Locus of:-

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Locus of- Octopus Corkscrew Giant
wheel Rides
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By- ????????? Kamel Puvanakumar Kamel
Puvanakumar
?????????
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Octopus Ride
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From the diagram, If we resolve horizontally
(i.e. along x axis) - (1) x2rcosqrcosa If
we resolve vertically (i.e. along y axis)
- (2) y2rsinqrsina Now, using (1)
(2), we could plot the graph of locus of the
octopus ride.
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Xrcos10t(r/2)cost , Yrsin10t(r/2)sint
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CorkScrew ride
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using cylindrical co-ordinates,
Vc(dr/dt)êgr(dq/dt)êgzc acd/dt(dr/dt)
-r(dq/dt)²êg rd/dt(dq/dt)2(dr/dt)(dq/dt)
êg (d/dt)(dz/dt)c however for this
problem, r is constant so, dr/dt0,
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, Vcr(dq/dt)êgzc ac
-r(dq/dt)²êg
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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
zÖ Vc²-r(dq/dt)² Ö
Vc²-4r² using integration we derive,
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1 zÖ(Vc²-4r²)t (where Vc is the constant
velocity along the track) 2 xrcost
Parametric Equation of a Circle 3 yrsint
using 1, 2 3, we can plot the graph of
the locus of corkscrew ride.
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zÖ(Vc²-4r²)t, xrcost, yrsint
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zÖ(Vc²-4r²)t, xrcost, yrsint
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zÖ(Vc²-4r²)t, xrcost, yrsint
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GIANT
WHEEL
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from the diagram, 1 xrcosq 2 yrsinq
using these equations, we could see the locus of
giant wheel ride.
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x rcosq, y rsinq
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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)
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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.
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otherwise - solving conservation of energy we
could gain K.E.at start1/2(mu²) K.E.at present
1/2(mV²) P.E.at start0 P.E.at present
mg(rrsinq) Energy at start Energy at
present so, 1/2(mu²) 01/2(mV²)mg(rrsinq) r
earranging this we get V Öu²-2g(rrsinq) w
e can use this, if we know the values of r, u and
q.
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E.G - The initial-velocity (u) 20m/s, the
radius of the wheel is (125/49) m and take
g9.8m/s². So find the velocity of the wheel at
the top. (p/2 to horizontal) "VÖu²-2g(rrsinq
)" so, V Ö20²-2g(125/49)(125/49)sin(p/2)
Ö140-40 Ö100 10m/s.
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• Summary
• Equation for the locus of octopus ride -
• (1) x2rcosqrcosa
• (2) y2rsinqsina
• Equation for the locus of corkscrew ride -
• 1 zÖ(Vc²-4r²)t
• 2 xrcost
• 3 yrsin(t)

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Equation for the locus of corkscrew ride -
1 xrcosq 2 yrsinq 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 VÖu²-2g(rr
sinq)
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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
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????? Thank You!! Thank You!! To-
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• 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!!!