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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=2rcos q +rcos a If we resolve vertically (i.e. along y axis) - - PowerPoint PPT Presentation
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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!!!