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Chapter 13: Problem 42

A roller coaster starts from rest at \(A,\) rolls down the track to \(B\), describes a circular loop of 40 -ft diameter, and moves up and down past point \(E .\) Knowing that \(h=60 \mathrm{ft}\) and assuming no energy loss due to friction, determine \((a)\) the force exerted by his seat on a 160 -lb rider at \(B\) and \(D,(b)\) the minimum value of the radius of curvature at \(E\) if the roller coaster is not to leave the track at that point.

Short Answer

Expert verified
(a) Calculate forces using centripetal force formulas. (b) Solve for minimum radius using energy conservation at point E.

Step by step solution

01

Understanding the Problem

We need to use the principles of energy conservation and centripetal force to determine the force on the rider at different points and the conditions required to keep the roller coaster on the track.
02

Calculate Potential Energy at Point A

At point A, the roller coaster has gravitational potential energy due to its height. The potential energy at A is calculated by \( PE_A = mgh \), where \( m \) is the mass (in this case equivalent to the rider's weight), \( g \) is the acceleration due to gravity \( (32.2 \ \text{ft/s}^2) \), and \( h \) is the height (60 ft). Thus, \( PE_A = 160 \times 60 \).
03

Calculate Velocity at Point B Using Energy Conservation

As the roller coaster descends to B (at the top of the loop, where we take height zero), all potential energy converts to kinetic energy, \( KE_B = \frac{1}{2} mv^2 \). Equating \( PE_A \) to \( KE_B \), calculate the velocity \( v_B \) using \( 160 \times 60 = \frac{1}{2} \times 160 \times v_B^2 \). Solve for \( v_B \).
04

Calculate Centripetal Force at Point B

At the top of the loop (B), the centripetal force required is provided by gravity and the seat force: \( F_{B} + mg = \frac{mv_B^2}{r} \), where \( r = 20 \) ft (half the loop diameter), and \( mg = 160 \). Substitute \( v_B \) from the previous step and solve for \( F_B \).
05

Calculate Velocity at Point D using Energy Conservation

Point D is at the same height as B on the loop, so the speed at D will be the same as at B due to energy conservation, hence \( v_D = v_B \).
06

Calculate Centripetal Force at Point D

Again using the formula for centripetal force, \( F_D - mg = \frac{mv_D^2}{r} \), with \( r = 20 \) ft. Since \( mg = 160 \) lbs and \( v_D = v_B \), we solve for \( F_D \).
07

Determine Minimum Radius of Curvature at Point E

At point E, the rider must not leave the track, meaning the gravitational force should not exceed the centripetal requirement. Use \( mg = \frac{mv_E^2}{R} \) to find the minimum \( R \), where \( m \) is the rider鈥檚 weight, \( g \) is gravity, and \( v_E \) is velocity at point E, derived from energy conservation from A to E assuming point E has a height \( h_E \) which needs to be calculated using \( v_E = \sqrt{ v_A^2 +2g(h-h_E)} \). Solve for \( R \).

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Key Concepts

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

Energy Conservation
In the context of a roller coaster, the principle of energy conservation is vital to understand. It helps predict how the coaster moves as it glides over various track sections.
  • At the start, a roller coaster has potential energy largely due to its elevated position. This energy is represented as gravitational potential energy, given by the formula \( PE = mgh \), where \( m \) is mass, \( g \) is gravitational acceleration, and \( h \) is height.
  • As the ride progresses, this potential energy converts into kinetic energy as the roller coaster descends and gains speed.
  • The total mechanical energy (potential plus kinetic) in the system remains constant if we neglect resistive forces like friction.
By applying this principle, one can calculate roller coaster speeds and forces at different points, maintaining safety and ensuring thrilling rides. It is crucial to assume no energy loss due to friction to apply energy conservation straightforwardly.
Centripetal Force
A circular loop in a roller coaster requires centripetal force to keep the cars moving in a curve rather than a straight line.
  • Centripetal force acts towards the center of the circle along which the coaster is moving.
  • The formula for centripetal force is \( F_c = \frac{mv^2}{r} \), where \( m \) is mass, \( v \) is velocity, and \( r \) is the radius of the loop.
  • This force is vital at the loop's peak to ensure that the roller coaster and its occupants do not fall away from the track.
At the top of a loop, centripetal force is satisfied by the combination of gravitational force and additional force exerted by the seat. This ensures that the rider remains safely seated while experiencing a thrilling upside-down twist.
Potential Energy
Potential energy in a roller coaster is primarily gravitational, dependent on the height of the track.
  • It can be calculated with the equation \( PE = mgh \).
  • This type of energy is highest when the coaster is at the peak of the track, such as the initial hill before it begins its descent or any subsequent high points.
  • As the coaster descends, this stored energy becomes kinetic, propelling the coaster forward.
Potential energy is essential to explain how roller coasters have enough energy to travel through loops and reach their final destination safely.
Kinetic Energy
Kinetic energy represents the energy of motion that a roller coaster has as it speeds along the track.
  • It is calculated using the equation \( KE = \frac{1}{2} mv^2 \), where \( m \) is mass and \( v \) is velocity.
  • As a roller coaster descends from a high point, potential energy converts to kinetic, causing the coaster to accelerate.
  • The highest kinetic energy occurs when the coaster is at the bottom of descents due to maximum speed.
Understanding kinetic energy is critical for engineers to design safe loops and thrilling dips without compromising the ride鈥檚 safety. It plays a crucial role in making sure the coaster completes its track with the right amount of speed and force.

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Most popular questions from this chapter

A 1-oz bullet is traveling with a velocity of 1400 ft/s when it impacts and becomes embedded in a 5-lb wooden block. The block can move vertically without friction. Determine (a) the velocity of the bullet and block immediately after the impact, (b) the horizontal and vertical components of the impulse exerted by the block on the bullet.

A 1 -lb stone is dropped down the "bottomless pit" at Carlsbad Caverns and strikes the ground with a speed of 95 fts. Neglecting air resistance, \((a)\) determine the kinetic energy of the stone as it strikes the ground and the height \(h\) from which it was dropped. (b) Solve part \(a\) assuming that the same stone is dropped down a hole on the moon. (Acceleration of gravity on the moon \(=5.31 \mathrm{ft}( \left\mathrm{s}^{2}\).) \right.

A 5 -kg sphere is dropped from a height of \(y=2 \mathrm{m}\) to test newly designed spring floors used in gymnastics. The mass of the floor section is \(10 \mathrm{kg}\), and the effective stiffness of the floor is \(k=120 \mathrm{kN} / \mathrm{m}\). Knowing that the coefficient of restitution between the ball and the platform is \(0.6,\) detrmine \((a)\) the height \(h\) reached by the sphere after rebound, \((b)\) the maximum force in the springs.

A \(1400-\mathrm{kg}\) automobile starts from rest and travels \(400 \mathrm{m}\) during a performance test. The motion of the automobile is defined by the relation \(x=4000 \mathrm{ln}(\cosh 0.03 t),\) where \(x\) and \(t\) are expressed in meters and seconds, respectively. The magnitude of the aerodynamic drag is \(D=0.35 v^{2},\) where \(D\) and \(v\) are expressed in newtons and \(\mathrm{m} / \mathrm{s},\) respectively. Determine the power dissipated by the aerodynamic drag when \((a) t=10 \mathrm{s},(b) t=15 \mathrm{s}\)

A package is projected up a \(15^{\circ}\) incline at \(A\) with an initial velocity of \(8 \mathrm{m} / \mathrm{s}\). Knowing that the coefficient of kinetic friction between the package and the incline is \(0.12,\) determine \((a)\) the maximum distance \(d\) that the package will move up the incline, \((b)\) the velocity of the package as it returns to its original position.
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