A carnival ride is designed to allow the general public to experience high accel

Question

A carnival ride is designed to allow the general public to experience high acceleration motion. The ride rotates about Point O in a horizontal circle such that the rider has a speed v0. The rider reclines on a platform A which rides on rollers such that friction is negligible. A mechanical stop prevents the platform from rolling down the incline.The carnival ride is modified so that 80 kg riders can move up and down the inclined wall as the speed of the ride increases. Knowing that the coefficient of static friction between the wall and the platform is 0.26, determine the range of values of the constant speed v0 for which the platform will remain in the position shown. (Round the minimum velocity, vmin to three decimal places and the maximum velocity, vmax to two decimal places.)

e minimum velocity is m/s and the maximum velocity is m/s.

Explanation / Answer

according to the given fiugure
radius of the revolutino at the given position = r
r = 5 + 1.5cos(70) = 5.51303 m

hence the range of the speeds can be found using the direction of fricotin as upwards and downwards
so for upward direction of friction

from force balance
let f be the friction, N be the normal reaction
then
Nsin(70) = mv^2/r + fcos(70)
where v is speed of the cart, m is its mass
Ncos(70) + fsin(70) = mg

also
f = mu*N where mu = 0.26
hence

Nsin(70) = mv^2/r + mu*N*cos(70)
Ncos(70) + mu*N*sin(70) = mg
N = mg/(cos(70) + mu*sin(70))

mg(sin(70) - mu*cos(70)) /(cos(70) + mu*sin(70))= mv^2/r
g(sin(70) - 0.26*cos(70)) /(cos(70) + 0.26*sin(70))= v^2/5.51303

v = 8.85850 m/s

similiarly for downward direction of friction the force balance equations become
Nsin(70) = mv^2/r - fcos(70)
Ncos(70) - fsin(70) = mg
hence
g(sin(70) + 0.26*cos(70)) /(cos(70) - 0.26*sin(70))= v^2/5.51303
v = 23.86213 m/s

hence
minimum velocity is 8.85850 m/s
maximum velocity is 23.86213 m/s

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