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In a carnival ride, passengers stand with their backs against the wall of a cylinder. The cylinder is set into rotation and the floor is lowered away from the passengers, but they remain stuck against the wall of the cylinder. For a cylinder with a 1.3-m radius, what is the minimum speed that the passengers can have so they do not fall if the coefficient of static friction between the passengers and the wall is 0.19?

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usmanbiu

Answer:

8.2 m/s

Explanation:

radius (r) = 1.3 m

coefficient of friction (u) = 0.19

acceleration due to gravity (g) = 9.8 m/s^{2}

for the passengers to remain stuck on the wall, the normal force must be equal to the centripetal force (while considering the coefficient of friction).

normal force = centripetal force x coefficient of friction

mg = [tex]\frac{mv^{2} }{r}[/tex] x u

g =  [tex]\frac{v^{2} }{r}[/tex] x u

v = [tex]\sqrt{\frac{gXr}{u} }[/tex]

v = [tex]\sqrt{\frac{9.8x1.3}{0.19} }[/tex]

v = 8.2 m/s

the minimum speed the passengers can have to stick to the wall = 8.2 m/s

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