Cycloid
A cycloid is the curve traced by a fixed point on the edge of a circle of radius \(r\) as it rolls without slipping along the \(x\)-axis.
Special Properties
- Brachistochrone:Brachistochrone: The cycloid is the fastest descent curve under gravity between two given points.
- Tautochrone:Tautochrone: Any particle placed on a cycloid reaches the lowest point at the same time, regardless of its starting position.
- Symmetry: The cycloid is symmetric about its midpoint in each arch.
Instructions
Move the sliders to change the circle position or change its properties.
Parametric Equations
Defining the Motion of the Circle
A circle of radius \( r \) rolls without slipping along the \(x\)-axis. When it has rotated by an angle \(\theta\), the arc length traveled by a point on the edge of the circle is equal to the distance traveled by the center of the circle along the \(x\)-axis:
$$s = r\theta$$
This means that the center of the circle is located at:
$$\left( x_{c},y_{c} \right) = \left( r\theta,r \right)$$
Determining the Position of the Point on the Circle
If we fix a point \(P\) on the edge of the circle, its motion is determined by:
- The point is at a distance \(r\) from the center.
- Its position relative to the center of the circle is given by the angle \(\theta\), which measures the rotation of the circle.
The horizontal component of the relative position is:
$$x_{P} = r \sin \theta$$
because the point traces a circular path around the center.
The vertical component of the relative position is:
$$y_{P} = r \cos \theta$$
Thus, the total position of the fixed point on the edge of the circle is:
$$x = x_{c} - x_{P} = r\theta - r\sin \theta$$
$$y = y_{c} - y_{P} = r - r \cos \theta$$
Parametric Equations of the Cycloid
$$x = r\left( \theta - \sin \theta \right)$$
$$y = r\left( 1 - \cos \theta \right)$$
Where:
- \(x\) is the horizontal coordinate of the point on the cycloid.
- \(y\) is the vertical coordinate of the point on the cycloid.
- \(\theta\) is the angle through which the circle has rotated.
This is the standard parametric form for describing the cycloid.
Arc Length of the Cycloid
To find the arc length of a single cycloid segment from \(\theta = 0\) to \(θ = 2\pi\), we use the general formula for the length of a parametric curve:
$$L = \int_{a}^{b}\sqrt{\left(\frac{dx}{d\theta}\right)^{2} + \left(\frac{dy}{d\theta}\right)^{2}}\ d\theta$$
We compute the derivatives of \(x\) and \(y\):
$$\frac{dx}{d\theta} = r\left( 1 - \cos \theta \right)$$
$$\frac{dy}{d\theta} = r \sin \theta$$
Substituting them into the arc length equation:
$$L = \int_{0}^{2\pi} \sqrt{r^{2} \left( 1 - \cos \theta \right)^{2} + r^{2} \sin^{2} \theta}\ d\theta$$
Factoring out \(r^{2}\) from the square root:
$$L = \int_{0}^{2\pi} r \sqrt{\left( 1 - \cos \theta \right)^{2} + \sin^{2} \theta}\ d\theta$$
Using the trigonometric identity \( \left( 1 - \cos \theta \right)^{2} + \sin^{2} \theta = 2\left( 1 - \cos \theta \right) \), we get:
$$L = \int_{0}^{2\pi} r \sqrt{2 \left(1 - \cos \theta \right)}\ d\theta$$
Since \(1 - \cos \theta = 2\sin^{2} \left(\theta / 2\right)\), we further simplify:
$$L = \int_{0}^{2\pi} r \cdot 2 \left\vert \sin \left( \theta /2 \right) \right\vert d\theta$$
Evaluating the integral, the arc length of one complete cycloid segment (one crest and one trough) is:
$$L = 8r$$
Area Under One Arc of the Cycloid
The area under one cycloid arch from \(θ = 0\) to \(\theta = 2\pi\) is found by integrating \(y\) with respect to \(x\):
$$A = \int y dx$$
$$A = \int_{0}^{2\pi}y \frac{dx}{d\theta} d\theta$$
where the parametric equations are:
$$x = r\left(\theta - \sin\theta\right)$$
$$y = r\left(1 - \cos\theta\right)$$
The derivative of \(x\) with respect to \(\theta\) is:
$$\frac{dx}{d\theta} = r \left( 1 - \cos \theta \right)$$
Substituting \(y\) and \(dx/d\theta\) into the integral:
$$A = \int_{0}^{2\pi} r\left( 1 - \cos \theta \right) \cdot r\left( 1 - \cos \theta \right) d\theta$$
$$A = r^{2} \int_{0}^{2\pi} \left( 1 - \cos \theta \right)^{2}d\theta$$
Expanding the Integral
Using the identity:
$$\left( 1 - \cos \theta \right)^{2} = 1 - 2 \cos \theta + \cos^{2}\theta$$
we substitute it into the integral:
$$A = r^{2} \int_{0}^{2\pi}\left( 1 - 2\cos\theta + \cos^{2}\theta \right) d\theta$$
Solving the Individual Integrals
- First integral:
- Second integral:
- Third integral:
$$\int_{0}^{2\pi} 1d\theta = 2\pi$$
$$\int_{0}^{2\pi} -2 \cos \theta \ d\theta = -2 \int_{0}^{2\pi} \cos \theta \ d\theta = 0$$
because the integral of \( \cos\theta\) over a full period is zero.
Using the identity
$$\cos^{2}\theta = \frac{1 + \cos 2\theta}{2}$$
we get:
$$\int_{0}^{2\pi} \cos^{2} \theta d\theta = \int_{0}^{2\pi}\frac{1 + \cos 2\theta}{2}d\theta$$
$$= \frac{1}{2} \int_{0}^{2\pi}1d\theta + \frac{1}{2} \int_{0}^{2\pi} \cos 2\theta d\theta$$
$$= \frac{1}{2}\left( 2\pi \right) + \frac{1}{2}\left( 0 \right) = \pi$$
Final Result
$$A = r^{2} \left( 2\pi + 0 + \pi \right) = r^{2}\left( 3\pi \right) = 3\pi r^{2}$$
Thus, the area under one complete arch of the cycloid is:
$$A = 3 \pi r^{2}$$
This means that the area under the cycloid is three times the area of the generating circle.