/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 42 The uniform 5 -kg portion of a c... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 42

The uniform 5 -kg portion of a circular hoop is released from rest while in the position shown where the torsional spring of stiffness \(k_{T}=15 \mathrm{N} \cdot \mathrm{m} / \mathrm{rad}\) has been twisted \(90^{\circ}\) clockwise from its undeformed position. Determine the magnitude of the pin force at \(O\) at the instant of release. Motion takes place in a vertical plane and the hoop radius is \(r=150 \mathrm{mm}\)

Short Answer

Expert verified
The magnitude of the pin force at O is determined using energy conservation and rotational dynamics, leading to a calculation involving angular velocity.

Step by step solution

01

Convert Angle and Radius to Proper Units

Convert the angle from degrees to radians: \(90^{\circ} = \frac{\pi}{2} \text{ radians}\). Also, convert the radius from millimeters to meters: \(r = 150 \text{ mm} = 0.15 \text{ m} \).
02

Determine the Restoring Torque from the Spring

Using the torsional spring stiffness, the restoring torque \( \tau \) at the instant of release is calculated as \(\tau = k_T \times \theta\). Substituting the values, \(\tau = 15 \text{ N} \cdot \text{m/rad} \times \frac{\pi}{2} \text{ rad} = 15\frac{\pi}{2} \text{ N}\,\text{m} \).
03

Calculate the Moment of Inertia of the Hoop

The moment of inertia \( I \) for a hoop about its center of mass is \(I = mr^2\). Here, \(m = 5\text{ kg}\) and \(r = 0.15\text{ m}\), so \(I = 5 \times (0.15)^2 = 0.1125 \text{ kg} \cdot \text{m}^2\).
04

Apply Energy Conservation (Initial Energy = Potential Energy in Spring)

Initial energy stored in the torsional spring is \(\frac{1}{2} k_T \theta^2\). This should equal the rotational kinetic energy at the point of release: \(\frac{1}{2} I \omega^2 \), where \(\omega\) is the angular velocity. Substituting \(\frac{1}{2} \times 15 \times (\frac{\pi}{2})^2 = \frac{1}{2} \times 0.1125 \times \omega^2\), solve for \(\omega\).
05

Solve for Angular Velocity \(\omega\)

Rearrange the energy equation to find \(\omega\): \( 15 \times \left(\frac{\pi}{2}\right)^2 = 0.1125\omega^2 \,\), and solve for \(\omega: \, \omega = \sqrt{ \frac{15 \times \left(\frac{\pi}{2}\right)^2}{0.1125} }\). Calculate \(\omega\).
06

Calculate the Linear Velocity of the Center of Mass

The linear velocity \(v\) of the hoop's center of mass is given by \(v = r\cdot \omega\). Substitute the found \(\omega \) and radius \(r=0.15\text{ m}\) to get \(v\).
07

Determine the Net Forces and Pin Force at O

At the point of release, use Newton's second law in circular motion: the sum of forces is equal to the mass times the linear acceleration of the center of mass. That force consists of the gravitational force and the force at pin O: \( F_O = mg + m a_{cent}\), with \(a_{cent} = r\cdot\omega^2 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Torsional Spring
A torsional spring is a device that exerts a torque when twisted, similar to how a linear spring exerts a force when compressed or stretched. The torsional spring in this problem has a stiffness, denoted as \(k_T\), which measures how resistant the spring is to being twisted. This stiffness has units of \(\text{N} \cdot \text{m/rad}\), meaning the torque produced per radian of twist.
  • In the exercise, the torsional spring was initially twisted \(90^{\circ}\), which is converted to \(\frac{\pi}{2}\) radians.
  • The restoring torque \(\tau\) produced by the spring is calculated using the formula \(\tau = k_T \times \theta\), where \(\theta\) is the angular displacement.
Understanding this relationship helps in analyzing how the energy stored in the spring influences the subsequent motion when the system is released. The energy stored in the spring is initially potential energy, which is later converted to kinetic energy of the hoop.
Moment of Inertia
The moment of inertia plays a crucial role in rotational dynamics, similar to mass in linear dynamics. It is a measure of an object's resistance to change in its rotational motion. For a circular hoop rotating about its center, the moment of inertia \(I\) can be calculated using the formula \(I = mr^2\), where \(m\) is the mass and \(r\) is the radius of the hoop.
  • In this problem, the hoop has a mass \(m = 5\, \text{kg}\) and a radius \(r = 0.15\, \text{m}\).
  • Substituting these values, the moment of inertia is \(I = 5 \times (0.15)^2 = 0.1125\, \text{kg} \cdot \text{m}^2\).
This moment of inertia serves as a key parameter in calculating other quantities like angular acceleration or energy associated with the hoop's rotation. It determines how much torque is necessary to produce a desired angular acceleration, just like mass serves this role in linear motion.
Energy Conservation
Energy conservation is a fundamental principle in physics stating that the total energy of an isolated system remains constant. In this exercise, energy conservation is applied to understand how the initial potential energy stored in the torsional spring converts into kinetic energy of the hoop when released.
  • The initial potential energy in the spring is given by \( \frac{1}{2} k_T \theta^2\), representing the work necessary to twist the spring initially.
  • Once released, this potential energy is converted entirely into the rotational kinetic energy of the hoop, expressed as \( \frac{1}{2} I \omega^2\), where \(\omega\) is the angular velocity.
Setting these two forms of energy equal allows us to solve for \(\omega\), showing how energy principles simplify the analysis of mechanical problems by linking potential and kinetic energy.
Angular Velocity
Angular velocity describes how fast an object rotates, telling us the rate of change of its angular position with respect to time. In this exercise, the angular velocity \(\omega\) of the hoop is an essential quantity derived from the energy conservation principle.
  • Initially, \(\omega\) is found by equating the potential energy in the spring to the hoop's kinetic energy, leading to the equation \( 15 \times \left(\frac{\pi}{2}\right)^2 = 0.1125\omega^2 \).
  • Solving for \(\omega\) provides insights into how the stored energy will dynamically affect the movement of the hoop.
The concept of angular velocity helps us understand the speed and acceleration of rotational systems, connecting different types of motion in a cohesive framework. It is a pivotal concept in determining not only the motion at the point of release but also consequential forces like those at the pin.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The uniform slender bar \(A B\) has a mass of \(8 \mathrm{kg}\) and swings in a vertical plane about the pivot at \(A\). If \(\dot{\theta}=2 \mathrm{rad} / \mathrm{s}\) when \(\theta=30^{\circ},\) compute the force supported by the pin at \(A\) at that instant.

The 165 -lb ice skater with arms extended horizontally spins about a vertical axis with a rotational speed of 1 rev/sec. Estimate his rotational speed \(N\) if he fully retracts his arms, bringing his hands very close to the centerline of his body. As a reasonable approximation, model the extended arms as uniform slender rods, each of which is 27 in. long and weighs 15 lb. Model the torso as a solid 135-lb cylinder 13 in. in diameter. Treat the man with arms retracted as a solid 165 -lb cylinder of 13-in. diameter. Neglect friction at the skate-ice interface.

The 75 -kg flywheel has a radius of gyration about its shaft axis of \(\bar{k}=0.50 \mathrm{m}\) and is subjected to the torque \(M=10\left(1-e^{-t}\right) \mathrm{N} \cdot \mathrm{m},\) where \(t\) is in seconds. If the flywheel is at rest at time \(t=0\) determine its angular velocity \(\omega\) at \(t=3\) s.

Each of the two 300 -mm uniform rods \(A\) has a mass of \(1.5 \mathrm{kg}\) and is hinged at its end to the rotating base \(B\). The 4 -kg base has a radius of gyration of \(40 \mathrm{mm}\) and is initially rotating freely about its vertical axis with a speed of 300 rev/min and with the rods latched in the vertical positions. If the latches are released and the rods assume the horizontal positions, calculate the new rotational speed \(N\) of the assembly

The 64.4 -lb solid circular disk is initially at rest on the horizontal surface when a 3 -lb force \(P\), constant in magnitude and direction, is applied to the cord wrapped securely around its periphery. Friction between the disk and the surface is negligible. Calculate the angular velocity \(\omega\) of the disk after the \(3-1 b\) force has been applied for 2 seconds and find the linear velocity \(v\) of the center of the disk after it has moved 3 feet from rest.
See all solutions

Recommended explanations on Physics Textbooks

Dynamics

Read Explanation

Energy Physics

Read Explanation

Quantum Physics

Read Explanation

Electrostatics

Read Explanation

Translational Dynamics

Read Explanation

Engineering Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.