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Chapter 13: Problem 83

The ball has a mass \(m\) and is attached to the cord of length \(l\). The cord is tied at the top to a swivel and the ball is given a velocity \(\mathbf{v}_{0} .\) Show that the angle \(\theta\) which the cord makes with the vertical as the ball travels around the circular path must satisfy the equation \(\tan \theta \sin \theta=v_{0}^{2} / g l\) Neglect air resistance and the size of the ball.

Short Answer

Expert verified
By considering the forces of tension and gravity acting on the ball, and ensuring their balance for the vertical components and match for the horizontal components (which is responsible for the circular motion), the equation \(\tan(\theta)\sin(\theta)=v_{0}^{2}/gl\) is derived, which is used to calculate the angle \(\theta\).

Step by step solution

01

Identify the Forces

First, you need to identify the forces acting on the ball. These are the tension \(T\) in the cord and the pull of gravity \(mg\), where \(m\) is the mass of the ball and \(g\) is the acceleration due to gravity.
02

Decompose the Forces

Once the acting forces are identified, they need to be decomposed into horizontal and vertical components. The vertical component of the tension in the cord is \(T cos(\theta)\) and the horizontal component is \(T sin(\theta)\). For the gravitational force, the vertical component is \(mg\). There is no horizontal component as gravity acts vertically downwards.
03

Balance the Forces

Since the ball is in a state of motion (circular path) and not falling, the vertical components of the forces should balance each other. Thus you set \(T cos(\theta) = mg\).
04

Centripetal Force Expression

The circular motion also enables us to say that the horizontal component of tension provides the necessary centripetal force for circular motion which is given by the equation \(T sin(\theta)=mv_{0}^{2} / l\).
05

Substitution and Solving

Next, you can substitute \(T\) from Step 3 into the equation from Step 4 which results in \(mg sin(\theta) / cos(\theta)=mv_{0}^{2} / l\). Solving for \(\theta\), you find \(tan(\theta) sin(\theta) = v_{0}^{2} / gl\). Thus proving that the angle the cord makes with the vertical must satisfy the equation given in the exercise.

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Key Concepts

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

Centripetal Force
Centripetal force is the invisible hand that ensures any object moving in a circular path remains on that path. Imagine tying a ball to a string and whirling it around; it's the force that tugs the ball inward, preventing it from flying off tangentially.

In mathematics, we express this as the force that equals mass times the velocity squared, divided by the radius of the circle (\( F_{c} = \frac{mv^{2}}{r} \)). In our exercise involving a swinging ball, this force is provided by the tension in the cord, which constantly pulls the ball towards the center of the circular path. But this isn't just a phenomenon for strings and balls鈥攃entripetal force is at play in car tires gripping a roundabout, planets orbiting the Sun, and even in the spin cycle of your washing machine!
Tension in a Cord
Next, let's knit together the concept of tension. Tension is a pulling force that acts along the length of a cord, string, or any material that can be pulled taut. Its role is crucial in scenarios where objects are suspended or swung around, like our ball on a cord.

In our exercise, tension does a dual dance. Firstly, it counters the downward tug of gravity, ensuring our ball doesn't plummet. Secondly, it provides the centripetal force needed to keep the ball on its circular path.

Decomposing Tension

Break it down, and you'll find that tension has two main characters鈥攙ertical and horizontal components. Vertical tension has one job: to balance the weight of the ball. Horizontal tension takes the lead in supplying the centripetal force. The beauty of physics lies in how these components interact, imposing a harmony of forces that allows the ball to hover at an angle, defying a dire descent.
Dynamics of Rigid Bodies
Lastly, we delve into the dynamics of rigid bodies. What are they? They're objects that don't deform under the forces applied to them鈥攊n essence, they maintain their shape and size. The ball in our exercise is considered a rigid body because it does not change shape as it swings.

Understanding the dynamics of such bodies involves looking at the forces involved and how they dictate motion. Even though our ball won't deform, it still must obey the laws of motion. It experiences gravity, tension, and it has mass鈥攁ll key players in the dynamical equation.

Balancing Act

The rigid body dynamics in our scenario require a balance between gravitational pull and the tension's vertical pull, alongside a harmony between the movement of the ball in the horizontal direction and the centripetal force needed. Physics is essentially a cosmic ballet, with each force choreographing the ball's path through space.

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Most popular questions from this chapter

Rod \(O A\) rotates counterclockwise at a constant angular rate \(\dot{\theta}=4 \mathrm{rad} / \mathrm{s} .\) The double collar \(B\) is pinconnected together such that one collar slides over the rotating rod and the other collar slides over the circular rod described by the equation \(r=(1.6 \cos \theta) \mathrm{m}\). If both collars have a mass of \(0.5 \mathrm{kg}\), determine the force which the circular rod exerts on one of the collars and the force that \(O A\) exerts on the other collar at the instant \(\theta=45^{\circ} .\) Motion is in the horizontal plane.

The cylindrical plug has a weight of 2 lb and it is free to move within the confines of the smooth pipe. The spring has a stiffness \(k=14 \mathrm{lb} / \mathrm{ft}\) and when no motion occurs the distance \(d=0.5 \mathrm{ft}\). Determine the force of the spring on the plug when the plug is at rest with respect to the pipe. The plug is traveling with a constant speed of \(15 \mathrm{ft} / \mathrm{s}\), which is caused by the rotation of the pipe about the vertical axis.

Determine the maximum constant speed at which the pilot can travel around the vertical curve having a radius of curvature \(\rho=800 \mathrm{m},\) so that he experiences a maximum acceleration \(a_{n}=8 g=78.5 \mathrm{m} / \mathrm{s}^{2} .\) If he has a mass of \(70 \mathrm{kg}\) determine the normal force he exerts on the seat of the airplane when the plane is traveling at this speed and is at its lowest point.

The conveyor belt is moving downward at \(4 \mathrm{m} / \mathrm{s}\) If the coefficient of static friction between the conveyor and the \(15-\mathrm{kg}\) package \(B\) is \(\mu_{s}=0.8,\) determine the shortest time the belt can stop so that the package does not slide on the belt.

If the 50 -kg crate starts from rest and achieves a velocity of \(v=4 \mathrm{m} / \mathrm{s}\) when it travels a distance of \(5 \mathrm{m}\) to the right, determine the magnitude of force \(\mathbf{P}\) acting on the crate. The coefficient of kinetic friction between the crate and the ground is \(\mu_{k}=0.3.\)
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