In a popular amusement park ride,guests enter a large cylindrical room and stand

Question

In a popular amusement park ride,guests enter a large cylindrical room and stand with their backs against the wall. The room then begins to spin and the centripetal force pins the guests to the wall. When the room spins fast enough, the floor of the room is dropped and the riders remain stuck against the wall. If the radius do the room is 3m and the riders have an average mass of 80kg each,how fast much the ride be spinning to prevent the riders from sliding down? The coefficient of static friction between the riders and the wall is 0.45.

Explanation / Answer

so sum forces in the x

N = m v^2/r

sum forces in the y

friction - mg = 0

u m v^2/r - mg = 0

u v^2/r - m g

0.45*v^2/3 - 80*9.81=0

v=72.33 m/s

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