?A circle of radius has its center at the origin. A circle of radius rolls without

Chapter 10, Problem 6

(choose chapter or problem)

A circle  of radius  has its center at the origin. A circle of radius  rolls without slipping in the counterclockwise direction around .  point  is located on a fixed radius of the rolling circle at a distance  from its center, \(0<b<r\). [See parts (i) and (ii) of the figure below.] Let  be the line from the center of  to the center of the rolling circle and let  be the angle that  makes with the positive -axis.

(a) Using  as a parameter, show that parametric equations of the path traced out by  are

\(x=b \cos 3 \theta+3 r \cos \theta\)        \(y=b \sin 3 \theta+3 r \sin \theta\)

Note: If \(b=0\), the path is a circle of radius ; if \(b=r\), the path is an . The path traced out by  for \(0<b<r\) is called an .

(b) Graph the curve for various values of  between  and .

(c) Show that an equilateral triangle can be inscribed in the epitrochoid and that its centroid is on the circle of radius  centered at the origin.

Note: This is the principle of the Wankel rotary engine. When the equilateral triangle rotates with its vertices on the epitrochoid, its centroid sweeps out a circle whose center is at the center of the curve.

(d) In most rotary engines the sides of the equilateral triangles are replaced by arcs of circles centered at the opposite vertices as in part (iii) of the figure. (Then the diameter of the rotor is constant.) Show that the rotor will fit in the epitrochoid if \(b \leqslant \frac{3}{4}(2-\sqrt{3}) r\).

Equation Transcription:

⩽

Text Transcription:

0 < b < r

x=b cos 3 theta + 3r cos theta

y=b sin 3 theta + 3r sin theta

b=0

b=r

b leqslant 3/4 (2-sqrt3)r