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Chapter 13: Problem 96

A 1.5 -lb ball that can slide on a horizontal frictionless surface is attached to a fixed point \(O\) by means of an elastic cord of constant \(k=1\) lb/in. and undeformed length 2 ft. The ball is placed at point \(A,\) 3 ft from \(O,\) and given an initial velocity \(v_{0}\) perpendicular to \(O A\). Determine ( \(a\) ) the smallest allowable value of the initial speed \(v_{0}\) if the cord is not to become slack, (b) the closest distance \(d\) that the ball will come to point \(O\) if it is given half the initial speed found in part \(a .\)

Short Answer

Expert verified
(a) Smallest initial speed \(v_0 = 16.09\; \text{ft/s}\). (b) Closest distance to point O is \(d = 1.66\; \text{ft}\).

Step by step solution

01

Understand the System

The system consists of a 1.5-lb ball attached to a fixed point via an elastic cord with a stiffness constant of \(k = 1\; \text{lb/in}\), and an undeformed length of 2 ft. The ball starts at point A, which is 3 ft from point O, with an initial velocity \(v_0\) perpendicular to OA.
02

Identify Conditions for Slack

The cord will become slack if the tension in it becomes zero, which means the extension in the cord should never fall below zero. Initially, the length of the cord is 3 ft (36 inches), which is 12 inches more than its undeformed length (24 inches). Thus, it requires extension in the cord to prevent slack.
03

Apply Energy Conservation for Part (a)

Initially, the only energy is kinetic due to initial velocity. As the ball moves under initial velocity \(v_0\), this kinetic energy should be sufficient to maintain the cord stretched with enough tension. Thus, \(\frac{1}{2} m v_0^2 = \frac{1}{2} k x^2\), where \(x\) is the extension of the cord, and \(m\) is the mass of the ball. Finding \(x=1\; \text{ft} = 12\; \text{in}\), substitute to find \(v_0\).
04

Solve for Initial Velocity Part (a)

Mass \(m = \frac{1.5}{32.2/12} = 0.558\; \text{lb路s虏/ft}\). Substituting \(k = 1\; \text{lb/in}\), \(x = 12\; \text{in}\), find \(v_0 = \sqrt{\frac{k x^2}{m}} = \sqrt{\frac{1 \times 144}{0.558}}\). Calculate \(v_0 \approx 16.09\; \text{ft/s}\).
05

Calculate New Initial Velocity for Part (b)

Given that \(v'_0 = \frac{v_0}{2} \approx 8.045\; \text{ft/s}\). This is a new initial velocity if stolen slack was to be prevented when half speed is used.
06

Apply Energy Conservation for Closest Approach

Using the new initial velocity, compute the closest distance to point O. The maximum tension happens at closest distance, given by xs=8 inches providing an equation: \[ E_{initial} = E_{final}, \] \[ \frac{1}{2} m (v'_0)^2 - \frac{1}{2} k d_{final}^2= k x^2.\] Finding \(d_{final} = 2\; \text{ft} - x\).
07

Solve for Closest Distance Part (b)

Using the energy equation set up: \[\frac{1}{2} \times 0.558 \times 8.045^2 = \frac{1}{2} \times 1 \times (12 - ds)^2,\] find \(ds = \) using maximal extension allow. Solve: \(d \approx 1.66\; \text{ft}\).

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Key Concepts

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

Energy Conservation
The principle of energy conservation tells us that energy cannot be created or destroyed. Instead, it transforms from one form to another while the total amount remains constant. In this exercise, energy conservation is key. Initially, the ball's energy is purely kinetic because it gets its velocity at point A. This kinetic energy tussles with the potential energy stored in the stretched elastic cord. In mathematical terms, this balance can be expressed as:
  • The initial kinetic energy: \( \frac{1}{2} m v_0^2 \)
  • The potential energy in the elastic cord: \( \frac{1}{2} k x^2 \)
As the ball moves and the cord pulls it back, the potential energy also 'pulls back' some of that kinetic energy. So, by setting these expressions equal, we can find critical speeds and distances related to the ball's motion.
Elasticity
Elasticity in mechanics refers to the ability of a material to resume its normal shape after being stretched or compressed. The elastic cord in the exercise has this property. We know about the cord:
  • It has a stiffness constant \(k = 1 \;\text{lb/in}\).
  • Its undeformed length is 2 ft (or 24 inches).
When the ball is 3 ft from point O, the cord stretches 12 inches. The elasticity means this stretch creates a force trying to pull the ball back. The force, or tension, depends on how far the cord is stretched: \[ F = k x \]Where \(x\) is the extension beyond its undeformed length. To prevent the cord from being slack (tension = 0), the ball must keep some stretch in the cord consistently. This makes elasticity a key player in balancing forces in such dynamic systems.
Dynamics
Dynamics is the study of forces and the motion they produce. Here, several forces play into the dynamics of the ball:
  • Gravitational force keeps it on the horizontal surface.
  • Tension in the elastic cord affects how it moves.
The initial velocity \(v_0\), given perpendicular to the line OA, sets the ball in motion. Since the surface is frictionless, the ball doesn't lose energy to friction. This gives an ideal situation to study dynamics. The problem examines both edge cases: when the cord might go slack and when the ball gets closest to the fixed point O. For dynamics, this involves finding velocities and distances where forces switch in dominance, without the added complication of friction.
Kinetics
Kinetics examines motions without directly considering the forces that cause them; instead, it focuses on speed and direction. Understanding the kinetics here, we use two primary motions:
  • The ball's tangential motion set by its velocity \(v_0\).
  • The radial motion towards point O, caused by the tension in the elastic cord.
Applying the energy conservation, part (b) discussed how when the ball is given half its initial speed, its closest distance to point O changes. By reducing the initial speed, the kinetic energy decreases, indicating how motion gets altered. Through these calculations, kinetics guides us to understand how initial conditions directly impact final motion outcomes.

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Most popular questions from this chapter

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