Question

(a) Using the Pythagorean Theorem, we can state that Show that d)d) and (b) We need the potential energy of the marble. This is due entirely to gravity, so U=mgy + C to be such that U-0 when ф 0. where we can choose C to be anything that makes our lives easier. We want What is C? Show that with this choice, 4b (using the expression for s found in (a).)

Answer 1

a) The cycloid's curve is given by

a) The cycloid's curve is given by x = b (phi + sin (phi)) and y = b (1 - cos (phi)) Differentiate x with respect to phi to obtain dx = b (1 + cos (phi)) d phi Differentiate y with respect y to obtain d y = b sin (phi) d phi Hence ds^2 = dx^2 + dy^2 = b^2 (1 + cos (phi))^2 d phi^2 + b^2 sin^2 (phi) d phi^2 This can be simplified to ds^2 = b^2 [(1 + cos (phi))^2 + sin^2 (phi)] d phi^2 = b^2 [1 + 2 cos (phi) + cos^2 phi + sin^2 phi] d phi^2 Now use the identity cos^2 phi + sin^2 phi = 1 to obtain ds^2 = b^2 [2 + 2 cos (phi)] d phi^2 = 2b^2 [1 + cos (phi)] d phi^1 Now use the identity 1 + cos (phi) = 2 cos^2 (phi/2)

and

$y=b (1-\cos (\phi ))$

Differentiate x with respect to phi to obtain

$dx=b (1+\cos (\phi ))d\phi$

Differentiate y with respect y to obtain

$dy= b \sin (\phi )\,d\phi$

Hence

$ds^2=dx^2+dy^2= b^2 (1+\cos (\phi ))^2\,d\phi^2+b^2 \sin ^2(\phi )\,d\phi^2$

This can be simplified to

Now use the identity

$\cos^2\phi+\sin^2\phi=1$

to obtain

$ds^2=b^2[2+2\cos(\phi)]d\phi^2=2b^2[1+\cos(\phi)]d\phi^2$

Now use the identity

$1+\cos(\phi)=2\cos^2(\phi/2)$

to obtain

$ds^2=2b^2[1+\cos(\phi)]d\phi^2=4b^2\cos^2(\phi/2)d\phi^2$

Now take the square root on both sides

$ds=2b\cos(\phi/2)\,d\phi$

and integrate to obtain s

$s=2b\int_0^\phi\cos(\phi/2)\,d\phi=4b\sin(\phi/2)$

b) The potential energy with C = 0 becomes

$U=mgy=b \, m \,g(1-\cos (\phi ))$

Now use the identity

$(1-\cos (\phi ))=2\sin^2(\phi/2)$

to obtain

Now plug in

$s=4b\sin(\phi/2)$

and write U in terms of s

$U=\frac{mg}{8b}\left(4b\sin(\phi/2) \right )^2=\frac{mg}{2\times 4 b }s^2=\frac{1}{2}\frac{mg}{4b}s^2$