fundamentally about how fast something spins or rotates. It's the rate at which an object travels around a circular path. It is usually denoted by the symbol \( \omega \) and is measured in radians per second. For a gyroscope, angular speed describes how quickly its spinning wheel rotates.

When you double the angular speed of a gyroscope's spinning flywheel, the spin rate increases. This results in the doubling of angular momentum, as both depend on angular speed. However, an increase in the angular momentum tends to decrease the precession angular speed. This is because the gyroscope can handle more of the weight of the applied torque owing to its increased momentum.

Angular speed can be influenced by external factors such as friction and resistance, but in theoretical exercises, it's most commonly adjusted by changing the input energy or force responsible for the motion of the rotating body.
Moment of Inertia
The moment of inertia is a measure of an object’s resistance to changes in its rotation. It depends on how the mass is distributed with respect to the axis of rotation. For instance, more mass further from the axis means a higher moment of inertia.

In a gyroscope, the moment of inertia plays a critical role. It quantifies how much torque is needed for a desired angular acceleration. When you double the moment of inertia of a gyroscope's flywheel, it becomes harder to change its rotational state. This leads to an increase in angular momentum when the angular speed remains constant. But increased angular momentum means the precession angular speed decreases, as it is inversely proportional to angular momentum.

Different shapes and mass distributions result in different moments of inertia. For example, a solid disc and a ring of the same mass and radius have different moments of inertia.
Torque
Torque is the rotational equivalent of linear force. It’s what causes an object to start spinning, stop spinning, or change its angle of rotation. It’s calculated as the product of force and the distance from the point of rotation, also known as the lever arm. For a gyroscope, it’s the force due to gravity acting at a distance from the pivot.

When you double the torque on a gyroscope by, say, doubling its mass or the distance from the pivot, you’re essentially increasing the rotational force acting upon it. This increase leads to faster precession if the angular momentum doesn’t change, as the precession angular speed is directly proportional to torque.

To visualize torque, think of using a wrench. The longer the wrench, the more torque you can apply with the same amount of force. This concept applies in rotational motion, affecting how quickly something can rotate or spin.
Angular Momentum
Angular momentum is the product of an object's moment of inertia and its angular speed. It’s a measure of how much rotation an object has and is conserved in a closed system without external torques. For any spinning object, maintaining angular momentum is key to its stability and predefined motion.

In a gyroscope, the angular momentum keeps it from tumbling over, thanks to the spinning flywheel. If you manipulate the components affecting the angular momentum, such as the flywheel's speed or its moment of inertia, you'll observe changes in how the gyroscope behaves.

When angular momentum is high, thanks to either an increased angular speed or a higher moment of inertia, the gyroscope becomes less responsive to external torques, slowing its precession angular speed. In environments where angular momentum is transferred or altered, such as a figure skater pulling in their arms to spin faster, different rotational dynamics can be achieved.

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Most popular questions from this chapter

An experimental bicycle wheel is placed on a test stand so that it is free to turn on its axe. If a constant net torque of 5.00 \(\mathrm{N} \cdot \mathrm{m}\) is applied to the tire for 2.00 \(\mathrm{s}\) , the angular speed of the tire increases from 0 to 100 rev/min. The external torque is then removed, and the wheel is brought to rest by friction in its bearings in 125 s. Compute (a) the moment of inertia of the wheel about the rotation axis; \((b)\) the friction torque; (c) the total number of revolutions made by the wheel in the 125 -s time interval.

A 392 -N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 25.0 rad/s. The radius of the wheel is \(0.600 \mathrm{m},\) and its moment of inertia about its rotation axis is 0.800\(M R^{2}\) . Friction does work on the wheel as it rolls up the hill to a stop, a height \(h\) above the bottom of the hill; this work has absolute value 3500 \(\mathrm{J}\) . Calculate \(h\) .

A small block with mass 0.250 \(\mathrm{kg}\) is attached to a string passing through a hole in a frictionless, horizontal surface (see Fig. 10.48 ). The block is originally revolving in a circle with a radius of 0.800 \(\mathrm{m}\) m about the hole with a tangential speed of 4.00 \(\mathrm{m} / \mathrm{s}\) . The string is then pulled slowly from below, shortening the radius of the circle in which the block revolves. The breaking strength of the string is 30.0 \(\mathrm{N}\) . What is the radius of the circle when the string breaks?

A 2.20-kg hoop 1.20 \(\mathrm{m}\) in diameter is rolling to the right without slipping on a horizontal floor at a steady 3.00 \(\mathrm{rad} / \mathrm{s}\) . (a) How fast is its center moving? (b) What is the total kinetic energy of the hoop? (c) Find the velocity vector of each of the following points, as viewed by a person at rest on the ground: (i) the highest point on the hoop; (ii) the lowest point on the hoop, (iii) a point on the right side of the hoop, midway between the top and the bottom. (d) Find the velocity vector for each of the points in part (c), except as viewed by someone moving along with same velocity as the hoop.

A stiff uniform wire of mass \(M_{0}\) and length \(L_{0}\) is cut, bent, and the parts soldered together so that it forms a circular wheel having four identical spokes coming out from the center. None of the wire is wasted, and you can neglect the mass of the solder. (a) What is the moment of inertia of this wheel about an axle through its center perpendicular to the plane of the wheel? (b) If the wheel is given an initial spin with angular velocity \(\omega_{0}\) and stops uniformly in time \(T\) , what is the frictional torque at its axle?

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