Question

The figure is part of a cycloid with parametric equations $\mathit{x}\mathbf{=}\mathit{a}\mathbf{(}\mathbf{\theta }\mathbf{+}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }\mathbf{)}\mathbf{,}\mathit{y}\mathbf{=}\mathit{a}\mathbf{(}\mathbf{1}\mathbf{-}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\mathbf{)}$ (The graph shown is like Figure 4.4 of Chapter 9 with the origin shifted to P2.) Show that the time for a particle to slide without friction along the curve from (x₁, y₁) to the origin is given by the differential equation for θ(t) is $\mathit{t}\mathbf{=}\sqrt{\frac{{\mathbf{a}}^{\mathbf{y}}}{\mathbf{g}}}\underset{\mathbf{0}}{\overset{\mathbf{1}}{\mathbf{\int }}}\frac{\mathbf{d}\mathbf{y}}{\sqrt{\mathbf{y}\left(y_1-y\right)}}$.

Hint: Show that the arc length element is $\mathit{d}\mathit{s}\mathbf{=}\sqrt{\frac{\mathbf{2}\mathbf{a}}{\mathbf{y}}}\mathit{d}\mathit{y}$. Evaluate the integral to show that the time is independent of the starting height y₁ .

Short answer

The time for a particle to slide without friction along the curve is $t=\sqrt{\frac{a}{g}}\mathrm{\pi }$ .

Step-by-step solution

Given Information

The equation of a cycloid $x=a(\mathrm{\theta }+\mathrm{sin\theta }),y=a(1-cos\theta )$.

Definition of Energy

An object's ability to accomplish work is defined as its energy. It can take many forms, including potential, kinetic, thermal, electrical, chemical, radioactive, and others.

The energy of the system is conserved, hence $\mathit{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{)}\mathbf{+}\mathit{m}\mathit{g}\mathit{y}$.

Find the value of x2 in terms of y and y-

It is given in the question that $x=a(\mathrm{\theta }+\mathrm{sin\theta }),y=a(1-cos\theta )$.

$\begin{array}{rcl}\stackrel{\bullet }{x}& =& a\left(1+\cos \theta \right)\stackrel{\bullet }{\theta }\\ \stackrel{\bullet }{y}& =& a\left(\sin \theta \right)\stackrel{\bullet }{\theta }\\ {\stackrel{\bullet }{\left(x\right)}}^2& =& a^2{\left(1+\cos \theta \right)}^2{\stackrel{\bullet }{\left(\theta \right)}}^2\\ {\stackrel{\bullet }{\left(x\right)}}^2& =& \frac{{\left(1+\cos \theta \right)}^2{\left(\stackrel{\bullet }{y}\right)}^2}{1-{\cos }^2\theta }\end{array}$

Simplify further.

$\stackrel{\bullet }{{\left(x\right)}^2}=\left(\frac{2a}{y}-1\right)\stackrel{\bullet }{{\left(y\right)}^2}$

Show that the time taken is independent of height

Particle is at rest at height y₁ hence E = mgy₁ .

Substitute the value of E in the equation mentioned below,

$\begin{array}{rcl}E& =& \frac{1}{2}m({\mathrm{x}}^2+{\mathrm{y}}^2)+mgy\\ mg{y}_1& =& \frac{a}{y}m{y}^2+mgy\\ y& =& -\sqrt{\frac{g}{a}\sqrt{y\left(y_1-y\right)}}\end{array}$

Integrate the above equation.

$t={\sqrt{\frac{a}{g}}}^y\underset{0}{\overset{1}{\int }}\frac{dy}{\sqrt{y\left(y_1-y\right)}}$

Hence, the statement is proved.