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The uniform 20 -kg slender bar is pivoted at \(O\) and swings freely in the vertical plane. If the bar is released from rest in the horizontal position, calculate the initial value of the force \(R\) exerted by the bearing on the bar an instant after release.

Short Answer

Expert verified
The initial force \( R = 0 \) because the bar’s center of mass starts dropping vertically under gravity; there's no centripetal component at the instant of release.

Step by step solution

01

Understand the Problem

We need to calculate the initial value of the force \( R \), which is the reaction force exerted by the pivot on the bar, right after it's released from a horizontal position.
02

Identify Relevant Forces and Equations

When the bar is released, the only forces acting on it are gravity and the reaction force at the pivot. The moment about the pivot due to gravity can be expressed as \( M = mg \frac{L}{2} \), where \( m = 20 \) kg and \( L \) is the length of the bar.
03

Determine Angular Acceleration

The bar begins to rotate about point \( O \) as soon as it is released. The angular acceleration \( \alpha \) can be determined from the equation \( I \alpha = M \), where \( I = \frac{1}{12}mL^2 \) is the moment of inertia of the bar about its center.
04

Calculate the Moment \( M \)

The moment \( M \) due to the gravitational force when the bar is horizontal is calculated as \( M = mg \frac{L}{2} \), substituting \( m = 20 \) kg, \( g = 9.81 \) m/s², and assuming \( L \) is known.
05

Apply Newton's Second Law for Rotation

Using \( I \alpha = M \), substitute for \( I \) and the previously calculated \( M \) to solve for \( \alpha \), the initial angular acceleration of the bar.
06

Calculate the Linear Acceleration of the Center of Mass

The linear acceleration \( a \) of the center of mass (located at \( L/2 \)) is \( a = \alpha \frac{L}{2} \).
07

Calculate the Required Reaction Force \( R \)

Using dynamics, the reaction at the pivot includes supporting the weight and the centripetal force due to the rotation: \( R = \, m( \alpha \frac{L}{2}) \). This accounts for the dynamic force felt by the pivot just after release.

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Key Concepts

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

Angular Acceleration
When an object begins to rotate, it experiences a change in rotational velocity, known as angular acceleration, denoted by \( \alpha \). For the slender bar in the problem, angular acceleration occurs because it is released from a horizontal position and starts to rotate about the pivot.

To mathematically determine \( \alpha \), we can apply the equation \( I \alpha = M \), where \( I \) is the moment of inertia and \( M \) is the moment caused by gravity.

The key point is that angular acceleration shows how quickly the rotational speed of the bar increases from rest. It is closely related to the forces causing the rotation.
Moment of Inertia
Moment of inertia, \( I \), is a critical concept in rotational motion. It analogously represents the mass distribution of an object, relative to its axis of rotation, similar to mass in linear motion.

For a slender bar, the moment of inertia can be calculated using the formula \( I = \frac{1}{12} m L^2 \), where \( m \) is the mass of the bar and \( L \) is its length.

This calculation tells us how much torque is needed for a desired angular acceleration. With a higher moment of inertia, an object will resist changes to its rotational motion more strongly.
Newton's Second Law for Rotation
Just as Newton's Second Law of motion describes the relationship between force, mass, and acceleration (\( F = ma \)), its rotational analog does the same for rotational motion. In this case, the law is expressed as \( I \alpha = M \), where torque (\( M \)) replaces force, and moment of inertia (\( I \)) and angular acceleration (\( \alpha \)) replace mass and linear acceleration, respectively.

In the exercise, this law helps us understand the dynamics of the bar as it begins to rotate around the pivot. It tells us that the moment about the pivot, created by the gravitational force, will result in an angular acceleration of the bar.

Understanding this relationship is essential for analyzing rotational systems, as it highlights the direct connection between applied torque and the resulting angular motion.
Centripetal Force
Centripetal force is necessary for any object in rotational motion to maintain its circular path. It acts inwardly towards the center of rotation.

In the scenario of the bar rotating about its pivot, centripetal force accounts for the inward force required to keep the center of mass moving in a circular path as it descends.

After the bar is released, the reacting force \( R \) at the pivot includes the component that provides centripetal force.
  • It ensures that despite gravity pulling the bar down, its path remains rotational around the pivot.
  • This force is critical in maintaining dynamic equilibrium as the bar begins its rotational motion.
Understanding centripetal force is key in describing the dynamic forces within rotating systems.

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