Spin and Barf Ride
Rotor diameter is 4.0 m. Minimum period of rotation is 1.7 s. The ride lasts 110 s.
1. What is the tangential velocity?
2. What is the centripetal acceleration is m/s2 and g’s?
To find the tangential velocity of the Spin and Barf ride, we need to calculate the distance traveled by the riders in one complete rotation, and then divide it by the time taken for that rotation.
1. The distance traveled in one complete rotation is equal to the circumference of the circle formed by the rotating ride. The circumference of a circle is given by the formula C = 2πr, where r is the radius (half the diameter). Therefore, the distance traveled in one rotation is C = 2π(4.0 m/2) = 4π m.
2. The time taken for one rotation is given as 1.7 seconds.
Now, we can calculate the tangential velocity by dividing the distance traveled in one rotation by the time taken for that rotation:
Tangential velocity = Distance traveled / Time taken
= 4π m / 1.7 s
To get the numerical value, we can use the approximation π ≈ 3.14:
Tangential velocity ≈ (4)(3.14) m / 1.7 s
≈ 7.52 m/s
Therefore, the tangential velocity of the Spin and Barf ride is approximately 7.52 m/s.
To find the centripetal acceleration, we can use the formula a = v^2 / r, where v is the tangential velocity and r is the radius of rotation.
3. Centripetal acceleration in m/s^2:
Centripetal acceleration = (Tangential velocity)^2 / Radius
= (7.52 m/s)^2 / 4.0 m
Calculating this value, we have:
Centripetal acceleration ≈ 14.1904 m^2/s^2
4. Centripetal acceleration in g's:
To convert centripetal acceleration to g's, we divide it by the acceleration due to gravity, which is approximately 9.8 m/s^2.
Centripetal acceleration in g's = Centripetal acceleration / Acceleration due to gravity
≈ 14.1904 m^2/s^2 / 9.8 m/s^2
≈ 1.45 g's
Therefore, the centripetal acceleration of the Spin and Barf ride is approximately 14.1904 m^2/s^2 or 1.45 g's.