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We want to hang a thin hoop on a horizontal nail and have the hoop make one complete small-angle oscillation each 2.0 \(\mathrm{s}\) What must the hoop's radius be?

Short Answer

Expert verified
The hoop's radius must be approximately 1.494 meters.

Step by step solution

01

Understand the Problem

We need to determine the radius of a thin hoop to make it oscillate with a period of 2.0 seconds while hanging on a nail. The hoop is undergoing small-angle oscillations, similar to a pendulum.
02

Recall the Formula for Period of Simple Pendulum

For small-angle oscillations, the period of a simple pendulum is given by the formula \( T = 2\pi \sqrt{\frac{L}{g}} \), where \( T \) is the period, \( L \) is the length of the pendulum, and \( g \) is the acceleration due to gravity (approximately \( 9.8\, \mathrm{m/s^2} \)). In this situation, \( L \) will be the equivalent rotational length related to the hoop.
03

Relate the Hoop's Rotation to a Simple Pendulum

When a hoop rotates about a point on its perimeter, its center of mass moves in a circle of radius equal to the hoop's radius \( R \). For small oscillations, the equivalent pendulum length is the diameter of the hoop minus the radius itself: \( L = 2R/3 \).
04

Substitute and Solve for Radius

Substitute \( L = \frac{2R}{3} \) into the formula for the period to solve for \( R \).\[2.0 = 2\pi \sqrt{\frac{\frac{2R}{3}}{9.8}}\]Simplify and isolate \( R \):\[1 = \pi \sqrt{\frac{2R}{3 \times 9.8}}\]Square both sides and solve for \( R \):\[1 = \pi^2 \times \frac{2R}{3 \times 9.8}\]\[R = \frac{3 \times 9.8}{2 \times \pi^2}\]Calculate \( R \): \( R \approx 1.494\, \mathrm{m} \).
05

Verify Dimensions

Re-check the calculation to ensure all dimensions align with a typical physics problem and confirm the radius found is within typical real-world dimensions for physics experiments or setups.

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Key Concepts

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

Small-Angle Oscillations
A small-angle oscillation occurs when the angle of displacement is sufficiently small, usually less than around 15 degrees. This condition simplifies the behavior of a pendulum-like system, such as a swinging hoop, as it can be treated using linear approximations.
This means the motion can be modeled similar to simple harmonic motion. Hence, for small angles, the sine of the angle \( \theta \approx \theta \), when measured in radians.
By using this simplification, the physics of pendulum systems can focus on its linear aspects rather than dealing with more complex trigonometric functions. This makes it easier to use formulas like \( T = 2\pi \sqrt{\frac{L}{g}} \) without dealing with higher complexities.
  • Small-angle oscillations help provide a clear path to approximate solutions, particularly useful in educational settings where precise angles are not used.
  • Systems remain periodic and predictable under the small-angle approximation.
Period of Oscillation
The period of oscillation (\( T \)) is the time taken for a pendulum to complete one full cycle of motion. For a simple pendulum, this period is derived from the formula \( T = 2\pi \sqrt{\frac{L}{g}} \), where \( L \) is the pendulum length, and \( g \) is the acceleration due to gravity.
This formula shows that the period depends on the pendulum length, simply:
  • The longer the pendulum, the longer the period.
  • The period is independent of the mass of the pendulum.
In oscillatory systems like hoops, the concept of equivalent length is used. Here, the rotational or physical characteristics of the hoop translate into a pendulum length to fit the formula for the period.
This formula is useful because it remains constant despite changes in amplitude, as long as the amplitude is small (small-angle assumption).
Center of Mass
The center of mass is a critical concept when analyzing the behavior of objects in motion, like a hoop on a nail. It represents the average position of all the mass in the object and serves as the point at which the force of gravity can be considered to act for motion calculations.
When a hoop is hung on a nail and allowed to swing, its center of mass travels along a circular trajectory, with a radius equivalent to the hoop's actual radius. This defines how we relate the oscillating hoop to a pendulum configuration.
  • By analyzing the center of mass, we understand the effective length of the pendulum, critical for calculating the period of oscillation.
  • The center of mass helps simplify complex objects into a more calculable point-mass system for motion analysis.
Understanding the center of mass allows one to predict the dynamic behavior of the hoop, especially in the context of rotation and gravitational forces at play.

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Most popular questions from this chapter

When a body of unknown mass is attached to an idealspring with force constant \(120 \mathrm{N} / \mathrm{m},\) it is found to vibrate with a frequency of 6.00 \(\mathrm{Hz}\) . Find (a) the period of the motion; (b) the angular frequency; (c) the mass of the body.

You pull a simple pendulum 0.240 m long to the side through an angle of \(3.50^{\circ}\) and release it. (a) How much time does it take the pendulum bob to reach its highest speed? (b) How much time does it take if the pendulum is released at an angle of \(1.75^{\circ}\) instead of \(3.50^{\circ} ?\)

A proud deep-sea fisherman hangs a 65.0 -kg fish from an ideal spring having negligible mass. The fish stretches the spring 0.120 \(\mathrm{m}\) . (a) Find the force constant of the spring. The fish is now pulled down 5.00 \(\mathrm{cm}\) and released. (b) What is the period of oscillation of the fish? (c) What is the maximum speed it will reach?

CP CALC An approximation for the potential energy of a KCl molecule is \(U=A\left[\left(R_{0}^{7} / 8 r^{8}\right)-1 / r\right],\) where \(R_{0}=2.67 \times\) \(10^{-10} \mathrm{m}, A=2.31 \times 10^{-28} \mathrm{J} \cdot \mathrm{m},\) and \(r\) is the distance between the two atoms. Using this approximation: (a) Show that the radial component of the force on each atom is \(F_{r}=A\left[\left(R_{0}^{7} / r^{9}\right)-1 / r^{2}\right].\) (b) Show that \(R_{0}\) is the equilibrium separation. (c) Find the minimum potential energy. (d) Use \(r=R_{0}+x\) and the first two terms of the binomial theorem \(\quad\) (Eq. 14.28\()\) to show that \(F_{r} \approx-\left(7 A / R_{0}^{3}\right) x,\) so that the molecule's force constant is \(k=7 A / R_{0}^{3} .\) (e) With both the \(K\) and \(C 1\) atoms vibrating in opposite directions on opposite sides of the molecule's center of mass, \(m_{1} m_{2} /\left(m_{1}+m_{2}\right)=3.06 \times 10^{-26} \mathrm{kg}\) is the mass to use in cal- culating the frequency. Calculate the frequency of small-amplitude vibrations.

You are watching an object that is moving in SHM. When the object is displaced 0.600 m to the right of its equilibrium position, it has a velocity of 2.20 \(\mathrm{m} / \mathrm{s}\) to the right and an acceleration of 8.40 \(\mathrm{m} / \mathrm{s}^{2}\) to the left. How much farther from this point will the object move before it stops momentarily and then starts to move back to the left?

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