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Chapter 11: Problem 80

During a manufacturing process, a conveyor belt starts from rest and travels a total of \(1.2 \mathrm{ft}\) befors temporarily coming to rest. Knowing that the jerk, or rate of change of acceleration, is limited to \(\pm 4.8 \mathrm{ft} / \mathrm{s}^{2}\) per second, determine (a) the shortest time required for the belt to move \(1.2 \mathrm{ft}\), (b) the maximum and average values of the velocity of the belt during that time.

Short Answer

Expert verified
The shortest time required is the duration of the acceleration profiles plus constant phase, calculated from jerk constraints.

Step by step solution

01

Understand the problem

The conveyor belt has a total displacement of 1.2 ft and starts and ends at rest, meaning its initial and final velocities are 0 ft/s. The jerk is limited to \( \pm 4.8 \text{ ft/s}^3 \). We need to find the shortest time for this motion and the maximum and average velocity.
02

Define motion profile

Given the initial and final conditions, the conveyor belt can use a jerk profile with three phases: acceleration increase, constant acceleration, and acceleration decrease. This symmetric jerk pattern helps in minimizing the time.
03

Determine acceleration profile duration

The problem allows jerk, \( j = \pm 4.8 \text{ ft/s}^3 \). Let's assume jerk lasts for time \( t_1 \) at inception and deceleration. Therefore, the peak acceleration \( a = j \cdot t_1 \), and the duration without jerk (constant acceleration phase) requires twice the jerk duration for total symmetry.
04

Calculate expressions for velocity and displacement

Using kinematic relations:\[ v = \,\int a \, dt\ ,\, x = \,\int v \, dt \]. \( v = \frac{1}{2} j t_1^2 \) after the jerk phase and \( x = \int ( \int j \, dt ) \, dt \footnotemark \), compute displacement equations for the phases.
05

Check boundary conditions and solve for minimal time

With symmetric jerk pattern, total area under acceleration (displacement curve) equals 1.2 ft. Set up equations for displacement: \( \frac{2}{3} jt_1^3 + aT = 1.2 \), and solve these with velocity continuity to find \( t_1 \). Then determine minimal time \( T + 2t_1 \).
06

Calculate velocity values

Compute max velocity reached using definite integration over jerk: \( v_{max} = jt_1^2 \). Use \( v_{average} = \frac{1.2}{T} \) for average velocity.

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Key Concepts

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

Acceleration
Acceleration is a key factor in understanding motion. It describes how quickly the velocity of an object changes over time. This change in velocity can be due to changes in speed or direction. Acceleration is usually measured in meters per second squared \(\text{m/s}^2\). For example, when a car speeds up when the light turns green, it's accelerating. When it slows down to stop, it's also accelerating, but in a negative direction.
  • Positive acceleration: Speeding up or moving faster in a given direction.
  • Negative acceleration (or deceleration): Slowing down.
In the context of the exercise, the belt starts from rest with an acceleration that depends on the jerk during different phases.
Velocity
Velocity is a vector quantity that refers to the speed of an object in a particular direction. Unlike speed, which is a scalar and only tells you how fast something is moving, velocity provides information about the direction of the motion as well.
  • Example: Driving north at 60 mph is a velocity because it specifies both speed and direction.
  • Velocity can change because of acceleration.
In the exercise, since the belt starts and ends at rest, the changes in velocity throughout the movement are due to the acceleration phases.
Kinematic Equations
Kinematic equations are used to describe the motion of objects. They relate the quantities of motion: displacement, initial velocity, final velocity, acceleration, and time. These equations help predict one quantity if other values are known.The basic kinematic equations include:
  • \( v = u + at \) - Final velocity after time \( t \).
  • \( s = ut + \frac{1}{2}at^2 \) - Displacement after time \( t \).
In the exercise, these equations help determine the displacement during each phase of movement for the conveyor belt, allowing us to find the shortest time necessary.
Jerk
Jerk is the rate at which acceleration changes. It's a derivative of acceleration and can be thought of as the "acceleration of acceleration." In physics, too much jerk can cause mechanical systems to wear out more quickly or lead to discomfort in vehicles.In the exercise, the constraint on jerk (\(\pm 4.8 \text{ ft/s}^3\)) ensures that the conveyor belt's acceleration changes smoothly. It allows for minimal wear and tear on the belt system while achieving the desired motion from rest to rest efficiently.
Understanding jerk is particularly crucial in applications requiring precise motion control, such as in manufacturing or robotics systems.

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Most popular questions from this chapter

The diameter of the eye of a stationary hurricane is \(20 \mathrm{mi}\) and the maximum wind speed is \(100 \mathrm{mi} / \mathrm{h}\) at the eye wall with \(r=10 \mathrm{mi}\). Assuming that the wind speed is constant for constant \(r\) and decreases uniformly with increasing \(r\) to \(40 \mathrm{mi} / \mathrm{h}\) at \(r=110 \mathrm{mi}\), determine the magnitude of the acceleration of the air at \((a) r=10 \mathrm{mi},(b) r=60 \mathrm{mi},\) (c) \(r=110 \mathrm{mi} .\)

An elevator is moving upward at a constant speed of \(4 \mathrm{m} / \mathrm{s}\). A man standing \(10 \mathrm{m}\) above the top of the elevator throws a ball upward with a speed of \(3 \mathrm{m} / \mathrm{s}\). Determine \((a)\) when the ball will hit the elevator, \((b)\) where the ball will hit the elevator with respect to the location of the man.

Conveyor belt \(A,\) which forms a \(20^{\circ}\) angle with the horizontal, moves at a constant speed of \(4 \mathrm{ft} / \mathrm{s}\) and is used to load an airplane. Knowing that a worker tosses duffel bag \(B\) with an initial velocity of \(2.5 \mathrm{ft} / \mathrm{s}\) at an angle of \(30^{\circ}\) with the horizontal, determine the velocity of the bag relative to the belt as it lands on the belt.

A particle moves in a straight line with a constant acceleration of \(-4\) ftls \(^{2}\) for 6 s, zero acceleration for the next 4 s, and a constant acceleration of \(+4 \mathrm{ft} / \mathrm{s}^{2}\) for the next \(4 \mathrm{s}\). Knowing that the particle starts from the origin and that its velocity is \(-8 \mathrm{ft} / \mathrm{s}\) during the zero acceleration time interval, (a) construct the \(v-t\) and \(x-t\) curves for \(0 \leq t \leq 14 \mathrm{s},(b)\) determine the position and the velocity of the particle and the total distance traveled when \(t=14 \mathrm{s}\).

The peripheral speed of the tooth of a 10 -in-diameter circular saw blade is \(150 \mathrm{ft} / \mathrm{s}\) when the power to the saw is turned off. The speed of the tooth decreases at a constant rate, and the blade comes to rest in \(9 \mathrm{s}\). Determine the time at which the total acceleration of the tooth is \(130 \mathrm{ft} / \mathrm{s}^{2}\).
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