Question

In a loop-the-loop ride a car goes around a vertical, circular loop at a constant speed. The car has a mass m = 228 kg and moves with speed v = 16.74 m/s. The loop-the-loop has a radius of R = 11.3 m.

1.What is the magnitude of the normal force on the care when it is at the bottom of the circle? (But as the car is accelerating upward.)
2.What is the magnitude of the normal force on the car when it is at the side of the circle (moving vertically upward)?
3.What is the magnitude of the normal force on the car when it is at the top of the circle?
4.Compare the magnitude of the cars acceleration at each of the above locations:

A. abottom = aside = atop
B. abottom < aside < atop
C. abottom > aside > atop
5.What is the minimum speed of the car so that it stays in contact with the track at the top of the loop?

Answer 1

In a loop-the-loop ride, the normal force varies depending on the car's position in the loop, being greatest at the bottom and least at the top. The car's acceleration due to gravity and its velocity together determine the necessary centripetal force. The minimum speed at the top of the loop is calculated based on the gravitational force equaling the centripetal force requirement.

Step-by-step explanation:

Loop-the-Loop Ride Forces and Speed

When dealing with a loop-the-loop ride in physics, the forces acting on the car can be analyzed using Newton's laws and concepts of circular motion. To answer your questions:

1. The magnitude of the normal force at the bottom of the circle can be found using the formula N = mg + mv²/R, where N is the normal force, m is the mass, g is the acceleration due to gravity, v is the speed, and R is the radius.
2. At the side of the circle, the normal force is still significant as it must provide the necessary centripetal force to keep the car moving in a circular path.
3. At the top of the circle, the normal force is at its minimum since gravity assists in providing the required centripetal force, resulting in N = m(v²/R - g).
4. The acceleration is the same at the bottom and the top but is maximum at the bottom due to the addition of gravitational acceleration. So, the correct comparison is C. abottom > aside > atop.
5. The minimum speed at the top of the loop for the car to stay in contact with the track is when the centripetal force equals the gravitational force, which gives v = √(gR), where v is the speed, g is the acceleration due to gravity, and R is the radius.

Remember that gravity always acts downward, which affects the normal force and therefore the centripetal force needed at various points in the loop.

Answer 2

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