Locus of:- Octopus Corkscrew Giant wheel Rides கமெல்றாஜ் By:- கமெல்றாஜ் Kamel Puvanakumar 

Presentation on theme: "Locus of:- Octopus Corkscrew Giant wheel Rides கமெல்றாஜ் By:- கமெல்றாஜ் Kamel Puvanakumar "— Presentation transcript:

Locus of:- Octopus Corkscrew Giant wheel Rides

கமெல்றாஜ் By:- கமெல்றாஜ் 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, V c =(dr/dt)ê g +r(d  dt)ê g +żĉ a c =[{d/dt(dr/dt)} -r(d  dt) ² ]ê g + [r{d/dt(dq/dt)}+{2(dr/dt)(d  dt)}] ê g + [{(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(d  dt)} =0 & {(d/dt)(dz/dt)}=0 hence, V c =r(d  dt)ê g +żĉ a c =-r{(d  dt) ² } ê g

but, a max. shouldn't exceed 4G due to safety issues. & the acceleration acts in the radial direction. therefore, a c =- 4, -4=-r(d  dt) ² so, (d  dt)=  (4/r) using this and sub. in V c ż=  [V c ² - {r(d  dt)} ² ] =  [V c ² - 4r ² ] using integration we derive,

1}z=[  (V c ² -4r ² )]t (where V c 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=[  (V c ² -4r ² )]t,x=rcost,y=rsint

z=[  (V c ² -4r ² )]t,x=rcost,y=rsint

z=[  (V c ² -4r ² )]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(V ² /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(mu ² )K.E. at present =1/2(mV ² ) P.E. at start =0 P.E. at present =mg(r+rsin  ) Energy at start = Energy at present so, 1/2(mu ² ) +0=1/2(mV ² )+mg(r+rsin  ) rearranging this we get V =  { u ² -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=[  (V c ² -4r ² )]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 –

நன்றி 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!!!

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