Question

If a circle of radius 12 rolls along the x-axis in quadrants I and II, what is an equation for the path of the center of the circle?

Answer 1

As the circle of radius 12 rolls along the x-axis in quadrants I and II, its center will trace a path along the x-axis and above it.

Let (x, y) be the coordinates of the center of the circle. As the circle rolls along the x-axis, its center will be a distance of 12 units above the x-axis, so we have:

y = 12

In quadrants I and II, the circle will rotate clockwise, so its center will move to the right. If we consider a point on the circumference of the circle that is in contact with the x-axis, we can see that its x-coordinate is equal to the length of the arc that the circle has rolled along. This arc length is equal to the radius of the circle times the angle in radians through which it has rotated. Since the angle of rotation is equal to the distance traveled along the x-axis divided by the radius, we have:

θ = x / 12

where θ is the angle of rotation in radians.

Now we can express the x-coordinate of the center of the circle in terms of θ:

x = 12θ

Substituting this into the equation for y, we get:

y = 12

x = 12θ

So the equation for the path of the center of the circle is:

(x, y) = (12θ, 12)

where θ is the angle in radians through which the circle has rotated as it rolls along the x-axis in quadrants I and II.

Answer 2

y = 3

Explanation:

Since we are talking about the motion of the center of the circle, the center will always be at the position (x, 3) ie, its y value will be 3 irrespective of the x-value in the I and II coordinates

Hence we can model the path of the center of origin as a straight horizontal line with the equation

y = 3