On a ride called the “gravitron,” riders board the ride and stand against a wall

Question

On a ride called the “gravitron,” riders board the ride and stand against a wall. The ride then starts spinning. Once it is going fast enough, the floor drops from under the riders, but the riders do not fall. If the radius of the ride is 15m and it makes one turn in 4.0 s, what is the velocity of a passenger on the ride, and what is her acceleration in terms of g? (use g = 10 m/s2)

The answer is supposed to be: 7.5 m/s for velocity and 2.5 g for acceleration

I'm using the equation v = 2 pi r / T to get 23.5 m/s velocity and the equation a=v^2/r to get 36.8 m/s^2 for acceleration and 3.7 acceleration in g. I don't know why I'm not getting the answer.

Explanation / Answer

Total distance travelled in 1 complete circle = 2*3.14*r
Total distance travelled in 1 complete circle = 2*3.14*15

Time taken to travel 1 turn = 4s

Speed = Distane / time = 2*3.14*15 / 4
v  = 23.5 m/s

Acceleration , a = v^2/r = 23.5^2/15
a = 36.8 m/s^2.
a = 3.75 g

Your Calculations are perfectly fine but what i feel is you are missing Friction here. There is friction between the rider and the wall which helps him to not fall inspite of floor being dropped. Please look out for friction value.

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