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A circle of radius 1 rolls ( without sliding ) along the x -axis so that its centre is of the form (t , 1 ) with t increasing. A certain point P touches the x-axis at the origin as the circle rolls. As the circle rolls further, the point P passes through the point ( x , 1/2) . Find x when it passes through ( x , 1/2 ) for first time.
ans---------> \frac{\pi }{3}-\frac{\sqrt{3}}{2}
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6 Answers
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Well , the curve described by a point lying on the circumference of a circle if the circle is rolling upon a straight line is called a CYCLOID .
Its parametric equation is , X = a ( z - sin z ) And Y = a ( 1 - cos z ) Where a is the radius of the circle and z is the angle which is shown in the figure as angle PAB . A is the center of the circle .
From your question , a = 1 cm . Suppose the point ( x , 1 / 2 ) is P .
According to the parametric equations ,
Y = 1 ( 1 - cos z ) = 1 / 2 , for the point P .
So t = pi / 3
Hence x for the point P = z - sin z = pi / 3 - sin ( pi / 3 ) = Ans
1Oh , and I forgot to add that the circle can only roll towards +ve X - axis as t is given increasing .
1I was going through my previous posts when I found out this one , a beauty . I tried to solve it another way , and goodness me , I found out a much simpler solution . Hope you like it -
Point X - { t , 1 } , i . e , center of the circle ;
Point A - { x , 1 / 2 } , whose x - co ordinate is to be found .
Point Y - { 0 , 1 / 2 } ,
Point O - { x , 0 } ,
XY = YZ = AO =1 / 2 { I am talking about distances }
XZ = 1 , AX = 1 { radius given 1 }
First , you should realize that the x - co ordinate , which we want to find , is ,
x = Length of arc AZ - Length of line AY ; ........................1
Now , from the triangle XYA ,
cos d = XY / AX = 1 / 2 ;
Hence , d = Î / 3 ;
Hence , length of arc AZ = R d { where R is the radius }
= { Î / 3 } . 1 ...............2
Again , sin d = AY / AX = AY ;
So , AY = √3 / 2 , ........................3
Hence , from 1 , 2 , and 3 ,
x = { Î / 3 } - { √3 / 2}
