Locus of:-

Octopus

Corkscrew

Giant wheel

Rides

கமெ�ல்றா�ஜ்

By:-

கமெ�ல்றா�ஜ்

Kamel Puvanakumar

From the diagram,

If we resolve horizontally (i.e. along x axis) -

(1) x=2rcos+rcos

If we resolve vertically (i.e. along y axis) -

(2) y=2rsin+rsin

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

using cylindrical co-ordinates,

Vc=(dr/dt)êg+r(ddt)êg+żĉ

ac=[{d/dt(dr/dt)} -r(ddt)²]êg + [r{d/dt(dq/dt)}+{2(dr/dt)(ddt)}] 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(ddt)} =0 & {(d/dt)(dz/dt)}=0 hence,

Vc=r(ddt)êg+żĉ

ac=-r{(ddt)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(ddt)²

so, (ddt)=(4/r)

using this and sub. in Vc

ż=[Vc²-{r(ddt)}²] =[Vc²-4r²]

using integration we derive,

1} z=[(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=[(Vc²-4r²)]t, x=rcost, y=rsint

z=[(Vc²-4r²)]t, x=rcost, y=rsint

z=[(Vc²-4r²)]t, x=rcost, y=rsint

from the diagram,

1] x=rcos

2] y=rsin

using these equations, we could see the locus of giant wheel ride.

x = rcos, y = rsin

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 = {(rcos)i ,(rsin)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-Mgcos= Ma = M(V²/r)

so,

V ={(T-Mgcos)r/M}

we can use this formula if we know the values of M , T and .

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=0 P.E.at present =mg(r+rsin)

Energy at start = Energy at present

so,1/2(mu²) +0=1/2(mV²)+mg(r+rsin)

rearranging this we get

V = {u²-2g(r+rsin)}

we can use this, if we know the values of r, u and .

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. (/2 to horizontal)

"V={u²-2g(r+rsin)}"

so,V= [20²-2g{(125/49)+(125/49)sin(/2)}] = {140-40}

= 100 = 10m/s.

Summary –

*Equation for the locus of octopus ride -

(1) x=2rcos+rcos

(2) y=2rsin+sin

*Equation for the locus of corkscrew ride -

1} z=[(Vc²-4r²)]t

2} x=rcost

3} y=rsin(t)

Equation for the locus of corkscrew ride -

1] x=rcos

2] y=rsin

formulae to find the velocity at different positions in a giant wheel ride (or in a vertical circle) -

1} V ={(T-Mgcos)r/M}

2} V={u²-2g(r+rsin)}

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 –

நன்றா

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!!!