A circular hoop of radius 65.3 cm and mass 2.16 kg is suspended on a horizontal nail. (a) Find its frequency of oscillation for small displacements from equilibrium. (b) What is the length of the equivalent simple pendulum?

To tackle this problem, we need to analyze the circular hoop's oscillation characteristics when it is displaced slightly from its equilibrium position. This involves some concepts from physics, particularly those related to rotational motion and simple harmonic motion.

Calculating the Frequency of Oscillation

First, let's find the frequency of oscillation for small displacements. The hoop can be treated as a physical pendulum. The formula for the frequency of a physical pendulum is given by:

f = (1/2π) * √(mgd/I)

Where:

• f = frequency of oscillation
• m = mass of the hoop (2.16 kg)
• g = acceleration due to gravity (approximately 9.81 m/s²)
• d = distance from the pivot to the center of mass (which is the radius of the hoop, 0.653 m)
• I = moment of inertia of the hoop about the pivot point

For a hoop, the moment of inertia about its center is given by:

I = mR²

However, since the hoop is pivoted at the edge, we need to use the parallel axis theorem to find the moment of inertia about the pivot point:

I = I_center + md²

Substituting the values:

• I_center = mR² = 2.16 kg * (0.653 m)² = 0.918 kg·m²
• d = R = 0.653 m
• I = 0.918 kg·m² + (2.16 kg * (0.653 m)²) = 0.918 kg·m² + 0.918 kg·m² = 1.836 kg·m²

Now we can substitute these values into the frequency formula:

f = (1/2π) * √((2.16 kg * 9.81 m/s² * 0.653 m) / 1.836 kg·m²)

Calculating the numerator:

mgd = 2.16 kg * 9.81 m/s² * 0.653 m ≈ 13.88 kg·m²/s²

Now substituting back into the frequency formula:

f = (1/2π) * √(13.88 / 1.836) ≈ (1/2π) * √(7.57) ≈ (1/2π) * 2.75 ≈ 0.438 Hz

Finding the Length of the Equivalent Simple Pendulum

Next, we need to determine the length of the equivalent simple pendulum that has the same frequency of oscillation. The frequency of a simple pendulum is given by:

f = (1/2π) * √(g/L)

Rearranging this formula to solve for L gives us:

L = g / (2πf)²

Substituting the values we have:

L = 9.81 m/s² / (2π * 0.438 Hz)²

Calculating the denominator:

(2π * 0.438)² ≈ (2.748)² ≈ 7.53

Now substituting back into the length formula:

L = 9.81 m/s² / 7.53 ≈ 1.303 m

Summary of Results

To summarize:

• The frequency of oscillation for the circular hoop is approximately 0.438 Hz.
• The length of the equivalent simple pendulum is approximately 1.303 m.

This analysis shows how we can apply principles of physics to understand the behavior of objects in motion, particularly in oscillatory systems. If you have any further questions or need clarification on any of the steps, feel free to ask!