/*! 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 13 A Scotch yoke is a mechanism tha... [FREE SOLUTION] | 91Ó°ÊÓ

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A Scotch yoke is a mechanism that transforms the circular motion of a crank into the reciprocating motion of a shaft (or vice versa). It has been used in a number of different internal combustion engines and in control valves. In the Scotch yoke shown, the acceleration of point \(A\) is defined by the relation \(a=-1.8 \sin k t,\) where \(a\) and \(t\) are expressed in \(\mathrm{m} / \mathrm{s}^{2}\) and seconds, respectively, and \(k=3 \mathrm{rad} / \mathrm{s}\). Knowing that \(x=0\) and \(v=0.6 \mathrm{m} / \mathrm{s}\) when \(t=0,\) determine the velocity and position of point \(A\) when \(t=0.5 \mathrm{s}\)

Short Answer

Expert verified
Velocity: \( v(0.5) = 0.6 \cos(1.5) \), Position: \( x(0.5) = 0.2 \sin(1.5) \).

Step by step solution

01

Understand the Problem

We need to find the velocity and position of point \( A \) on a Scotch yoke mechanism at \( t=0.5 \text{ s} \), given that the acceleration \( a=-1.8 \sin(3t) \), position \( x=0 \), and velocity \( v=0.6 \text{ m/s} \) when \( t=0 \).
02

Integrate Acceleration to Find Velocity

To find velocity, integrate the acceleration function: \[ v(t) = \int a(t) \, dt = \int -1.8 \sin(3t) \, dt. \]The integral of \( -1.8 \sin(3t) \) is:\[ v(t) = 0.6 \cos(3t) + C_1. \]Apply the initial condition \( v(0) = 0.6 \) to find \( C_1 \):\[ 0.6 = 0.6 \cos(0) + C_1 \Rightarrow C_1 = 0. \] Thus, the velocity equation is:\[ v(t) = 0.6 \cos(3t). \]
03

Integrate Velocity to Find Position

Next, integrate the velocity function to find position:\[ x(t) = \int v(t) \, dt = \int 0.6 \cos(3t) \, dt. \]The integral of \( 0.6 \cos(3t) \) is:\[ x(t) = 0.2 \sin(3t) + C_2. \]Apply the initial condition \( x(0) = 0 \) to find \( C_2 \):\[ 0 = 0.2 \sin(0) + C_2 \Rightarrow C_2 = 0. \]Thus, the position equation is:\[ x(t) = 0.2 \sin(3t). \]
04

Calculate Velocity at t=0.5 s

Plug \( t=0.5 \text{ s} \) into the velocity equation:\[ v(0.5) = 0.6 \cos(3 \times 0.5) = 0.6 \cos(1.5). \]Calculate \( \cos(1.5) \) using a calculator to find the velocity value.
05

Calculate Position at t=0.5 s

Plug \( t=0.5 \text{ s} \) into the position equation:\[ x(0.5) = 0.2 \sin(3 \times 0.5) = 0.2 \sin(1.5). \]Calculate \( \sin(1.5) \) using a calculator to find the position value.

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Key Concepts

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

Kinematics
Kinematics is the branch of mechanics that focuses on the movement of objects without considering the forces that cause this movement. It deals with understanding how positions, velocities, and accelerations change over time. In the context of a Scotch yoke mechanism, kinematics is essential because it helps us determine how the position and velocity of point A change as the mechanism operates.

In the exercise, the given parameters are the initial conditions of position and velocity at time zero, with the acceleration as a function of time. This setup is typical for kinematics problems where the primary goal is to trace the motion of a point through time using fundamental concepts.
  • We start with known initial conditions, such as the initial position and velocity.
  • We use the acceleration function to determine how these initial conditions change over time.
With kinematics, the tools are mathematics and calculus, which allow us to see that even without external forces, we can predict future states of motion.
Integration of Functions
Integration is a mathematical process that helps us find quantities like velocity and position from acceleration. When we are given acceleration as a function of time, we can use integration to uncover the change in velocity, and further integration will provide the change in position.

In our exercise, we began by integrating the function for acceleration to find the velocity. The process of integration reversed the differentiation that, in physics, typically brings these relationships together:
  • First integral of acceleration gives velocity: \[ v(t) = \int -1.8 \sin(3t) \, dt. \]
  • Second integral of velocity provides position:\[ x(t) = \int 0.6 \cos(3t) \, dt. \]
Using integration constants, which we find using initial conditions at a specific time (in this case, time zero), we complete a picture of function over time. This helps to create precise models of movement for mechanics like the Scotch yoke.
Acceleration and Velocity Relationship
Acceleration and velocity are closely linked through calculus. Acceleration is the rate of change of velocity over time. When we want to determine velocity from a known acceleration function, we integrate.

For the Scotch yoke exercise, we understand that:
  • Velocity is the integral of acceleration: the formula \( v(t) = 0.6 \cos(3t) \) results from integrating. It describes how velocity varies with respect to time.
  • Sometimes initial velocity affects the constant term from integration, as we saw when we derived \( v(0) = 0.6 \).
Similarly, we understand position as the integral of velocity, telling us specifically how a point like A moves spatially over time. The time component helps create a full motion narrative for the point involved.
Mechanical Engineering Concepts
The Scotch yoke mechanism is one among many mechanical systems used to convert rotational motion into reciprocating motion. Understanding how such mechanisms work is fundamental in mechanical engineering, where predicting movement and function is key to design and analysis.

Mechanical engineering involves incorporating physics and mathematics to predict how mechanisms like the Scotch yoke will behave over time. It looks into:
  • The dynamics of movement, which include how different forces and motions interact.
  • The role of components like cranks and yokes in influencing motion and functionality.
In practical terms, the ability to analyze mechanisms like the Scotch yoke using these principles allows engineers to design efficient, functional machines and devices that work based on reliable predictions of motion and force interactions.

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Most popular questions from this chapter

A snowboarder starts from rest at the top of a double black diamond hill. As she rides down the slope, GPS coordinates are used to determine her displacement as a function of time: \(x=0.5 t^{3}+t^{2}+2 t\) where \(x\) and \(t\) are expressed in feet and seconds, respectively. Determine the position, velocity, and acceleration of the boarder when \(t=5\) seconds.

A commuter train traveling at \(40 \mathrm{mi} / \mathrm{h}\) is \(3 \mathrm{mi}\) from a station. The train then decelerates so that its speed is \(20 \mathrm{mi} / \mathrm{h}\) when it is \(0.5 \mathrm{mi}\) from the station. Knowing that the train arrives at the station \(7.5 \mathrm{min}\) after beginning to decelerate and assuming constant decelerations, determine (a) the time required for the train to travel the first \(2.5 \mathrm{mi}\), \((b)\) the speed of the train as it arrives at the station, ( \(c\) ) the final constant deceleration of the train.

Car \(A\) is traveling at 40 milh when it enters a 30 milh speed zone. The driver of car \(A\) decelerates at a rate of \(16 \mathrm{ft} / \mathrm{s}^{2}\) until reaching a speed of \(30 \mathrm{mil}\), which she then maintains. When car \(B\), which was initially \(60 \mathrm{ft}\) behind car \(A\) and traveling at a constant speed of \(45 \mathrm{mil}\), enters the speed zone, its driver decelerates at a rate of \(20 \mathrm{ft}\) s " until reaching a speed of \(28 \mathrm{mi} / \mathrm{h}\). Knowing that the driver of car \(B\) maintains a speed of \(28 \mathrm{mi} / \mathrm{h}\), determine \((a)\) the closest that car \(B\) comes to car \(A,(b)\) the time at which car \(A\) is \(70 \mathrm{ft}\) in front of car \(B .\)

A brass (nonmagnetic) block \(A\) and a steel magnet \(B\) are in equilibrium in a brass tube under the magnetic repelling force of another steel magnet \(C\) located at a distance \(x=0.004 \mathrm{m}\) from \(B\). The force is inversely proportional to the square of the distance between \(B\) and \(C .\) If block \(A\) is suddenly removed, the acceleration of block \(B\) is \(a=-9.81+k / x^{2},\) where \(a\) and \(x\) are expressed in \(\mathrm{m} / \mathrm{s}^{2}\) and meters, respectively, and \(k=4 \times 10^{-4} \mathrm{m}^{3} / \mathrm{s}^{2} .\) Determine the maximum velocity and acceleration of \(B .\)

Steep safety ramps are built beside mountain highways to enable vehicles with defective brakes to stop safely. A truck enters a 750 -ft ramp at a high speed \(v_{0}\) and travels \(540 \mathrm{ft}\) in \(6 \mathrm{s}\) at constant deceleration before its speed is reduced to \(v_{0} / 2 .\) Assuming the same constant deceleration, determine ( \(a\) ) the additional time required for the truck to stop \((b)\) the additional distance traveled by the truck.

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