/*! 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 107 A small \(2-\) kg sphere \(B\) i... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A small \(2-\) kg sphere \(B\) is attached to the bar \(A B\) of negligible mass that is supported at \(A\) by a pin and bracket and connected at \(C\) to a moving support \(D\) by mans of a spring of constant \(k=3.6 \mathrm{kN} / \mathrm{m}\). Knowing that support \(D\) undergoes a vertical displacement \(\delta=\delta_{m} \sin \omega_{f} t,\) where \(\delta_{m}=3 \mathrm{mm}\) and \(\omega_{f}=15 \mathrm{rad} / \mathrm{s}\), determine \((a)\) the magnitude of the maximum angular velocity of bar \(A B,\) (b) the magnitude of the maximum acceleration of sphere \(B .\)

Short Answer

Expert verified
Maximum angular velocity is found using dynamic analysis and maximum acceleration involves the radius and angular acceleration relations.

Step by step solution

01

Convert Units

First convert the displacement amplitude \(\delta_m=3 \, \text{mm}\) to meters \(\delta_m=0.003 \, \text{m}\). The spring constant is already given in kN/m, so convert it to N/m: \(k = 3600 \, \text{N/m}\).
02

Calculate Maximum Spring Force

The maximum displacement of the spring will be \(\delta_m=0.003 \, \text{m}\). Calculate the maximum spring force using Hooke's Law: \(F_{\text{spring}} = k \cdot \delta_m = 3600 \, \text{N/m} \cdot 0.003 \, \text{m} = 10.8 \, \text{N}\).
03

Determine Angular Displacement

Use the balance of forces and consider the bar as a dynamic system. The force from the spring causes a dynamic response where the angular displacement \(\theta\) can be found. Since it's a complex dynamic analysis, assume small angles and use \(\theta = \frac{F_{\text{spring}}}{m_{B} \cdot g}\). Here, \(m_B = 2 \, \text{kg}\) and \(g = 9.81 \, \text{m/s}^2\). Substitute in for a detailed derivation.
04

Calculate Maximum Angular Velocity

To find the maximum angular velocity \(\omega\), use energy conservation or dynamic analysis. With \(\theta\) known, often \(\omega_{\text{max}} = \omega_{f} \cdot \theta_{\text{max}}\). Substitute \(\theta_{\text{max}}\) from the previous step.
05

Compute Maximum Acceleration of Sphere B

The maximum acceleration of the sphere can be found using \(a = r \cdot \alpha\), where \(r\) is the distance from the rotation point to the sphere and \(\alpha = a_{\text{max}}/r\). With maximum angular acceleration related to \(\omega_{\text{max}}\), solve for \(a_{\text{max}} = r \cdot \alpha_{\text{max}}\).

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 91Ó°ÊÓ!

Key Concepts

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

Spring Force
When discussing spring force, we are referring to the force exerted by a spring when it is compressed or stretched. This is typically calculated using Hooke's Law, which states:
  • Force exerted by the spring is proportional to the distance it is stretched or compressed, denoted as \( F = -k \cdot x \).
  • Here, \( k \) is the spring constant (in N/m), and \( x \) is the deformation from the spring's equilibrium position (in meters).
The negative sign indicates the direction of the force opposes the direction of displacement, aiming to restore the spring system to its original state. In our exercise, the spring has a constant of \( 3600 \, \text{N/m} \) and undergoes a maximum displacement of \( 0.003 \, \text{m} \), resulting in a maximum force of \( 10.8 \, \text{N} \). Understanding this principle helps in visualizing how forces within springs influence connected objects, such as the sphere in our scenario.
Angular Velocity
Angular velocity is a measure of how fast an object rotates around a particular point or axis. It is typically expressed in radians per second (rad/s). In dynamic systems involving rotation, determining the maximum angular velocity of a component, like bar AB, is crucial. To achieve this, one may utilize energy conservation methods or dynamic analysis to relate angular velocity to other dynamic variables.
For our exercise, we detect the influence of support D through a sinusoidal vertical displacement, which affects the rotation speed of the bar. We consider the maximum angular velocity to be connected through a relation with angular displacement \( \theta \) as \( \omega_{\text{max}} = \omega_{f} \cdot \theta_{\text{max}} \). Here, \( \omega_{f} \) is the external frequency, and \( \theta_{\text{max}} \) could be derived from the dynamics or energy calculations executed beforehand.
Understanding these dynamics offers insights into how rotational speeds are calculated and controlled in systems involving periodic motions.
Angular Acceleration
Angular acceleration refers to the rate at which angular velocity changes over time. It provides information on how quickly an object is speeding up or slowing down in its rotational motion. For a comprehensive analysis of a system involving a rotating element like the sphere at the end of a bar, understanding angular acceleration is critical.In mathematical terms, angular acceleration (\( \alpha \)) can be expressed as the change in angular velocity (\( \Delta \omega \)) over the change in time (\( \Delta t \)), or \( \alpha = \frac{\Delta \omega}{\Delta t} \). This connects to linear acceleration through the relationship \( a = r \cdot \alpha \), where \( r \) is the radius or distance from the pivot point to the point of interest (e.g., the sphere).In the given exercise, determining the maximum acceleration involves understanding how these quantities interact, allowing us to identify the forces on the sphere B as it moves due to the angular dynamics involved.
Energy Conservation
Energy conservation is a fundamental principle in dynamics, particularly in analyzing systems undergoing motion like our bar-sphere setup. It states that, within a closed system, total energy remains constant over time, barring exclusions like external forces or dissipative forces such as friction.
  • Kinetic energy associated with linear and rotational movements of a system is given by \( KE = \frac{1}{2}mv^2 \) for linear or \( KE = \frac{1}{2}I\omega^2 \) for rotational motion.
  • Potential energy, particularly from the spring, can be expressed as \( PE = \frac{1}{2}kx^2 \).
In our exercise, energy conservation assists in calculating maximum velocities or accelerations by setting up energy exchanges between potential and kinetic forms. During such movements, the mechanical energy exchange helps attain unknown variables like maximum angular velocity by equating the respective kinetic and potential energy forms.

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 8 -kg uniform bar \(A B\) is hinged at \(C\) and is attached at \(A\) to a spring of constant \(k=500 \mathrm{N} / \mathrm{m}\). If end \(A\) is given a small displacement and released, determine \((a)\) the frequency of small oscillations, (b) the smallest value of the spring constant \(k\) for which oscillations will occur.

Three collars each have a mass \(m\) and are connected by pins to bars \(A C\) and \(B C,\) each having length \(l\) and negligible mass. Collars \(A\) and \(B\) can slide without friction on a horizontal rod and are connected by a spring of constant \(k\) Collar \(C\) can slide without friction on a vertical rod and the system is in equilibrium in the position shown. Knowing that collar \(C\) is given a small displacement and released, determine the frequency of the resulting motion of the system.

A 150 -kg electromagnet is at rest and is holding \(100 \mathrm{kg}\) of scrap steel when the current is turned off and the steel is dropped. Knowing that the cable and the supporting crane have a total stiffiness equivalent to a spring of constant \(200 \mathrm{kN} / \mathrm{m}\), determine \((a)\) the frequency, the amplitude, and the maximum velocity of the resulting motion, \((b)\) the minimum tension that will occur in the cable during the motion, \((c)\) the velocity of the magnet 0.03 s after the current is turned off.

A motor of mass \(M\) is supported by springs with an equivalent spring constant \(k .\) The unbalance of its rotor is equivalent to a mass \(m\) located at a distance \(r\) from the axis of rotation. Show that when the angular velocity of the motor is \(\omega_{f}\), the amplitude \(x_{m}\) of the motion of the motor is $$ x_{m}=\frac{r(m / M)\left(\omega_{f} / \omega_{n}\right)^{2}}{1-\left(\omega_{f} / \omega_{n}\right)^{2}} $$ where \(\omega_{n}=\sqrt{k / M}\)

A 2 -kg block is supported by a spring with a constant of \(k=128 \mathrm{N} / \mathrm{m}\) and a dashpot with a coefficient of viscous damping of \(c=0.6 \mathrm{N} \cdot \mathrm{s} / \mathrm{m}\). The block is in equilibrium when it is struck from below by a hammer that imparts to the block an upward velocity of \(0.4 \mathrm{m} / \mathrm{m} / \mathrm{s}\). Determine \((a)\) the logarithmic decrement, (b) the maximum upward displacement of the block from equilibrium after two cycles.

See all solutions

Recommended explanations on Physics Textbooks

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.