Locus of:-
Octopus
Corkscrew
Giant wheel
Rides
By:-
கமெல்றாஜ்
கமெல்றாஜ்
Kamel Puvanakumar
Kamel Puvanakumar
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
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, a =-4,
c
-4=-r(dq/dt)²
so,
(dq/dt)= (4/r)
using this and sub. in V c
ż= [V ²-{r(dq/dt)}²]
c
= [V ²-4r²]
c
using integration we derive,
1} z=[(V ²-4r²)]t
c (where V is the constant velocity along
c
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=[(V ²-4r²)]t,
c x=rcost, y=rsint
z=[(V ²-4r²)]t,
c x=rcost, y=rsint
z=[(V ²-4r²)]t,
c x=rcost, y=rsint
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 ={(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 =0 P.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 = {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={u²-2g(r+rsinq)}"
so,
V= [20²-2g{(125/49)+(125/49)sin(p/2)}]
= {140-40}
= 100
= 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=[(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 ={(T-Mgcosq)r/M}
2} V={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!!!