Problem 102 - University Physics with Modern Physics

Question

A demonstration gyroscope wheel is constructed by removing the tire from a bicycle wheel 0.650 \(\mathrm{m}\) in diameter, wrapping lead wire around the rim, and taping it in place. The shaft projects 0.200 \(\mathrm{m}\) at each side of the wheel, and a woman holds the ends of the shaft in her hands. The mass of the system is 8.00 \(\mathrm{kg}\) ; its entire mass may be assumed to be located at its rim. The shaft is horizontal, and the wheel is spinning about the shaft at 5.00 \(\mathrm{rev} / \mathrm{s}\) . Find the magnitude and direction of the force each hand exerts on the shaft (a) when the shaft is at rest; (b) when the shaft is rotating in a horizontal plane about its center at \(0.050 \mathrm{rev} / \mathrm{s} ;\) (c) when the shaft is rotating in a horizontal plane about its center at 0.300 \(\mathrm{rev} / \mathrm{s}\) . (d) At what rate must the shaft rotate in order that it may be supported at one end only?

Step-by-Step Solution

Answer

(a) 39.24 N; (b) & (c) Calculate torque to support precession; (d) Solve rotating balance equation.

Step 1: Determine the Radius of the Wheel

The diameter of the wheel is 0.650 m. Hence, the radius \( r \) is half of the diameter. \[ r = \frac{0.650 \ m}{2} = 0.325 \ m \]

Step 2: Calculate the Moment of Inertia

Since the entire mass is assumed to be located at the rim, the moment of inertia \( I \) of the wheel is given by:\[ I = m r^2 \]Substituting the given values:\[ I = 8.00 \ kg \times (0.325 \ m)^2 = 0.845 \ kg \cdot m^2 \]

Step 3: Calculate Initial Angular Velocity in Radians per Second

The wheel spin at \(5.00\) revolutions per second, first convert this to radians per second using: \[ \omega = 5.00 \frac{rev}{s} \times 2\pi \frac{rad}{rev} = 10\pi \frac{rad}{s} \]

Step 4: Calculate Angular Momentum

The angular momentum \( L \) of the system can be calculated using:\[ L = I \omega \]Substituting the values:\[ L = 0.845 \ kg \cdot m^2 \times 10\pi \ \frac{rad}{s} = 8.45\pi \ kg \cdot m^2/s \]

Step 5: Analyze Forces when Shaft is at Rest

When the shaft is at rest (\(a\)), the forces exerted by the hands are simply supporting the weight of the system, which is evenly distributed due to symmetrical configuration. Thus, each hand exerts a force equal to half the weight:\[ F = \frac{8.00 \ kg \times 9.81 \frac{m}{s^2}}{2} = 39.24 \ N \]

Step 6: Calculate Precessional Angular Velocity for Rotating Shaft

The precessional angular velocity \( \Omega \) can be related to the torque exerted. For cases \(b\) and \(c\), calculate \( \Omega \) using:\[ \tau = I \Omega \omega \]Given that the wheel's rotation causes a horizontal torque, solve:\(\Omega_b = 0.050 \frac{rev}{s} \times 2\pi \frac{rad}{rev} = 0.1\pi \frac{rad}{s}\) \(\Omega_c = 0.300 \frac{rev}{s} \times 2\pi \frac{rad}{rev} = 0.6\pi \frac{rad}{s} \)

Step 7: Calculate Hand Forces With Precession

For \(b\) and \(c\): Each hand must provide enough torque to offset the precession forces. Use: \[ F_{hand} = \frac{2 \cdot I \cdot \omega \cdot \Omega_{b or c}}{shaft\ length} \]CalculateFor \(b\):\[ F_{hand} = \frac{2 \cdot 0.845 \ kg \cdot m^2 \cdot 10\pi \ \frac{rad}{s} \cdot 0.1\pi \ \frac{rad}{s}}{0.4 \ m} \approx 42.35 \ N\] The forces solve as much force support must be initiated by each arm.

Step 8: Determine Rotational Rate for Single Hand Support

To find the rate at which one end needs to fully support the gyroscope:Set gravity for single support using \( \Omega_r \), defined by the balance of gravitational torque with gyroscope's precession:\[ L \cdot \Omega_r = m \cdot g \cdot \frac{shaft length}{2}\]Solve:\[ \Omega_r = \frac{8.00 \ kg \cdot 9.81 \ ms^{-2} \cdot 0.2 \ m}{8.45 \pi \ kg \cdot m^2 \]

Key Concepts

Moment of Inertia

The moment of inertia is a crucial concept when dealing with rotational motion, similar to how mass affects linear motion. It quantifies the resistance of an object to changes in its rotational motion. In simple terms, it's like the rotational equivalent of mass. For a point mass at a distance from the axis of rotation, the moment of inertia \( I \) is given by the formula:

• \( I = m r^2 \)

Here, \( m \) is the mass of the object, and \( r \) is the radius from the axis of rotation. In our gyroscope problem, the mass, \( 8.00 \ kg \), is concentrated at a radius of \( 0.325 \ m \). This results in a moment of inertia of \( 0.845 \ kg \cdot m^2 \).
The larger the moment of inertia, the more torque is required to alter the object's angular velocity. It emphasizes that, due to the mass being distributed farther from the axis, it becomes harder to spin the wheel.

Angular Momentum

Angular momentum is like the momentum we know in linear terms but for rotating objects. It tells us how much motion the object has in its spin, and it is conserved unless acted upon by an external torque. The formula for calculating angular momentum \( L \) of a rotating object is:

• \( L = I \omega \)

where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. In the example given, our gyroscope spins at \( 5.00 \ rev/s \), which converts to \( 10\pi \ rad/s \). Given the moment of inertia is \(0.845 \ kg \cdot m^2\), the angular momentum is computed to be \(8.45\pi \ kg \cdot m^2/s\).
Angular momentum helps determine how stable the gyroscope's spin is and how much force is needed to change its spinning direction or rate.

Precession

Precession is the phenomenon where the axis of a spinning object moves in space, often creating a circular motion. This happens when an external torque is applied, causing the spin axis to change direction. In gyroscopes, precession occurs due to gravitational torque acting on the spinning wheel.
The rate of precession can be calculated and relates to the torque (\( \tau \)) and the already existing angular momentum (\( L \)). It can be expressed as:

• \( \Omega = \frac{\tau}{L} \)

In our case, by knowing the gravitational force and the geometry of the setup, we can calculate how fast the axis precesses. For instance, given angular velocities of \(0.1\pi \ rad/s\) and \(0.6\pi \ rad/s\) for different cases (\(b\) and \(c\)), it shows how the speed at which the gyroscope's axis precesses changes based on these values. Understanding precession allows calculation of how the gyroscope behaves in different scenarios, such as being supported at one end.

Torque

Torque is the force that causes objects to rotate about an axis. It's similar to how a force causes linear motion. Torque \( \tau \) depends on how large the force is, the distance from the axis at which it acts, and the angle at which it is applied. Mathematically:

• \( \tau = r F \sin\theta \)

where \( F \) is the force applied, \( r \) is the distance from the pivot point, and \( \theta \) is the angle between the force vector and the lever arm. In gyroscopes, when the torque is applied due to external forces like gravity, it leads to precession.
In the exercise, when precession happens, it changes how the applied forces by the hands balance out the gyroscope's motion. For example, calculation of \( F_{hand} \) to counterbalance precession would involve determining the resulting torque. Here, each hand must provide sufficient torque on the gyroscope to stabilize it when external forces intend to tip it over.