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Chapter 3: Problem 69

The hollow tube assembly rotates about a vertical axis with angular velocity \(\omega=\dot{\theta}=4 \quad \mathrm{rad} / \mathrm{s}\) and \(\dot{\omega}=\ddot{\theta}=-2 \operatorname{rad} / \mathrm{s}^{2} .\) A small \(0.2-\mathrm{kg}\) slider \(P\) moves inside the horizontal tube portion under the control of the string which passes out the bottom of the assembly. If \(r=0.8 \mathrm{m}, \dot{r}=-2 \mathrm{m} / \mathrm{s},\) and \(\ddot{r}=4 \mathrm{m} / \mathrm{s}^{2}\) determine the tension \(T\) in the string and the horizontal force \(F_{\theta}\) exerted on the slider by the tube.

Short Answer

Expert verified
Tension: 1.76 N; Horizontal Force: -3.52 N.

Step by step solution

01

Understand the Dynamics of Rotational Motion

The problem involves a rotating assembly with a slider moving within a tube. We need to account for the centripetal and Coriolis accelerations affecting the slider due to its radial and tangential velocities and accelerations. These forces are influenced by the given angular velocities \(\omega = 4 \text{ rad/s}\) and their derivatives \(\dot{\omega} = -2 \text{ rad/s}^2\).
02

Identify Forces and Accelerations

Forces acting on the slider include the tension in the string \(T\) and the horizontal force \(F_\theta\) due to the tube. The accelerations involved are radial acceleration \(\ddot{r} = 4 \text{ m/s}^2\) and tangential acceleration components related to \(\ddot{\theta} = \dot{\omega} = -2 \text{ rad/s}^2\). The radial distance \(r\) changes at \(\dot{r} = -2 \text{ m/s}\).
03

Calculate the Radial Forces

Use the equation for radial force balance in a rotating system: \[ F_r = m(\ddot{r} - r\omega^2) \] Plugging in the values, \[ F_r = 0.2(4 - 0.8\times4^2) = 0.2(4 - 12.8) = 0.2(-8.8) \] \[ F_r = -1.76 \text{ N} \] Since \(F_r = T\), the tension is the inward force required to keep slider P in circular motion.
04

Calculate the Horizontal Force Exerted by the Tube

For the horizontal force \(F_\theta\), consider the Coriolis force and acceleration components influenced by \(\ddot{\theta}\). The tangential force balance is:\[ F_\theta = m(r \dot{\omega} + 2\dot{r} \omega) \] Substitute in given values:\[ F_\theta = 0.2(0.8(-2) + 2(-2)4) = 0.2(-1.6 - 16) \] \[ F_\theta = 0.2(-17.6) = -3.52 \text{ N} \]
05

Interpret and Conclude

The tension \(T\) in the string is \(1.76 \text{ N}\). This tension is a result of the slider's radial motion and is directed inward. The negative sign for \(F_\theta = -3.52 \text{ N}\) indicates the horizontal force exerted by the tube opposes the slider's tangential velocity, effectively acting as a counteracting force to its movement.

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Key Concepts

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

Centripetal Force
Anything moving in a circle is subject to a centripetal force. This force is always directed toward the center of the circle or rotation path. For Circular Motion, without this force, the object would continue to move in a straight line, due to inertia.In our exercise, the tension in the string (\( T \)) acts as the centripetal force keeping the slider in its path. Even though the slider appears to want to travel in a straight line, the string pulls it inwards, maintaining its circular pathway. The centripetal force is calculated using the formula:\[F_c = m(\ddot{r} - r \omega^2)\]
  • \( m \): mass of the object (0.2 kg for this slider)
  • \( \ddot{r} \): radial acceleration (4 m/s虏)
  • \( r \): radial distance (0.8 m)
  • \( \omega \): angular velocity (4 rad/s)
For this problem, the calculation results in a tension (i.e., centripetal force) of 1.76 N, directed toward the center of rotation.
Coriolis Effect
When dealing with rotating systems, the Coriolis Effect is a critical concept to grasp. This pseudo-force arises due to the rotation of the reference frame when an object moves within it. It significantly influences how we perceive the motion of the object, causing an apparent deflection from its path.In the horizontal force calculation for this problem, we consider the Coriolis force on the slider. The Coriolis acceleration affects objects moving within a rotating system, like our slider in the hollow tube:\[F_{\theta} = m(r \dot{\omega} + 2\dot{r} \omega)\]This formula combines the 'extra' accelerations due to the slider's velocity and the system's rotation. While not a real force like tension, the Coriolis force has real effects on how objects move within rotating frames.
Angular Motion
Angular Motion describes how an object spins or rotates about a central point. It involves several key aspects like angular velocity and angular acceleration, which tell us how fast and in what manner the rotation is changing.In our slider problem, the angular velocity (\( \omega = 4 \, \text{rad/s} \)) indicates how fast the assembly rotates. A constant change in this rate of rotation is defined by angular acceleration (\( \ddot{\theta} = -2 \, \text{rad/s}^2 \)). Angular acceleration tells us that rotation is slowing down.
  • Angular velocity (\( \omega \)): the speed of rotation
  • Angular acceleration (\( \ddot{\theta} \)): the rate of change of angular velocity
Understanding how these parameters change helps in analyzing the forces and motions within the rotating system, like how and why tension and Coriolis effects are present.

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Most popular questions from this chapter

The car is moving with a speed \(v_{0}=65 \mathrm{mi} / \mathrm{hr}\) up the 6 -percent grade, and the driver applies the brakes at point \(A,\) causing all wheels to skid. The coefficient of kinetic friction for the rain-slicked road is \(\mu_{k}=0.60 .\) Determine the stopping distance \(s_{A B} .\) Repeat your calculations for the case when the car is moving downhill from \(B\) to \(A\).

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The slotted arm \(O A\) rotates about a fixed axis through \(O\). At the instant under consideration, \(\theta=\) \(30^{\circ}, \dot{\theta}=45 \operatorname{deg} / \mathrm{s},\) and \(\ddot{\theta}=20 \mathrm{deg} / \mathrm{s}^{2} .\) Determine the forces applied by both arm \(O A\) and the sides of the slot to the 0.2 -kg slider \(B\). Neglect all friction, and let \(L=0.6 \mathrm{m} .\) The motion occurs in a vertical plane.
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