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Chapter 6: Problem 157

Each of the uniform bars \(O A\) and \(O B\) has a mass of \(2 \mathrm{kg}\) and is freely hinged at \(O\) to the vertical shaft, which is given an upward acceleration \(a=g / 2\). The links which connect the light collar \(C\) to the bars have negligible mass, and the collar slides freely on the shaft. The spring has a stiffness \(k=130 \mathrm{N} / \mathrm{m}\) and is uncompressed for the position equivalent to \(\theta=0 .\) Calculate the angle \(\theta\) assumed by the bars under conditions of steady acceleration.

Short Answer

Expert verified
The bars assume an angle of approximately \(0.019\) radians.

Step by step solution

01

Define the Forces Involved

Consider the forces acting on the system. The upward acceleration of the shaft is \(a = \frac{g}{2}\). Given each bar has a mass of \(2 \text{ kg}\), its weight would normally be \(mg = 2g\). But due to the additional upward acceleration of the shaft, the effective downward force on each bar becomes \(mg' = m(g - a) = 2g - g = g\) because \(a = \frac{g}{2}\). Thus the effective downward force is \(2 \times g = 19.6 \text{ N}\).
02

Calculate the Centrifugal Force

When the bars rotate to an angle \(\theta\) around the shaft, a centrifugal force acts outward. Since each bar is 1 meter long while uniform, center of each bar is at 0.5 meters from the origin. Centrifugal force can be expressed as \(F_c = \frac{1}{2}ma\theta = \frac{1}{2}(2)\left(\frac{g}{2}\right)(\frac{1}{2})\theta = \frac{g\theta}{4}\).
03

Balance the Forces to Determine \(\theta\)

The net force that tends to push collar C away from the shaft is the centrifugal force. This force should balance the spring force when compressed to equilibrium angle \(\theta\). This spring force is \(F_s = kx = k(1-\cos(\theta))\) where stiffness \(k = 130\, N/m\). Equilibrium condition implies \(\frac{g\theta}{4} = 130(1-\cos(\theta))\). This implies \[\frac{9.8\theta}{4} = 130(1-\cos(\theta))\].
04

Solve for \(\theta\)

Rearrange and solve the equation: \[9.8\theta = 520(1 - \cos(\theta)) \]. Approximating \(1 - \cos(\theta) \approx \frac{\theta^2}{2}\) for small \(\theta\), we solve for \(\theta\). \[9.8\theta = 520 \times \frac{\theta^2}{2} \]. This simplifies to \[\theta = \frac{19.6}{520} \cdot \theta^2\], leading to \(520\theta - 9.8 = 0\). Solving, \(\theta = 0.019\) rad for small angles.

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Key Concepts

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

Uniform Bars
Uniform bars are objects with consistent mass distribution throughout their length. This means that any segment of the bar has the same mass and size characteristics as any other segment of equal length. Such bars simplify the analysis of forces since their center of mass is always at the midpoint. In the problem given, bars OA and OB, each with a mass of 2 kg, can be treated as uniform because their mass and geometry remain constant across their span.
  • The center of mass for each bar is exactly halfway along its length.
  • This constancy allows for straightforward calculations of forces like weight and centrifugal forces.
Understanding this concept helps in analyzing how forces like weight and centrifugal force act on these bars during their motion.
Centrifugal Force
Centrifugal force is a pseudo force perceived in a rotating frame of reference, pushing radial outward. It is due to the tendency of the object to move in a straight line, while the reference frame rotates. In our exercise, this force becomes significant when the bars rotate around the shaft, creating an outward force from the center of rotation.
  • Expressed as: \( F_c = m \cdot r \cdot \omega^2 \)
  • Where \( m \) is the mass, \( r \) is the radius (distance from the axis of rotation), and \( \omega \) is the angular velocity.
In this specific scenario, the centrifugal force contributes to the system's dynamics by attempting to push the bars away from the shaft as they rotate. This force has to be balanced with other forces to maintain equilibrium.
Spring Stiffness
Spring stiffness, represented by the constant \( k \), measures a spring's resistance to deformation. It defines the relationship between the force exerted on the spring and its displacement, expressed in \( N/m \). In the given problem, as bars rotate, the spring connected to them experiences compression, which generates a restoring force opposing this motion.
  • Hooke's Law: \( F_s = k \cdot x \)
  • \( F_s \) is the force exerted by the spring and \( x \) is the displacement from its rest position.
Determining the angle \( \theta \), when the system is in equilibrium, requires balancing this spring force with other forces like centrifugal force. Understanding spring stiffness is crucial to solving the problem, as this force provides stability and resistance to the rotating bars.
Upward Acceleration
Upward acceleration in this problem is referred to as the shaft's acceleration, which is equivalent to \( a = \frac{g}{2} \). This translates to the shaft moving upwards with half the gravitational acceleration. It significantly influences how the overall forces act on the uniform bars.
  • The effective gravitational force is altered due to upward acceleration.
  • This changed force impacts how the system's dynamic balance is achieved.
The resulting decrease in the effective gravitational force ensures the bars and the entire system react differently compared to when only gravity is in effect. This needs to be accounted for when analyzing how forces act and are balanced in order to find the angle \( \theta \) that the bars maintain due to this complex interplay of forces.

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Most popular questions from this chapter

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