Problem 7 - An Introduction to Mechanics

Question

Suspended hoop A thin hoop of mass \(M\) and radius \(R\) is suspended from a string through a point on the rim of the hoop. If the support is turned with high angular velocity \(\omega\), the hoop will spin as shown, with its plane nearly horizontal and its center nearly on the axis of the support. The string makes angle \(\alpha\) with the vertical. (a) Find, approximately, the small angle \(\beta\) between the plane of the hoop and the horizontal. Assume that the center of mass is at rest. (b) Find, approximately, the radius of the small circle traced out by the center of mass about the vertical axis. (c) Find a criterion for the validity of the assumption that motion of the center of mass can be neglected. (With skill you can demonstrate this motion with a rope. It is a favorite cowboy lariat trick.)

Step-by-Step Solution

Answer

(a) \(\beta = \frac{g}{\omega^2 R} \cos(\alpha)\); (b) \(r = R\alpha\); (c) \(\omega^2 R \gg g\).

Step 1: Analyze the Forces and Torque

For this suspended hoop, we need to consider the forces acting on it. The main forces involved are tension in the string, gravitational force, and centrifugal force due to rotation. With these forces, the torque about the center of mass must be zero for equilibrium.

Step 2: Establish the Geometry of the Problem

For small angle approximations, use geometry to relate angles. The angle \( \beta \) between the plane of the hoop and the horizontal can be found using the forces involved. By balancing torques around the hoop's center, we can identify relations between \( \alpha \) (angle with vertical) and \( \beta \).

Step 3: Apply Small Angle Approximation

Since \( \beta \) is small, use the approximation that \( \sin \beta \approx \beta \) and \( \cos \beta \approx 1 \). This makes the equations simpler and allows for more straightforward solutions.

Step 4: Solve for Angle \(\beta\)

Use the torque equilibrium to find that \( M g R \sin \beta = M R \omega^2 R \cos(\alpha) \). Assuming small angles \( \sin \beta \approx \beta \) and \( \cos \alpha \approx 1 \), the equation simplifies to \( \beta = \frac{g}{\omega^2 R} \cos(\alpha) \).

Step 5: Determine the Radius of the Circle Traced by the COM

The center of mass traces a small circular path about the vertical axis due to the rotating hoop. The radius \( r \) of this path is found using \( r = R \sin(\alpha) \). Assuming small angles, this simplifies to \( r \approx R \alpha \).

Step 6: Establish Criterion for Neglecting CM Motion

The assumption that the motion of the center of mass can be neglected relies on \( \omega^2 R \gg g \), meaning the centrifugal force due to rotation must be much greater than the gravitational force. This ensures that the hoop spins nearly horizontally, maintaining the center of mass almost stationary.

Key Concepts

Angular Velocity

Angular velocity, denoted as \( \omega \), is a core concept in rotational dynamics. It refers to how fast an object spins around a predefined axis. In this scenario, the hoop rotates rapidly due to its angular velocity, causing it to nearly level out horizontally. Think of it like twirling a frisbee; the faster it spins, the more stable its motion looks.
Angular velocity is measured in radians per second (rad/s). High angular velocity implies a swift rotation which, for the hoop in this exercise, means it can spin near horizontally without wobbling too much. With high \( \omega \), the centrifugal force intensifies, balancing closely with gravitational pull, allowing the hoop to maintain an inclined plane at a small angle with respect to the horizontal.
Understanding how angular velocity interacts with forces like gravity and tension helps predict the hoop's behavior as it spins, indicating the shape and stability of its motion.

Torque Equilibrium

In rotational dynamics, torque equilibrium occurs when all torques acting on a system cancel each other out. This condition is crucial for a spinning hoop because it maintains steady motion without tipping over.
To have torque equilibrium, the sum of the torques around the hoop's center of mass must be zero. Torque here is the product of force and the distance from the pivot point (lever arm) at which the force is applied, mathematically expressed as \( \tau = rF \sin\theta \).
For the hoop:

• Gravitational force pulls it downward.
• Centrifugal force arises from its fast spin.
• Tension from the string supports it.

Balancing these forces and their respective torques ensures that the hoop spins in place without significant up-and-down motion, creating a stable equilibrium and allowing the small angle \( \beta \) to be calculated easily.

Small Angle Approximation

The small angle approximation simplifies complex trigonometric equations when the angles involved are tiny. This approximation allows us to consider \( \sin\theta \approx \theta \) and \( \cos\theta \approx 1 \) since when \( \theta \) is small, its sine and tangent values closely match the actual angle measurement.
In this exercise, the small angle \( \beta \), the angle between the hoop's plane and the horizontal, benefits from such simplification. By assuming \( \beta \) is small, calculations become more straightforward. Specifically, approximating the sine of \( \beta \) with \( \beta \) itself streamlines the process of determining how the hoop's orientation influences its rotation pattern.
This approach is helpful when exploring minute angles in dynamic systems, where they rear precise, yet manageable solutions, without requiring advanced trigonometric computations. It's a standard technique widely used in physics to enable easier handling of rotational and oscillatory systems.