6 Engineering Example 4

6.1 Crank used to drive a piston

Introduction

A crank is used to drive a piston as in Figure 20.

Figure 20 :

{ Crank used to drive a piston}

Problem

The angular velocity of the crankshaft is the rate of change of the angle [maths rendering] The piston moves horizontally with velocity [maths rendering] and acceleration [maths rendering] ; [maths rendering] is the length of the crank and [maths rendering] is the length of the connecting rod. The crankpin performs circular motion with a velocity of [maths rendering] and centripetal acceleration of [maths rendering] . The acceleration [maths rendering] of the piston varies with [maths rendering] and is related by

[maths rendering]

Find the maximum and minimum values of the acceleration [maths rendering] when [maths rendering] mm and [maths rendering] mm.

Mathematical statement of the problem

We need to find the stationary values of [maths rendering] when [maths rendering] mm and [maths rendering] mm. We do this by solving [maths rendering] and then analysing the stationary points to decide whether they are a maximum, minimum or point of inflexion.

Mathematical analysis .

[maths rendering] so [maths rendering] .

To find the maximum and minimum acceleration we need to solve

[maths rendering]

[maths rendering]

[maths rendering]

[maths rendering] and as [maths rendering] mm and [maths rendering] mm

[maths rendering] or [maths rendering]

CASE 1: [maths rendering]

If [maths rendering] then [maths rendering] or [maths rendering] . If [maths rendering] then [maths rendering]

so [maths rendering]

If [maths rendering] then [maths rendering] so

[maths rendering]

In order to classify the stationary points, we differentiate [maths rendering] with respect to [maths rendering] to find the second derivative:

[maths rendering]

At [maths rendering] we get [maths rendering] which is negative.

So [maths rendering] gives a maximum value and [maths rendering] is the value at the maximum.

At [maths rendering] we get [maths rendering] which is negative.

So [maths rendering] gives a maximum value and [maths rendering]

CASE 2: [maths rendering]

If [maths rendering] then [maths rendering] so [maths rendering] .

[maths rendering]

At [maths rendering] we get [maths rendering] which is positive.

So [maths rendering] gives a minimum value and [maths rendering]

Thus the values of [maths rendering] at the stationary points are:-

[maths rendering] (maximum), [maths rendering] (maximum) and [maths rendering] (minimum).

So the overall maximum value is [maths rendering] and the minimum value is

[maths rendering] where we have substituted [maths rendering] mm ( [maths rendering] 0.15 m) and [maths rendering] mm ( [maths rendering] m).

Interpretation

The maximum acceleration occurs when [maths rendering] and [maths rendering] .

The minimum acceleration occurs when [maths rendering] and [maths rendering] .

Exercises
  1. Locate the stationary points of the following functions and distinguish among them as maxima, minima and points of inflection.
    1. [maths rendering] . [Remember [maths rendering] ]
    2. [maths rendering]
    3. [maths rendering]
  2. A perturbation in the temperature of a stream leaving a chemical reactor follows a decaying sinusoidal variation, according to

    [maths rendering]

    where [maths rendering] and [maths rendering] are positive constants.

    1. Sketch the variation of temperature with time.
    2. By examining the behaviour of [maths rendering] , show that the maximum temperatures occur at times of [maths rendering] .
    1. [maths rendering] when [maths rendering]

      [maths rendering]

      [maths rendering] locates a local minimum.

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    2. [maths rendering] when [maths rendering] [maths rendering] when [maths rendering]

      However, [maths rendering] on either side of [maths rendering] so [maths rendering] is a point of inflection.

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    3. [maths rendering]

      This is zero when [maths rendering] i.e. [maths rendering]

      However, this equation has no real roots (since [maths rendering] ) and so [maths rendering] has no stationary points. The graph of this function confirms this:

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      Nevertheless [maths rendering] does have a point of inflection at [maths rendering] as the graph shows, although at that point [maths rendering]

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    2. [maths rendering] implies [maths rendering] , so [maths rendering] and

      [maths rendering] , [maths rendering]  integer

      Examination of [maths rendering] reveals that only even values of [maths rendering] give [maths rendering] for a maximum so setting [maths rendering] gives the required answer.