Locus of:- Octopus Corkscrew Giant wheel Rides

Locus of:- Octopus Corkscrew Giant wheel Rides

By:- கமெல்றாஜ் Kamel Puvanakumar Kamel Puvanakumar கமெல்றாஜ்

Octopus Ride

From the diagram, If we resolve horizontally (i.e. along x axis) - (1) x=2rcosq+rcosa If we resolve vertically (i.e. along y axis) - (2) y=2rsinq+rsina Now, using (1) & (2), we could plot the graph of locus of the octopus ride.

X=rcos10t+(r/2)cost , Y=rsin10t+(r/2)sint

CorkScrew ride

using cylindrical co-ordinates, Vc=(dr/dt)êg+r(dq/dt)êg+żĉ ac=[{d/dt(dr/dt)} -r(dq/dt)²]êg + [r{d/dt(dq/dt)}+{2(dr/dt)(dq/dt)}] êg+ [{(d/dt)(dz/dt)}]ĉ however for this problem, r is constant so, dr/dt=0, {d/dt(dr/dt)} = 0 and velocity is constant (as it is maximum) , so, {d/dt(dq/dt)} =0 & {(d/dt)(dz/dt)}=0 hence, Vc=r(dq/dt)êg+żĉ ac=-r{(dq/dt)²}êg

but, a max. shouldn't exceed 4G due to safety issues. & the acceleration acts in the radial direction. therefore,ac=-4, -4=-r(dq/dt)² so, (dq/dt)=Ö (4/r) using this and sub. in Vc ż=Ö [Vc²-{r(dq/dt)}²] =Ö [Vc²-4r²] using integration we derive,

1} z=[Ö(Vc²-4r²)]t(where Vc is the constant velocity along the track) 2} x=rcost Parametric Equation of a Circle 3} y=rsint using 1}, 2} & 3}, we can plot the graph of the locus of corkscrew ride.

z=[Ö(Vc²-4r²)]t, x=rcost, y=rsint

GIANT WHEEL

from the diagram, 1] x=rcosq 2] y=rsinq using these equations, we could see the locus of giant wheel ride.

x = rcosq, y = rsinq

As giant wheel acts in a vertical circle, we could find the velocity at different positions. in a vertical circle, from above equation, we know that displacement = r = {(rcosq)i ,(rsinq)j } using variable acceleration, we know velocity = V = (dr/dt) acceleration = a = (dV/dt) = {d/dt( dr/dt)} solving {d/dt( dr/dt)}, we will get |a| = (V²/r)

so, solving N2nd law radially (assuming wind resistance and other forces are negligible)- T-Mgcosq = Ma = M(V²/r) so, V =Ö{(T-Mgcosq)r/M} we can use this formula if we know the values of M , T and q.

otherwise - solving conservation of energy we could gain K.E.at start=1/2(mu²)K.E.at present =1/2(mV²) P.E.at start=0P.E.at present =mg(r+rsinq) Energy at start = Energy at present so, 1/2(mu²) +0=1/2(mV²)+mg(r+rsinq) rearranging this we get V = Ö{u²-2g(r+rsinq)} we can use this, if we know the values of r, u and q.

E.G - The initial-velocity (u) = 20m/s, the radius of the wheel is (125/49) m and take g=9.8m/s². So find the velocity of the wheel at the top. (p/2 to horizontal) "V=Ö{u²-2g(r+rsinq)}" so, V= Ö[20²-2g{(125/49)+(125/49)sin(p/2)}] = Ö{140-40} = Ö100 = 10m/s.

Summary – • Equation for the locus of octopus ride - • (1) x=2rcosq+rcosa • (2) y=2rsinq+sina • Equation for the locus of corkscrew ride - • 1} z=[Ö(Vc²-4r²)]t • 2} x=rcost • 3} y=rsin(t)

Equation for the locus of corkscrew ride - 1] x=rcosq 2] y=rsinq formulae to find the velocity at different positions in a giant wheel ride (or in a vertical circle) - 1} V =Ö{(T-Mgcosq)r/M} 2} V=Ö{u²-2g(r+rsinq)}

Try and draw these equations using autograph, change the values of r, v. you will find some nice graphs. Way to get on to autograph –

நன்றி Thank You!! Thank You!! To-

Loyd Pryor ( Load analysis of a vertical corkscrew roller coaster track) • Mr. David Harding (maths tutor OSFC) • Dr. Andrew Preston (maths tutor OSFC) • And all my friends involved in this!!!

Locus of:- Octopus Corkscrew Giant wheel Rides