Concept explainers
A Ferris wheel with a radius of 15 m makes one complete rotation every 12 seconds.
- a. Using the fact that the distance traveled by a rider in one rotation is 2πr, the circumference of the wheel, find the speed with which the riders are moving.
- b. What is the magnitude of their centripetal acceleration?
- c. For a rider with a mass of 50 kg, what is the magnitude of the
centripetal force required to keep that rider moving in a circle? Is the weight of the rider large enough to provide this centripetal force at the top of the cycle? - d. What is the magnitude of the normal force exerted by the seat on the rider at the top of the cycle?
- e. What will happen if the Ferris wheel is going so fast that the weight of the rider is not sufficient to provide the centripetal force at the top of the cycle?
(a)
The speed at which the riders are moving.
Answer to Problem 2SP
The speed at which the riders are moving is
Explanation of Solution
Given Info: The radius of the Ferris wheel is
Write the equation for the speed.
Here,
Rewrite the above equation for the riders in the Ferris wheel.
Here,
Substitute
Conclusion:
Thus the speed at which the riders are moving is
(b)
The magnitude of centripetal acceleration.
Answer to Problem 2SP
The magnitude of centripetal acceleration is
Explanation of Solution
Given Info: The radius of the Ferris wheel is
Write the equation for centripetal acceleration.
Here,
Substitute
Conclusion:
Thus the magnitude of centripetal acceleration is
(c)
The magnitude of the centripetal force required to keep the rider moving in a circle and whether the weight of the rider is large enough to provide the centripetal force at the top of the cycle.
Answer to Problem 2SP
The magnitude of the centripetal force required to keep the rider moving in a circle is
Explanation of Solution
Given Info: The mass of the rider is
Write the equation for the centripetal force.
Here,
Substitute
Write the equation for the weight of the ride.
Here,
The value of
Substitute
Conclusion:
Thus the magnitude of the centripetal force required to keep the rider moving in a circle is
(d)
The magnitude of the normal force exerted by the seat on the rider at the top of the cycle.
Answer to Problem 2SP
The magnitude of the normal force exerted by the seat on the rider at the top of the cycle is
Explanation of Solution
The forces acting on the rider at the top of the cycle are the normal force and the weight of the rider. The normal force acts upward and the weight of the rider acts downward. The net force is downward and it is equal to the centripetal force. This implies the difference between the weight of the rider and the centripetal force gives the magnitude of the normal force.
Here,
Substitute
Conclusion:
Thus the magnitude of the normal force exerted by the seat on the rider at the top of the cycle is
(e)
What will happen if the Ferris wheel moves so fast that the weight of the rider is not sufficient to provide the centripetal force at the top of the cycle.
Answer to Problem 2SP
The rider will let go of the safety bar and he will fly out at a trajectory tangent to the Ferris wheel.
Explanation of Solution
On a Ferris wheel, the circular motion is vertical. The forces acting on the rider in the Ferris wheel are normal force and the weight of the rider.
At the bottom of the cycle the centripetal force needed for the vertical motion of the rider is provided by the weight of the rider. At the top of the cycle the centripetal acceleration is provided by the weight of the rider.
When the Ferris wheel moves so fast, the weight of the rider becomes not sufficient to provide the centripetal force at the top of the cycle. Then the rider will let go of the safety bar and he will fly out of at a trajectory tangent to the Ferris wheel. At this moment, gravity will also accelerate the rider to the ground.
Conclusion:
Thus the rider will let go of the safety bar and he will fly out at a trajectory tangent to the Ferris wheel.
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Chapter 5 Solutions
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