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Determine the bending moment \(\mathbf{M}\) at the tangency point \(A\) in the semicircular rod of radius \(r\) and mass \(m\) as it rotates about the tangent axis with a constant and large angular velocity \(\omega .\) Neglect the moment \(m g r\) produced by the weight of the rod.

Short Answer

Expert verified
The bending moment \( \mathbf{M} \) at point \( A \) is zero.

Step by step solution

01

Understand the Problem

We are given a semicircular rod with radius \( r \) and mass \( m \) rotating around its tangent point with angular velocity \( \omega \). We need to find the bending moment \( \mathbf{M} \) at the tangency point \( A \). The effect of gravity is negligible in this scenario.
02

Identify the Relevant Equations

The bending moment \( \mathbf{M} \) is affected by the centrifugal forces due to the rotation. We can use the equation for the bending moment caused by a distributed mass in rotation: \( \mathbf{M} = \int r \, dF \), where \( dF = \omega^2 \rho \, dV \) is the differential centrifugal force.
03

Set Up the Integral

The rod can be discretized into differential elements. Consider the element \( d\theta \) at an angle \( \theta \) from the vertical. The differential volume is given by \( dV = \rho r \, d\theta \times \text{(arc length)} \). The arc length for each differential element is \( r \, d\theta \), so \( dV = \rho r^2 \, d\theta \).
04

Calculate the Differential Centrifugal Force

Substitute into the equation: \( dF = \omega^2 \rho r^2 \, d\theta \). The distance from \( A \) is \( r \cos\theta \), so the moment arm is \( r \cos\theta \). Therefore, \( dM = \omega^2 \rho r^3 \cos\theta \, d\theta \).
05

Integrate Over the Semicircle

Integrate the expression for \( dM \) from \( \theta = 0 \) to \( \theta = \pi \): \[ \mathbf{M} = \omega^2 \rho r^3 \int_0^\pi \cos\theta \, d\theta = \omega^2 \rho r^3 \left[\sin\theta\right]_0^\pi = \omega^2 \rho r^3 (0 - 0) = 0. \]
06

Analyze the Result

Despite the centrifugal force, the symmetrical nature of the semicircular rod means that the net moment about the point of tangency is zero. Thus, the bending moment \( \mathbf{M} \) is indeed zero.

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Key Concepts

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

Semicircular Rod
A semicircular rod is a fascinating object often studied in physics and engineering. Imagine taking a perfect circle and dividing it in half; the resulting shape is a semicircle. A rod bent into this shape forms a semicircular rod. Its characteristics are defined by two main factors: its radius, which is the distance from the center to any point on the curve, and its mass, which is the weight distributed along the curve.
In practical applications, semicircular rods might be used in structures where both stability and aesthetics are important. The semicircular shape allows for even distribution of forces, making it a useful element in certain mechanical systems. When considering a rotating semicircular rod, like the one described in the exercise, it’s important to understand how the unique geometry affects physical properties like bending moments.
Angular Velocity
Angular velocity is a measure of how quickly an object rotates or revolves around a central point or axis. It is commonly represented by the symbol \( \omega \) (omega) and is expressed in radians per second.
For a rotating semicircular rod, angular velocity is crucial because it determines how fast the rod spins around its tangent axis. This speed of rotation influences other forces acting on the rod, such as centrifugal force, which in turn affects the bending moment at any given point. Understanding angular velocity helps in calculating these dynamic forces to ensure the rod's stability and function in mechanical systems.
Centrifugal Force
Centrifugal force is an apparent force that acts outward on a body moving around a center, arising from the body's inertia. In the context of a semicircular rod rotating with angular velocity \( \omega \), centrifugal force is key to determining the forces experienced along the rod.
This force depends on the mass of the rod segment and its distance from the rotational center. As each part of the rod spins, it experiences an outward force trying to "fling" it away from the center of rotation. This outward push is what we refer to as centrifugal force. The exercise solution demonstrates how these forces are integrated along the semicircular rod to help calculate the overall bending moment.
Tangent Axis
The tangent axis is the straight line around which the semicircular rod is rotating in the exercise. This line is tangent to the semicircle at the point of contact, known as the tangency point. It serves as the pivot or rotational axis for the rod.
Understanding the tangent axis is critical because it is the point where bending moments are calculated. As the rod rotates around this axis with constant angular velocity, the forces and resultant moments change depending on the distribution of mass and the rod's geometry. Knowing the location and properties of the tangent axis allows engineers to predict how the rod will behave under rotational forces.
Mechanical Rotation
Mechanical rotation refers to the motion where an object spins around a fixed axis, such as the semicircular rod in this exercise. This process is governed by principles of mechanics and rotational dynamics.
When a system involving mechanical rotation is studied, several factors are considered, including angular velocity, the shape of the rotating object, and the axis of rotation. For the semicircular rod, these aspects interplay to influence how centrifugal forces cause bending moments. Mechanical rotation not only helps in analyzing the stresses and strains within rotating objects but also informs design choices in engineering different rotational systems.

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Most popular questions from this chapter

The circular disk of 120 -mm radius rotates about the \(z\) -axis at the constant rate \(\omega_{z}=20 \mathrm{rad} / \mathrm{s}\), and the entire assembly rotates about the fixed \(x\) -axis at the constant rate \(\omega_{x}=10 \mathrm{rad} / \mathrm{s}\). Calculate the magnitudes of the velocity \(\mathbf{v}\) and acceleration a of point \(B\) for the instant when \(\theta=30^{\circ}\)

The pendulum oscillates about the \(x\) -axis according to \(\theta=\frac{\pi}{6} \sin 3 \pi t\) radians, where \(t\) is the time in seconds. Simultaneously, the shaft \(O A\) revolves about the vertical \(z\) -axis at the constant rate \(\omega_{z}=2 \pi \mathrm{rad} / \mathrm{sec}\) Determine the velocity \(\mathbf{v}\) and acceleration a of the center \(B\) of the pendulum as well as its angular acceleration \(\alpha\) for the instant when \(t=0\)

The assembly, consisting of the solid sphere of mass \(m\) and the uniform rod of length \(2 c\) and equal mass \(m,\) revolves about the vertical \(z\) -axis with an angular velocity \(\omega .\) The rod of length \(2 c\) has a diameter which is small compared with its length and is perpendicular to the horizontal rod to which it is welded with the inclination \(\beta\) shown. Determine the combined angular momentum \(\mathbf{H}_{O}\) of the sphere and inclined rod.

The collar and clevis \(A\) are given a constant upward velocity of 8 in./sec for an interval of motion and cause the ball end of the bar to slide in the radial slot in the rotating disk. Determine the angular acceleration of the bar when the bar passes the position for which \(z=3\) in. The disk turns at the constant rate of 2 rad/sec.

The 5 -kg disk and hub \(A\) have a radius of gyration of \(85 \mathrm{mm}\) about the \(z_{0}\) -axis and spin at the rate \(p=1250\) rev/min. Simultaneously, the assembly rotates about the vertical \(z\) -axis at the rate \(\Omega=400\) rev/min. Calculate the gyroscopic moment \(\mathbf{M}\) exerted on the shaft at \(C\) by the disk and the bending moment \(M_{O}\) in the shaft at \(O .\) Neglect the mass of the shaft but otherwise account for all forces acting on it.

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