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Chapter 5: Problem 36

A roller coaster at an amusement park has a dip that bottoms out in a vertical circle of radius \(r\). A passenger feels the seat of the car pushing upward on her with a force equal to twice her weight as she goes through the dip. If \(r=20.0 \mathrm{~m},\) how fast is the roller coaster traveling at the bottom of the dip?

Short Answer

Expert verified
The roller coaster is traveling at 14 m/s at the bottom of the dip.

Step by step solution

01

Understand the Forces at the Bottom of the Dip

At the bottom of the dip, the forces acting on the passenger are the gravitational force downward ( mg ) and the normal force ( N ) upward, exerted by the seat, which is noted to be twice the passenger's weight. Thus, N = 2mg .
02

Apply Newton's Second Law in the Vertical Direction

At the bottom of the dip, the net force acting on the passenger is the centripetal force, which is the difference between the normal force and the gravitational force:\[ N - mg = m\frac{v^2}{r} \]Here, we'll substitute N = 2mg from our earlier understanding.
03

Substitute the Known Values into the Equation

Substitute N = 2mg and the radius r = 20.0 \, \mathrm{m} into the centripetal force equation:\[ 2mg - mg = m\frac{v^2}{20} \]This simplifies to:\[ mg = m\frac{v^2}{20} \]
04

Simplify and Solve for Velocity

Cancel the mass m from both sides of the equation:\[ g = \frac{v^2}{20} \]Rearrange to solve for v^2:\[ v^2 = 20g \]Taking the square root gives:\[ v = \sqrt{20g} \]Substitute g = 9.8 \, \mathrm{m/s^2}:\[ v = \sqrt{20 \times 9.8} = \sqrt{196} \]
05

Calculate the Final Result

Compute the square root to find the velocity:\[ v = 14 \, \mathrm{m/s} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Centripetal Force
When you're on a roller coaster, you might have noticed how it feels like something is pulling you towards the center of the curve, especially at the bottom of a dip. This inward force is known as centripetal force. It is essential for keeping objects moving in a circular path.

To define it simply:
  • The centripetal force is directed towards the center of the circle.
  • It is calculated using the formula: \( F_c = m \frac{v^2}{r} \), where \( m \) is mass, \( v \) is velocity, and \( r \) is the radius of the circle.
  • Without sufficient centripetal force, the object would fly off in a straight line.
Understanding this force helps us solve problems involving objects in circular motion, like roller coasters.
Newton's Second Law
Newton's Second Law is crucial for understanding motion. It relates force, mass, and acceleration, and can be expressed as \( F = ma \). This means the force acting on an object is equal to the mass of the object multiplied by its acceleration.

When applying this law:
  • It helps us determine how forces affect motion.
  • In the context of a roller coaster: the net force at the bottom of the dip is the difference between the normal force (the push from the seat) and the gravitational force.
  • This difference is what provides the necessary centripetal force for circular motion.
Understanding this law is key to analyzing any situation where forces result in movement.
Gravitational Force
Gravitational force is the attraction between two masses. On Earth, it gives us weight and is why objects fall when dropped. For anyone in a roller coaster, it acts downwards, opposing the normal force from the seat.

Key points about gravitational force:
  • The force is calculated as \( F_g = mg \), where \( g \) is the acceleration due to gravity (9.8 m/s虏 on Earth).
  • In our problem, \( F_g \) is one of the forces to consider at the bottom of the roller coaster dip.
  • This force is constant and acts vertically downwards.
Gravitational force is vital in understanding how different forces interact at play during roller coaster dynamics.
Roller Coaster Dynamics
Roller coaster dynamics involve understanding how different forces interact to create thrilling rides. These forces must be carefully calculated to ensure safety and enjoyment.

In the dynamics:
  • Centripetal force keeps passengers safely in the loop and dips.
  • Newton's Second Law helps us calculate the relationship between speed, forces, and motion.
  • Gravitational force is countered by the normal force from the seat.
  • The sensation of weightlessness or increased weight comes from the interplay of these forces.
By understanding these dynamics, engineers design coasters that provide the desired thrills without compromising on safety.

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Most popular questions from this chapter

A satellite is in a circular orbit around an unknown planet. The satellite has a speed of \(1.70 \times 10^{4} \mathrm{~m} / \mathrm{s}\), and the radius of the orbit is \(5.25 \times 10^{6} \mathrm{~m}\). A second satellite also has a circular orbit around this same planet. The orbit of this second satellite has a radius of \(8.60 \times 10^{6} \mathrm{~m}\). What is the orbital speed of the second satellite?

A rigid massless rod is rotated about one end in a horizontal circle. There is a mass \(m_{1}\) attached to the center of the rod and a mass \(m_{2}\) attached to the outer end of the rod. The inner section of the rod sustains three times as much tension as the outer section. Find the ratio \(m_{2} / m_{1}\)

A \(2100-\mathrm{kg}\) demolition ball swings at the end of a \(15-\mathrm{m}\) cable on the arc of a vertical circle. At the lowest point of the swing, the ball is moving at a speed of \(7.6 \mathrm{~m} / \mathrm{s}\). Determine the tension in the cable. (Hint: The tension serves the same purpose as the normal force at point 1 in Figure \(5-21 .)\)

Concept Simulation 5.1 at reviews the concepts that are involved in this problem. A child is twirling a \(0.0120-\mathrm{kg}\) ball on a string in a horizontal circle whose radius is 0.100 \(\mathrm{m}\). The ball travels once around the circle in \(0.500 \mathrm{~s}\). (a) Determine the centripetal force acting on the ball. (b) If the speed is doubled, does the centripetal force double? If not, by what factor does the centripetal force increase?

A satellite moves on a circular earth orbit that has a radius of \(6.7 \times 10^{6} \mathrm{~m}\). A model airplane is flying on a 15 -m guideline in a horizontal circle. The guideline is parallel to the ground. Find the speed of the plane such that the plane and the satellite have the same centripetal acceleration.
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