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A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow has an 8.00-m radius, at how many revolutions per minute are the riders subjected to a centripetal acceleration equal to that of gravity?

a) 15.3 rpm
b) 17.7 rpm
c) 21.1 rpm
d) 24.5 rpm

1 Answer

2 votes

Final answer:

The correct number of revolutions per minute at which riders experience a centripetal acceleration equal to gravity, given an 8.00 m radius ride, is approximately 17.7 rpm.

Step-by-step explanation:

To find out at how many revolutions per minute (rpm) the riders on a fairground ride are subjected to a centripetal acceleration equal to that of gravity, we need to establish a relationship between centripetal acceleration and angular velocity. The centripetal acceleration (ac) experienced by an object moving in a circular path of radius r at a constant speed v is given by ac = v2 / r. The acceleration due to gravity (g) is 9.81 m/s2. As we want the centripetal acceleration to be equal to g, we have: v2 = gr. To find the speed v, we take the square root giving us v = √(gr).

The relationship between linear velocity v and angular velocity ω (in radians per second) is v = ωr. We can express angular velocity in revolutions per minute by using the conversion 1 revolution = 2π radians and 1 minute = 60 seconds. Solving for ω in rpm, we get: ω = v / (2πr) • 60.

Substituting the values r = 8.00 m and g = 9.81 m/s2 into the equations, first to find v, then ω yields the number of revolutions per minute at which the ride must spin to create a centripetal acceleration equal to the acceleration due to gravity. The calculation results in a speed of approximately 17.7 rpm, which means the correct answer is b) 17.7 rpm.

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