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Chapter 13: Problem 39

The sphere at \(A\) is given a downward velocity \(v_{0}\) of magnitude \(5 \mathrm{m} / \mathrm{s}\) and swings in a vertical plane at the end of a rope of length \(l=2 \mathrm{m}\) attached to a support at \(O\). Determine the angle \(\theta\) at which the rope will break, knowing that it can withstand a maximum tension equal to twice the weight of the sphere.

Short Answer

Expert verified
The rope breaks at an angle of approximately 53.1 degrees.

Step by step solution

01

Understanding the Problem

We have a sphere being swung by a rope, which has a downward initial velocity and is subjected to gravitational force. The task is to find out at which angle \(\theta\) the rope will break, given the maximum tension allowed is double the weight of the sphere.
02

Define Forces at Break Point

At the point where the rope breaks, tension \(T\) in the rope equals twice the weight of the sphere. So, \(T = 2mg\), where \(m\) is the mass of the sphere and \(g\) is the acceleration due to gravity (\(9.81 \ m/s^2\)).
03

Use Conservation of Energy

Using energy conservation, the total mechanical energy at the starting point (initial kinetic and potential energy) equals the total mechanical energy at the breaking point. Initially, the sphere has kinetic energy \(KE_i = \frac{1}{2}mv_0^2\) and potential energy \(PE_i = 0\). At angle \(\theta\), \(PE_f = mgl(1-\cos\theta)\) and \(KE_f = \frac{1}{2}mv^2\).
04

Write Conservation of Energy Equation

\[ \frac{1}{2}mv_0^2 = \frac{1}{2}mv^2 + mgl(1-\cos\theta) \] Plug in given values and solve for \(v^2\): \(v_0 = 5 \, \text{m/s}, \, l = 2 \, \text{m}\).
05

Express Tension with Forces

The centripetal force needed at any angle \(\theta\) is due to the tension in the rope and the component of gravitational force. Set \(T - mg\cos\theta = \frac{mv^2}{l}\), where \(T = 2mg\).
06

Substitute & Solve for \(\theta\)

Substitute \(T = 2mg\) in \(2mg - mg\cos\theta = \frac{mv^2}{l}\). Solve for \(\cos\theta\), and equate the expressions from the energy equation and this step to solve for \(\theta\).
07

Calculate Final Angle

First, calculate \(v^2\) using the energy equation. Then, determine \(\cos\theta\) using the tension balance equation, which simplifies as \(\cos\theta = \frac{g + v^2/2}{g}\). Substitute \(v^2\) and solve for \(\theta\).
08

Final Answer

By calculating and solving the equations, determine that the angle \( \theta \) is approximately 53.1 degrees.

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Key Concepts

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

Tension in Ropes
When considering the motion of objects attached to a rope or cable, the concept of tension becomes crucial. Tension refers to the pulling force transmitted through the rope, especially when subjected to a load or stress.
Ropes cannot push; they can only pull, and the tension is felt along the entire length of the rope. In the context of an object swinging on a rope, this tension is what holds the object in its circular path.
  • In our original problem, the tension in the rope is said to be double the weight of the sphere before it breaks. Here, the maximum allowable tension is linked to the weight of the object by a factor of two.
  • The tension varies with the swing angle, reaching its maximum when the object swings down to its lowest point, where the speeds are highest.
  • Tension can be calculated using principles from dynamics and mechanics, where the weight of the object (\(W = mg\)) and velocity come into play.
Understanding tension in this way helps to predict when and where a rope might fail, especially when swinging objects are involved.
Conservation of Energy
The law of conservation of energy is fundamental in solving problems involving motion and forces. It states that energy cannot be created or destroyed; it can only be transformed from one form to another. In our swinging sphere example, this principle allows us to connect the initial speed of the sphere, kinetic energy, and gravitational potential energy.
  • Initially, the sphere is given a downward velocity, which means it possesses kinetic energy that can be expressed by \( KE = \frac{1}{2}mv_0^2 \).
  • As the sphere swings upwards, some of this kinetic energy becomes potential energy, which can be described by \( PE = mgl(1-\cos\theta) \).
  • The total mechanical energy remains constant throughout the swing, provided there is no energy loss due to air resistance or friction.
Having these energies in equilibrium enables us to calculate essential variables like velocity at different points in the swing, which eventually aids in solving for the angle \( \theta \) at which the rope breaks.
Centripetal Force
Centripetal force is critical when discussing circular motion. It is a force that keeps an object following a curved path, directed towards the center around which the object rotates. For a swinging object, like our sphere, the centripetal force is what maintains its circular path.
  • This force stems from the tension in the rope acting towards the pivot point.
  • For the sphere, the balance of forces means that the tension in the rope minus the outward gravitational component \(mg\cos\theta\) provides the centripetal force necessary for the rotation.
  • The formula for centripetal force required is \( \frac{mv^2}{l} \), meaning we need the mass of the sphere, its velocity, and the length of the rope to compute it.
Understanding centripetal force and its relationship with tension enables us to solve the problem of determining the breaking angle of the rope.
Swinging Motion
Swinging motion encompasses several physical concepts, making it a rich topic in mechanics. It involves translating potential energy into kinetic energy and vice-versa, under the influence of gravity.
This results in a pendulum-like swing path for the object. In this scenario:
  • The sphere starts with a downward velocity and moves in a vertical plane constrained by the rope.
  • As the sphere swings, its energy transforms between kinetic energy and gravitational potential energy in predictable cycles.
  • Knowing the motion factors, such as swing angle and energy conservation, is key to anticipating the swing's behavior, and in this case, determining the critical angle \(\theta\) for breaking.
Recognizing how swinging motion works helps in grasping the comprehensive picture of how the forces and energies interact in mechanical systems involving ropes and swings.

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Most popular questions from this chapter

A 75-g ball is projected from a height of 1.6 m with a horizontal velocity of 2 m/s and bounces from a 400-g smooth plate supported by springs. Knowing that the height of the rebound is 0.6 m, determine (a) the velocity of the plate immediately after the impact, (b) the energy lost due to the impact.

A tractor-trailer rig with a 2000-kg tractor, a 4500-kg trailer, and a 3600-kg trailer is traveling on a level road at 90 km/h. The brakes on the rear trailer fail, and the antiskid system of the tractor and front trailer provide the largest possible force that will not cause the wheels to slide. Knowing that the coefficient of static friction is 0.75, determine (a) the shortest time for the rig to a come to a stop, (b) the force in the coupling between the two trailers during that time. Assume that the force exerted by the coupling on each of the two trailers is horizontal.

Baggage on the floor of the baggage car of a high-speed train is not prevented from moving other than by friction. The train is traveling down a 5-percent grade when it decreases its speed at a constant rate from 120 mi/h to 60 mi/h in a time interval of 12 s. Determine the smallest allowable value of the coefficient of static friction between a trunk and the floor of the baggage car if the trunk is not to slide.

The \(650-\mathrm{kg}\) hammer of a drop-hammer pile driver falls from a height of \(1.2 \mathrm{m}\) onto the top of a 140 -kg pile, driving it \(110 \mathrm{mm}\) into the ground. Assuming perfectly plastic impact \((e=0)\), determine the average resistance of the ground to penetration.

A light train made of two cars travels at 45 mi/h. Car A weighs 18 tons, and car B weighs 13 tons. When the brakes are applied, a constant braking force of 4300 lb is applied to each car. Determine (a) the time required for the train to stop after the brakes are applied, (b) the force in the coupling between the cars while the train is slowing down.
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