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In engineering, a crank-and-slider mechanism can be used to change circular motion into motion back and forth in a straight line - Leaving Cert Mathematics - Question 9 - 2018

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In engineering, a crank-and-slider mechanism can be used to change circular motion into motion back and forth in a straight line. In the diagrams below, the crank [... show full transcript

Worked Solution & Example Answer:In engineering, a crank-and-slider mechanism can be used to change circular motion into motion back and forth in a straight line - Leaving Cert Mathematics - Question 9 - 2018

Step 1

Find \( LCO \), correct to the nearest degree.

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Answer

To find ( LCO ), we can apply the sine rule in triangle ( LCO ):

[ \frac{10}{\sin 15^\circ} = \frac{LCO}{\sin x} ]

This can be rearranged to:

[ LCO = \frac{10 \cdot \sin x}{\sin 15^\circ} ]

Calculating for ( x ) gives:

[ \sin x = \frac{10 \cdot \sin 15^\circ}{30} \approx 0.77645 ]

Taking the arcsine yields:

[ x \approx 51^\circ ]

Therefore, ( LCO ) is approximately ( 51^\circ ).

Step 2

Write down the period and range of \( f \).

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Answer

The period of function ( f(\alpha) ) is determined by the angular movement of the crank, which travels fully around in a circle. Thus:

  • Period = ( 2\pi , \text{radians} ) or ( 360^\circ )
  • Range = ( [10, 30] , cm )

Step 3

Complete the table below for \( f(\alpha) \).

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( \alpha )( 0^\circ )( 90^\circ )( 180^\circ )( 270^\circ )( 360^\circ )
( f(\alpha) ) (cm)( 30 )( 18.28 )( 10 )( 18.28 )( 30 )

Step 4

Draw a rough sketch of \( f \) in the domain \( 0^\circ \leq \alpha \leq 360^\circ \).

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Answer

Using the values from the table, we would plot the points (0, 30), (90, 18.28), (180, 10), (270, 18.28), and (360, 30) on a graph. The graph will illustrate the behavior of ( f(\alpha) ), which increases and decreases according to the sinusoidal nature of the crank's motion.

Step 5

for which of the three positions of the mechanism will a 1 degree change in \( \alpha \) cause the greatest change in the position of [C]? Explain your answer.

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Answer

Referring to the steepness of the graph plotted from table values, Diagram 2 shows the steepest slope at ( \alpha = 90^\circ ). This indicates that a 1 degree change in angle ( \alpha ) will result in a substantial change in the position of [C]. Therefore, Diagram 2 is the correct position.

Step 6

Find \( r \), correct to the nearest cm.

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Answer

From the cosine rule applied to triangle ( AOB ):

[ r^2 = |AB|^2 + |AX|^2 - 2 |AB| |AX| \cos(10^\circ) ]

Substituting the values:

[ r^2 = 36^2 + 31^2 - 2 \times 36 \times 31 \times \cos(10^\circ) ]

Calculating yields: [ r \approx 7 , cm \text{ (to the nearest cm)} ]

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