Question

A frictionless wire is bent into the shape of a cycloid curve, with coordinates given by the parametric equations
? = ?(? + sin ?), ? = ?(1 − cos ?), for −? < ? < ?. The x axis is horizontal, and y is vertically upwards. A bead of
mass m slides freely on the wire. Show that the distance s, measured along the wire from the origin, is given by
? = 4? sin. Write out the potential energy as a function of s, and find the tangential component (i.e., the component
parallel to the wire) Ft of the force on the bead. Write out a differential equation for s(t), and find the period of the
oscillations. (Start from ?? = sqrt(??^2 + ??^2), write dx and dy in terms of ? and d?, and use trigonometric identities
relating cos ? to cos^2 ?/2 or sin^2 ?/2 to simplify the square root.)

4. A frictionless wire is bent into the shape of a cycloid curve, with coordinates given by the parametric equations x-R (θ + sin θ). y-R (1-cos θ), for-r < θ < π. The x axis is horizontal, and y İs vertically upwards. A bead of mass m slides freely on the wire. Show that the distance s, measured along the wire from the origin, is given by s = 4R sin Write out the potential energy as a function of s, and find the tangential component (ie, the component parallel to the wire) Ft of the force on the bead. Write out a differential equation for s(f), and find the period of the oscillations. (Start from ds- /dx2 + dy2, write dr and dv in terms of θ and da and use trigonometric identities relating cos θ to cos2 θ or sin 2 θ to simplify the square root.)

Answer 1

Given that: A friction less wire is bent into the shape of a cycloid curve, with coordinates given by the parametric equation x = R(theta + sin theta) y = R(1 - cos theta) S rightarrow distance of bead measured along the wire from the origin Rightarrow ds squareroot dx^2 + dy^2 Therefore dx = R(d theta + cos theta d theta) dy L sin theta d theta Therefore ds = squareroot R^2 d theta^2 (1 + cos theta)^2 + R^2 d theta^2 sin^2 theta ds = R d theta squareroot 1 + cos^2 theta + 2 cos theta + sin^2 theta ds = R d theta squareroot 2 + 2 cos theta [cos^2 theta + sin^2 theta = 1] ds = R d theta squareroot 2(1 + cos theta) times 2/2 [1 + cos theta/2 = cos^2 theta/2] \r\n R d theta squareroot 4 - cos^2 theta/2 = 2 R cos theta/2 d theta ds = 2 R cos theta/2 d theta Rightarrow Integrating S = 2R sin theta/2/1/2 Rightarrow S = 4 R sin theta/2 potential energy: v = mgh[Here R = y axis] Therefore y = R[1 - cos theta) rightarrow v = mg y = mg R(1 - cos theta) [Because 1 - cos theta/2 sin^2 theta/2]