/*! 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 101 The ball of mass \(m\) is guided... [FREE SOLUTION] | 91Ó°ÊÓ

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The ball of mass \(m\) is guided along the vertical circular path \(r=2 r_{c} \cos \theta\) using the arm \(O A\). If the arm has a constant angular velocity \(\dot{\theta}_{0},\) determine the angle \(\theta \leq 45^{\circ}\) at which the ball starts to leave the surface of the semicylinder. Neglect friction and the size of the ball.

Short Answer

Expert verified
The angle \(\theta\) at which the ball starts to leave the surface of the semicylinder is \(\theta = \cos^{-1} (\frac{r \dot{\theta}^{2}}{g})\).

Step by step solution

01

Identify the Forces

The forces acting on the ball are the gravitational force \(m g \cos \theta \) acting downwards and the normal reaction force \(N\) acting upwards when the ball is on the semicylinder. In the frame of the ball, there also exists a centrifugal force \(m r \dot{\theta}^{2}\) acting upwards.
02

Apply Newton's Second Law

In the rotating frame, the ball appears at rest. Therefore, by Newton's second law, the sum of forces on the ball must be zero. That means the upwards and downwards forces must balance out.\nHence, \(m g \cos \theta = N + m r \dot{\theta}^{2}\) where \(N\) is the normal force at the contact point.
03

Determine at What Angle the Ball Starts to Leave the Surface

The ball starts to leave the surface when the normal reaction force (\(N\)) is zero. Then, the equation from Step 2 becomes: \( m g \cos \theta = m r \dot{\theta}^{2} \). Solving for \( \theta \), gives \( \theta = \cos^{-1} (\frac{r \dot{\theta}^{2}}{g}) \).

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Key Concepts

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

Centrifugal Force
In dynamics engineering mechanics, 'centrifugal force' is often discussed when analyzing objects in a rotating reference frame. Imagine you are sitting in a car rounding a curve, the force you feel pushing you to the outside of the curve is similar to what is described as centrifugal force.

However, in a strict sense, centrifugal force is not a real force; it is a perceived force that appears only within a rotating frame of reference. Specifically, when an object is in circular motion, inertia tends to maintain the object's straight-line path. As the object is instead forced to follow a curved path, this resistance to the change in motion is mistaken as an outward force—that is the centrifugal force.

In the exercise given, the ball rotates along a vertical circular path due to the arm's constant angular velocity. From the ball's perspective, there is a force that 'pushes' it outward against the semicylinder. Mathematically, the centrifugal force can be represented as \( m r \theta^2 \), where \( m \) is the mass, \( r \) is the radius from the arm's pivot to the ball, and \( \theta \) is the angular velocity. When this force exceeds the restraining forces, such as gravity or friction, the object can break away from its path, which is how the exercise determines the angle at which the ball starts to leave the semicylinder's surface.
Newton's Second Law
Newton's second law is central to understanding the motion of objects in all areas of mechanics, including dynamics engineering mechanics. It states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This relationship can be succinctly expressed by the iconic equation \( F = m \times a \), where \( F \) is the net force, \( m \) is the mass of the object, and \( a \) is the acceleration.

In our scenario with the ball, Newton's second law beautifully ties into how we describe forces in a rotating system. While the ball undergoes rotational motion, it appears to be at rest in its own rotating reference frame, yet we know it is accelerating due to the change in direction of its velocity. According to Newton's second law, this means that forces are at play. In the exercise solution, we equate the gravitational force affected by the cosine of angle \( \theta \) and the centrifugal 'force' to find the angle where the ball ceases to maintain contact with the semicylinder - effectively when the net force is zero, and the ball begins to leave the surface because the required centripetal force for circular motion is not provided anymore.
Rotational Motion
Rotational motion refers to the movement of an object around a center or an axis. It is a complex type of motion as different parts of the object can have different velocities, but they all have the same angular velocity, \( \theta \). Common examples include the rotation of planets, wheels, or any round object along an axis.

When dealing with rotational motion, one must consider the angular equivalent concepts to those in linear motion—angular velocity, angular acceleration, and moment of inertia (rotational mass). In the case of our exercise, angular velocity \( \theta \) is a pivotal variable. It is defined as the rate at which the object rotates and is typically measured in radians per second.

The exercise involves a ball guided along a vertical circular path by an arm with a constant angular velocity. The constant rotational motion affects the net forces at play on the ball, specifically the magnitude of the centrifugal force, which changes with the cosine of angle \( \theta \). This effect determines the critical point when the ball loses contact with the semicylinder, highlighting the close relationship between linear and rotational dynamics in engineering mechanics.

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Most popular questions from this chapter

The smooth block \(B\) of negligible size has a mass \(m\) and rests on the horizontal plane. If the board \(A C\) pushes on the block at an angle \(\theta\) with a constant acceleration \(\mathbf{a}_{0}\) determine the velocity of the block along the board and the distance \(s\) the block moves along the board as a function of time \(t .\) The block starts from rest when \(s=0, t=0.\)

A motorcyclist in a circus rides his motorcycle within the confines of the hollow sphere. If the coefficient of static friction between the wheels of the motorcycle and the sphere is \(\mu_{s}=0.4,\) determine the minimum speed at which he must travel if he is to ride along the wall when \(\theta=90^{\circ}\) The mass of the motorcycle and rider is \(250 \mathrm{kg}\), and the radius of curvature to the center of gravity is \(\rho=20 \mathrm{ft}\) Neglect the size of the motorcycle for the calculation.

The airplane, traveling at a constant speed of \(50 \mathrm{m} / \mathrm{s}\) is executing a horizontal turn. If the plane is banked at \(\theta=15^{\circ},\) when the pilot experiences only a normal force on the seat of the plane, determine the radius of curvature \(\rho\) of the turn. Also, what is the normal force of the seat on the pilot if he has a mass of \(70 \mathrm{kg}.\)

The particle has a mass of \(0.5 \mathrm{kg}\) and is confined to move along the smooth vertical slot due to the rotation of the arm \(O A .\) Determine the force of the rod on the particle and the normal force of the slot on the particle when \(\theta=30^{\circ} .\) The rod is rotating with a constant angular velocity \(\dot{\theta}=2 \mathrm{rad} / \mathrm{s} .\) Assume the particle contacts only one side of the slot at any instant.

The boy of mass 40 kg is sliding down the spiral slide at a constant speed such that his position, measured from the top of the chute, has components \(r=1.5 \mathrm{m}, \theta=(0.7 t) \mathrm{rad}\) and \(z=(-0.5 t) \mathrm{m},\) where \(t\) is in seconds. Determine the components of force \(\mathbf{F}_{r}, \mathbf{F}_{\theta},\) and \(\mathbf{F}_{z}\) which the slide exerts on him at the instant \(t=2\) s. Neglect the size of the boy.

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