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Chapter 17: Problem 91

A small 4 -lb collar \(C\) can slide freely on a thin ring of weight 6 lb and radius 10 in. The ring is welded to a short vertical shaft, which can rotate freely in a fixed bearing. Initially, the ring has an angular velocity of \(35 \mathrm{rad} / \mathrm{s}\) and the collar is at the top of the ring \((\theta=0)\) when it is given a slight nudge. Neglecting the effect of friction, determine ( \(a\) ) the angular velocity of the ring as the collar passes through the position \(\theta=90^{\circ},(b)\) the corresponding velocity of the collar relative to the ring.

Short Answer

Expert verified
(a) The angular velocity of the ring is reduced due to collar displacement. (b) Collar's relative velocity depends on ring movement and its own displacement.

Step by step solution

01

Initial Setup and Conservation of Angular Momentum

The system consists of a rotating ring and a collar. The angular momentum must be conserved because there's no external torque. Initially, the ring (with collar at the top) has an angular velocity \( \omega_0 = 35 \, \text{rad/s} \). The total initial angular momentum \( L_i \) is given by \( L_i = I_r \omega_0 \), where \( I_r \) is the moment of inertia of the ring with the collar at the top.
02

Calculate Initial Moment of Inertia

The initial moment of inertia \( I_r \) is the sum of the ring's and collar's moments of inertia. For the ring (regarded as a hoop), \( I_r = m_r R^2 \), where \( m_r \) is the mass of the ring and \( R \) is the radius. The collar starts at the center, adding no initial inertia. Thus, \( I_r = m_r R^2 = (6\, \text{lb} / g) (10\, \text{in})^2\). Use \( g = 32.2 \, \text{ft/s}^2 \) to convert weights to mass.
03

Apply Angular Momentum Conservation

As the collar moves to \( \theta = 90^\circ \), both the collar and the ring rotate together. The final angular momentum \( L_f = I_f \omega_f \) must equal \( L_i \). The total moment of inertia \( I_f \) now includes the collar's rotational inertia, \( I_f = m_r R^2 + m_c R^2 \), where \( m_c \) is the collar's mass.
04

Solve for Final Angular Velocity

With \( L_f = L_i \), set \( (m_r R^2 + m_c R^2) \omega_f = m_r R^2 \omega_0 \). Solve for \( \omega_f \), \( \omega_f = \frac{m_r R^2}{m_r R^2 + m_c R^2} \omega_0 \). Substituting \( m_r = \frac{6}{32.2} \text{lb路s}^2\,/\,\text{ft} \), \( m_c = \frac{4}{32.2} \text{lb路s}^2\,/\,\text{ft}\), and converting inches to feet for \( R \), solve for \( \omega_f \).
05

Calculate Collar's Relative Velocity

The collar moves with the combination of the ring's rotation and its own original position along the ring. Its relative velocity \( v_{cr} \) to the ring can be expressed as \( v_{cr} = \omega_f \times R \times \theta \) for angular displacement \( \theta = 90^\circ \). Convert \( R \) to feet in calculations.

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Key Concepts

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

Moment of Inertia
Understanding moment of inertia is crucial to analyzing rotational dynamics problems, such as the one presented with the rotating ring and collar. The moment of inertia is essentially the rotational equivalent of mass in linear motion. It represents an object's resistance to changes in its angular velocity. For a given body, moment of inertia depends on its shape, mass distribution, and the chosen axis of rotation. In the exercise, we have:
  • The ring, considered as a hoop, has a constant shape and mass distribution at a radius of 10 inches. Its moment of inertia, when the collar is at the top, is given by the formula: \[ I_r = m_r R^2 \] where \( m_r \) is the ring鈥檚 mass and \( R \) is the radius.
  • The collar, when it moves, adds more moment of inertia since it too rotates along the ring. Thus, when it reaches \( \theta = 90^\circ \), its contribution is additive to the system.
This system demonstrates how dynamic and positional changes alter the overall moment of inertia, affecting rotational behaviors.
Rotational Dynamics
Rotational dynamics is the study of rotational motion propelled by forces. It focuses on how torque affects the angular velocity and acceleration of a system.In scenarios without external torques, like this exercise, the principle of angular momentum conservation becomes key. Here鈥檚 why it helps:
  • Angular momentum \( L \) is the product of moment of inertia \( I \) and angular velocity \( \omega \): \[ L = I \omega \]
  • With no external torques, the initial and final angular momentum remain equal. This is represented as: \[ L_i = L_f \] Where the subscript \( i \) denotes initial states, and \( f \) denotes final states.
  • Applying such conservation allows computations for unknowns, like the final angular velocity \( \omega_f \), using known initial conditions.
By setting up the relating equations with clear initial and final parameters, we solve for variables like \( \omega_f \) even when complex movements, like collar shifts, occur.
Angular Velocity
Angular velocity represents the rate of change of an object鈥檚 rotational angle per unit time, typically measured in radians per second. In rotational systems, changes in angular conditions typically reflect shifts in angular velocity.In this exercise:
  • The initial angular velocity \( \omega_0 = 35 \, \text{rad/s} \) when the collar is at the top indicates the system's primary movement.
  • As the collar slides to \( \theta = 90^\circ \), the changes in distribution of system mass dictate a new angular velocity \( \omega_f \). Since the overall angular momentum doesn't change, applying: \[ (m_r R^2 + m_c R^2) \omega_f = m_r R^2 \omega_0 \] helps solve for \( \omega_f \), reflecting the system's dynamic change.
  • The collar鈥檚 relative velocity is crucial, which is calculated using: \[ v_{cr} = \omega_f \times R \times \theta \] This describes how fast the collar moves with respect to its position on the rotating ring.
Understanding these concepts aids in predicting and calculating dynamics in rotating systems with variable mass distributions.

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Most popular questions from this chapter

The rotor of an electric motor has a mass of 25 kg, and it is observed that 4.2 min is required for the rotor to coast to rest from an angular velocity of 3600 rpm. Knowing that kinetic friction produces a couple of magnitude 1.2 N?m, determine the centroidal radius of gyration for the rotor.

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A 9-in.-radius cylinder of weight 18 lb rests on a 6-lb carriage. The system is at rest when a force P of magnitude 2.5 lb is applied as shown for 1.2 s. Knowing that the cylinder rolls without sliding on the carriage and neglecting the mass of the wheels of the carriage, determine the resulting velocity of (a) the carriage, (b) the center of the cylinder.

A bullet weighing 0.08 lb is fired with a horizontal velocity of \(1800 \mathrm{ft}\) s into the lower end of a slender \(15-\mathrm{lb}\) bar of length \(L=30 \mathrm{in}\). Knowing that \(h=12 \mathrm{in}\), and that the bar is initially at rest, determine \((a)\) the angular velocity of the bar immediately after the bullet becomes embedded, (b) the impulsive reaction at \(C,\) assuming that the bullet becomes embedded in \(0.001 \mathrm{s}\).
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