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A uniform equilateral triangular plate with a side \(b\) is suspended from three vertical wires of the same length \(l\). Determine the period of small oscillations of the plate when \((a)\) it is rotated through a small angle about a vertical axis through its mass center \(G,(b)\) it is given a small horizontal displacement in a direction perpendicular to \(A B .\)

Short Answer

Expert verified
For both cases, the period of oscillation is \(T = 2\pi \sqrt{\frac{l}{g}}\).

Step by step solution

01

Identify system properties and geometry

Consider an equilateral triangular plate of mass \(m\) and side \(b\). The center of mass \(G\) is at the centroid of the triangle, which is about \(\frac{b}{\sqrt{3}}\) from any vertex in the plane of the triangle.
02

Moment of inertia of the system

The moment of inertia \(I\) of the equilateral triangle about a vertical axis through its centroid is \(I = \frac{m b^2}{12}\).
03

Calculate restoring torque

When the plate is rotated about its vertical axis through \(G\) by a small angle \(\theta\), a restoring torque \(\tau = -C\theta\) is exerted, where \(C\) is a constant. Since it's a small oscillation, \(C = \frac{mgb^2}{12l}\) given by the product of force and perpendicular distance.
04

Determine the angular frequency

The angular frequency \(\omega\) is given by: \(\omega^2 = \frac{C}{I}\). Substituting, \(\omega^2 = \frac{mgb^2}{12l} \times \frac{12}{m b^2} = \frac{g}{l}\).
05

Find the period of oscillation for rotation

The period \(T\) is related to angular frequency by \(T = 2\pi / \omega\). Therefore, \(T = 2\pi \sqrt{\frac{l}{g}}\).
06

Calculate horizontal displacement force

When displaced horizontally, it's similar to a pendulum, due to gravity and suspension, restoring force leads to \(F = -mg\theta\).
07

Frequency of horizontal oscillation

The mass experiences a restoring force similar to the vertical pendular motion, with frequency \(\omega^2 = \frac{g}{l}\).
08

Determine the period of oscillation for horizontal displacement

The period remains the same for small oscillations displaced horizontally due to equivalent restoring force, \(T = 2\pi \sqrt{\frac{l}{g}}\). Thus, the period is equal for rotations and horizontal displacements.

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Key Concepts

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

Moment of Inertia
The concept of moment of inertia is fundamental when studying mechanical oscillations, particularly in systems involving rotational motion. Moment of inertia, often denoted as \( I \), is a measure of an object's resistance to angular acceleration about a particular axis. It depends on the mass distribution concerning the axis of rotation. For a uniform equilateral triangle, which has sides of length \( b \) and a consistent mass distribution, the moment of inertia about its centroid is special. Mathematically, it is given by:\[I = \frac{m b^2}{12}\]Here, \( m \) represents the mass of the triangular plate. This formula shows how the shape and mass of an object affect its tendency to resist changes in its rotational motion. The further the mass is from the axis of rotation, the higher the moment of inertia, making it harder to rotate.
Angular Frequency
Angular frequency, denoted by \( \omega \), is a key concept in examining oscillatory systems. It represents how many oscillations occur per unit of time and is directly linked to the speed of the oscillation. In our triangle case, after determining the restoring torque and using the moment of inertia, we find the expression for angular frequency.Using the relation:\[\omega^2 = \frac{C}{I}\]And substituting in the values for our system, it simplifies to:\[\omega^2 = \frac{g}{l}\]where \( g \) is the acceleration due to gravity, and \( l \) is the length of the wires from which the triangle is suspended. This formula indicates that the angular frequency depends only on the acceleration due to gravity and the length of the wire, showing that this property is purely dependent on external factors and not the properties of the triangle itself.
Restoring Torque
Restoring torque is the pivotal force responsible for bringing an oscillating system back to its equilibrium position. In this triangular setup, when the plate is rotated by a small angle \( \theta \), a torque is generated. This is due to the gravitational force acting on the center of mass of the triangle, which tries to bring it back to its initial vertical orientation.The restoring torque \( \tau \) is expressed as:\[\tau = -C\theta\]Where \( C \) is a constant; specifically for this system:\[C = \frac{mgb^2}{12l}\]Here, \( mg \) represents the gravitational force acting downward, \( b \) is the side length of the triangle, and \( l \) is the wire length. The negative sign indicates that the torque works in the opposite direction of the displacement angle \( \theta \), ensuring the triangle moves back to its original position.
Equilateral Triangle Geometry
Understanding the geometry of an equilateral triangle is crucial for deriving formulas related to oscillations. An equilateral triangle has all sides equal, with interior angles of 60 degrees. This symmetry aids in simplifying calculations, especially when dealing with physical properties like the center of mass and moment of inertia.In this situation, the centroid of the triangle (point \( G \)), where its mass is considered to act, is positioned at a distance of \( \frac{b}{\sqrt{3}} \) from any vertex. This geometric property is pivotal to calculations in the mechanical oscillations of the triangular plate.
  • Each side is equal, simplifying rotational symmetry considerations.
  • The distance from the centroid to a vertex being \( \frac{b}{\sqrt{3}} \) helps simplify torque and inertia calculations.
  • These properties make equilateral triangles ideal for symmetrical oscillatory studies.
Understanding these geometric aspects is crucial for comprehending how such shapes behave dynamically in systems.

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Most popular questions from this chapter

A uniform rod \(A B\) can rotate in a vertical plane about a horizontal axis at \(C\) located at a distance \(c\) above the mass center \(G\) of the rod. For small oscillations determine the value of \(c\) for which the frequency of the motion will be maximum.

A 4 -kg collar can slide on a frictionless horizontal rod and is attached to a spring with constant \(k\). It is acted upon by a periodic force of magnitude \(P=P_{m} \sin \omega_{f} t,\) where \(P_{m}=9 \mathrm{N}\) and \(\omega_{f}=5 \mathrm{rad} / \mathrm{s}\). Determine the value of the spring constant \(k\) knowing that the motion of the collar has an amplitude of \(150 \mathrm{mm}\) and is \((a)\) in phase with the applied force, \((b)\) out of phase with the applied force.

A horizontal platform \(P\) is held by several rigid bars that are connected to a vertical wire. The period of oscillation of the platform is found to be 2.2 s when the platform is empty and 3.8 s when an object \(A\) of unknown moment of inertia is placed on the platform with its mass center directly above the center of the plate. Knowing that the wire has a torsional constant \(K=27 \mathrm{N} \cdot \mathrm{m} / \mathrm{rad}\), determine the centroidal moment of inertia of object \(A\).

A machine element weighing 800 lb is supported by two springs, each having a constant of 200 lb/in. A periodic force of maximum value 30 lb is applied to the element with a frequency of 2.5 cycles per second. Knowing that the coefficient of damping is 8 lb-s/n. determine the amplitude of the steady- state vibration of the element.

In the case of light damping \(\left(c

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