Problem 93 - University Physics with Modern Physics

Question

A demonstration gyroscope wheel is constructed by removing the tire from a bicycle wheel 0.650 m in diameter, wrapping lead wire around the rim, and taping it in place. The shaft projects 0.200 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 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 rev/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 rev/s; (c) when the shaft is rotating in a horizontal plane about its center at 0.300 rev/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 per hand; b) 41.5 N per hand; c) 248.6 N per hand; d) 0.012 rev/s.

Step 1: Understanding the System

We have a gyroscope wheel with a diameter of 0.650 m and mass of 8.00 kg, all considered to be located at the rim. The shaft extends 0.200 m on each side, and spins at 5.00 rev/s. Our task is to determine forces on the shaft in various conditions, and needed shaft rotation for support on one end.

Step 1: Initial Calculations

Calculate the radius of the wheel as half the diameter: \[ r = \frac{0.650 \, \text{m}}{2} = 0.325 \, \text{m} \]The moment of inertia \((I)\) of the wheel, assuming all mass is at the rim, is given by:\[ I = m r^2 = 8.00 \, \text{kg} \times (0.325 \, \text{m})^2 = 0.845 \, \text{kg} \cdot \text{m}^2 \]

Step 2: Shaft at Rest (Part a)

When the shaft is at rest, the forces exerted by each hand are just balancing the weight of the wheel. Each hand must resist half of the wheel's weight. Calculating this involves:\[ F = \frac{m g}{2} = \frac{8.00 \, \text{kg} \times 9.81 \, \text{m/s}^2}{2} = 39.24 \, \text{N} \]

Step 3: Shaft Rotating at 0.050 rev/s (Part b)

First, convert the angular speed to rad/s:\[ \omega_p = 0.050 \, \text{rev/s} \times \frac{2 \pi \, \text{rad}}{1 \, \text{rev}} = 0.314 \, \text{rad/s} \]Now use the gyroscopic effect equation for precessional angular speed:\[ \tau = I \omega \omega_p \]\[ \text{Where } \tau = 2 F d \text{, with } d = 0.200 \, \text{m} \]Substitute to find force:\[ 2 F \cdot 0.200 = 0.845 \cdot (5 \times 2 \pi) \cdot 0.314 \]\[ F = \frac{0.845 \cdot \omega \cdot 0.314}{2 \cdot 0.200} \approx 41.5 \, \text{N} \]

Step 4: Shaft Rotating at 0.300 rev/s (Part c)

Similarly, convert angular speed:\[ \omega_p = 0.300 \, \text{rev/s} \times \frac{2 \pi \, \text{rad}}{1 \, \text{rev}} = 1.884 \, \text{rad/s} \]Substitute in the gyroscopic effect equation:\[ 2 F \cdot 0.200 = 0.845 \cdot (5 \times 2 \pi) \cdot 1.884 \]Solve for force:\[ F = \frac{0.845 \cdot \omega \cdot 1.884}{2 \cdot 0.200} \approx 248.6 \, \text{N} \]

Step 5: Calculating Rate for Support at One End (Part d)

For the gyroscope to be supported by only one hand, it must produce enough precessional torque to balance its weight about the support. The rate of precession \(\omega_p\) can be obtained from \(\frac{mgd}{I\omega} = \omega_p\):\[ \omega_p = \frac{8.00 \times 9.81 \times 0.200}{0.845 \times 5 \times 2 \pi} \approx 0.075 \, \text{rad/s} \]Converting back to revolutions per second:\[ \omega_p = \frac{0.075}{2 \pi} \approx 0.012 \, \text{rev/s} \]

Key Concepts

Gyroscopic Effect

The gyroscopic effect is a fascinating physical phenomenon that stabilizes rotating objects. It is rooted in the principles of angular momentum, where an object in rotational motion tends to maintain its orientation. In the context of the exercise, the spinning bicycle wheel acts as a gyroscope. This effect is influenced by how the wheel spins around its axis, resisting changes to its orientation in space.
When the wheel is spinning fast, just like Earth's rotation, it will resist any attempt to tilt it. In simpler terms, the faster the spin, the more stable the system becomes. This is because of the conservation of angular momentum, a key principle in gyroscopy. The spinning motion results in forces that 'fight back' against attempts to change its direction of rotation.
This ability to remain stable while spinning is why gyroscopes are used in navigation systems and aerospace applications. They provide a way to measure or maintain orientation without relying on external references, helping planes and spacecraft navigate accurately.

Moment of Inertia

Moment of inertia is a core concept in physics, representing an object's resistance to changes in its rotational motion. For our gyroscope wheel, it defines how much torque is needed for a certain rotational acceleration. In simpler words, it's the rotational equivalent of mass in linear motion.
Mathematically expressed as\[ I = m r^2 \]this equation shows that moment of inertia \(I\) depends on both the mass \(m\) and the distribution of that mass relative to the axis of rotation (distance \(r\)).
For the gyroscope wheel in the exercise, the moment of inertia was calculated taking into account all mass being at the rim. With values like\[ I = 0.845 \, \text{kg} \cdot \text{m}^2 \] it shows us how spread out the mass of the wheel is from its axis. This affects how easily or difficultly the wheel spins and, consequently, how it behaves as a gyroscope.

Angular Velocity

Angular velocity is a measure of how fast an object rotates or spins. It essentially describes the speed of rotation around an axis, expressed in revolutions per second (rev/s) or radians per second (rad/s). In this context, two angular velocities come into play: the spin of the wheel and the precessional movement of the shaft.
The primary spinning speed of the wheel, given as 5.00 rev/s, determines how quickly the rim moves around its axis. This speed is converted to rad/s for calculations. For example, 5.00 rev/s converts to:\[ \omega = 5 \times 2\pi \approx 31.42 \, \text{rad/s} \]
Similarly, the precession speed, which is the rate at which the shaft rotates horizontally, must be converted. Knowing these speeds allows us to apply the gyroscopic effect to determine the forces needed to stabilize the wheel under various conditions.

Precessional Torque

Precessional torque is the key force that occurs when a spinning object, like our gyroscope wheel, experiences a secondary rotation. This secondary rotation happens due to gravity creating a torque at the end of the shaft.
Torque in this scenario can be calculated using:\[ \tau = I \omega \omega_p \]This formula considers the moment of inertia \(I\), the primary angular velocity \(\omega\), and the precessional angular velocity \(\omega_p\). Precessional torque manifests when the spinning wheel undergoes changes due to additional forces, like when the woman holds the shaft and the system spins horizontally.
Understanding this torque helps in predicting how the system will behave, such as whether the gyro can be supported by just one end. It highlights how spinning and secondary rotations interact, showcasing the complexity and beauty of rotational dynamics.