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 :
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
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Locate the stationary points of the following functions and distinguish among them as maxima, minima and points of inflection.
- [maths rendering] . [Remember [maths rendering] ]
- [maths rendering]
- [maths rendering]
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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.
- Sketch the variation of temperature with time.
- By examining the behaviour of [maths rendering] , show that the maximum temperatures occur at times of [maths rendering] .
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[maths rendering]
when
[maths rendering]
[maths rendering]
[maths rendering] locates a local minimum.
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[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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[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:
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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[maths rendering]
when
[maths rendering]
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[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.
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