/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 84 The ball has a mass \(m\) and is... [FREE SOLUTION] | 91Ó°ÊÓ

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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
The angle \( \theta \) that the string makes with the vertical, as the ball travels around a circular path satisfies the equation \( \tan \theta \sin \theta = \frac{v_{0}^{2}}{gl} \) for the motion of the pendulum.

Step by step solution

01

Identifying Forces

Draw a free body diagram for the moving ball. Label the forces acting on the ball i.e., gravity acting vertically downward and the tension in the string acting along the direction of the string. The tension can be split into two components, one vertical and one horizontal. The vertical part of the tension balances the ball's weight, and the horizontal part provides the centripetal force necessary for the circular motion.
02

Setting up the equations

Set up the equations of motion. The vertical part can be set up from the forces along the vertical direction, i.e., \(T \cos \theta = mg\), where T is the total tension. The horizontal component can be set up from the forces along the horizontal direction, i.e., \(T \sin \theta = m \frac{v_{0}^{2}}{l}\). Dividing the horizontal component by the vertical component gives \( \tan \theta = \frac{v_{0}^{2}}{gl} \)
03

Finding \( \sin \theta \)

Express \( \sin \theta \) from the equation found above, \( \sin \theta = \frac{v_{0}}{\sqrt{gl}} \)
04

Finding \( \tan \theta \sin \theta \)

Multiply the equations found in Step 2 and 3 to find \( \tan \theta \sin \theta \) which results in \( \tan \theta \sin \theta = \frac{v_{0}^{2}}{gl} \)

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Key Concepts

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

Circular Motion
In circular motion, an object moves along a circular path due to a central force. This type of motion is crucial to understand when discussing objects tied to a fixed point and moving in circles, like the ball attached to a cord in this exercise. Unlike linear motion, circular motion continuously changes the direction of the velocity vector, even if the speed remains constant. This continual change in direction requires a force, known as the centripetal force, to act towards the center of the circle. Without this force, the object would move in a straight line due to inertia. Key aspects of circular motion include:
  • Constant speed but changing velocity.
  • Acceleration directed towards the center of the circular path.
  • Dependent on both the object's speed and radius of the circle.
Understanding these principles helps in tackling problems where objects move in a circular path, ensuring a robust grasp of dynamics and motion.
Free Body Diagram
A free body diagram is a graphical representation used to visualize the forces acting on an object. It helps in breaking down the forces into components, making it easier to solve dynamics problems. Let's consider the ball: when it is in motion, there are typically two forces at play - the gravitational force pulling it downwards and the tension in the cord pulling it towards the center of the circular path. By identifying these forces on a free body diagram, you can:
- Clarify each force's direction and magnitude.
- Simplify complex interactions into single points of analysis.
- Understand how forces contribute to the object's motion.
For circular motion, a free body diagram typically includes:
  • The object's weight, acting downwards.
  • The cord's tension, often partitioned into horizontal and vertical components.
This breakdown is essential for setting up equations that describe the behavior of the system.
Centripetal Force
Centripetal force is the key force that keeps an object moving in circular motion. Its role is to continuously change the direction of the object's velocity, ensuring it follows a circular path. In our exercise, the horizontal component of the tension in the cord is the centripetal force. It's important to note that while the force acts towards the circle's center, it doesn't change the speed of the object, only its direction. Characteristics of centripetal force include:
- Acting perpendicular to the object's velocity.
- Dependent on the object's mass, speed, and the circle's radius.
- Calculated using the formula: \( F_c = m \cdot \frac{v^2}{r} \)
In the exercise, where \( m \) is the mass of the ball, \( v \) is its constant speed, and \( r \) is the radius of the path determined by the cord's length. Recognizing the role of centripetal force is crucial for understanding how objects maintain circular motion.
Equations of Motion
Equations of motion are mathematical expressions that describe the movement of objects under the influence of various forces. In this exercise, we analyze vertical and horizontal forces to derive relations that define the ball's motion. The tension in the string, divided into components, leads to the key equations:
- The vertical component: \( T \cos \theta = mg \), balances the gravitational force.
- The horizontal component: \( T \sin \theta = m \frac{v_{0}^{2}}{l} \), provides the centripetal force.
By solving these equations, we gain insights into the motion's dynamics. Dividing the horizontal equation by the vertical one gives the relation \( \tan \theta = \frac{v_{0}^{2}}{gl} \), linking the angle \( \theta \) to initial velocity \( v_0 \), gravity \( g \), and cord length \( l \). These relations allow predicting behaviors such as angle changes with varying speeds. Exploring equations of motion not only helps solve this particular problem but also enhances comprehension of dynamics in broader contexts.

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Most popular questions from this chapter

The vehicle is designed to combine the feel of a motorcycle with the comfort and safety of an automobile. If the vehicle is traveling at a constant speed of \(80 \mathrm{~km} / \mathrm{h}\) along a circular curved road of radius \(100 \mathrm{~m}\), determine the tilt angle \(\theta\) of the vehicle so that only a normal force from the seat acts on the driver. Neglect the size of the driver.

The \(400-\mathrm{kg}\) mine car is hoisted up the incline using the cable and motor \(M\). For a short time, the force in the cable is \(F=\left(3200 t^{2}\right) \mathrm{N}\) where \(t\) is in seconds. If the car has an initial velocity \(v_{1}=2 \mathrm{~m} / \mathrm{s}\) at \(s=0\) and \(t=0\), determine the distance it moves up the plane when \(t=2 \mathrm{~s}\).

The tractor is used to lift the \(150-\mathrm{kg}\) load \(B\) with the 24-m-long rope, boom, and pulley system. If the tractor travels to the right at a constant speed of \(4 \mathrm{~m} / \mathrm{s}\), determine the tension in the rope when \(s_{A}=5 \mathrm{~m}\). When \(s_{A}=0, s_{B}=0\).

The tube rotates in the horizontal plane at a constant rate of \(\dot{\theta}=4 \mathrm{rad} / \mathrm{s}\). If a \(0.2-\mathrm{kg}\) ball \(B\) starts at the origin \(O\) with an initial radial velocity \(\dot{r}=1.5 \mathrm{~m} / \mathrm{s}\) and moves outward through the tube, determine the radial and transverse components of the ball's velocity at the instant it leaves the outer end at \(C, r=0.5 \mathrm{~m}\). Hint: Show that the equation of motion in the \(r\) direction is \(\ddot{r}-16 r=0\). The solution is of the form \(r=A e^{-4 t}+B e^{4 t} .\) Evaluate the integration constants \(A\) and \(B,\) and determine the time \(t\) when \(r=0.5 \mathrm{~m}\). Proceed to obtain \(v_{r}\) and \(v_{\theta}\).

The conveyor belt delivers each \(12-\mathrm{kg}\) crate to the ramp at \(A\) such that the crate's speed is \(v_{A}=2.5 \mathrm{~m} / \mathrm{s}\) directed down along the ramp. If the coefficient of kinetic friction between each crate and the ramp is \(\mu_{k}=0.3\), determine the smallest incline \(\theta\) of the ramp so that the crates will slide off and fall into the cart.

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