/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 101 An accelerometer \(C\) is mounte... [FREE SOLUTION] | 91Ó°ÊÓ

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An accelerometer \(C\) is mounted to the side of the roller-coaster car and records a total acceleration of \(3.5 g\) as the empty car passes the bottommost position of the track as shown. If the speed of the car at this position is \(215 \mathrm{km} / \mathrm{h}\) and is decreasing at the rate of \(18 \mathrm{km} / \mathrm{h}\) every second, determine the radius of curvature \(\rho\) of the track at the position shown.

Short Answer

Expert verified
The radius of curvature \(\rho\) at the position shown is approximately 105.08 meters.

Step by step solution

01

Convert Units

First, we need to convert the speed of the roller-coaster car from \(215 \mathrm{km/h}\) to \(\mathrm{m/s}\). Use the conversion factor \(1 \mathrm{km/h} = \frac{1000}{3600} \mathrm{m/s}\), thus:\[215 \mathrm{km/h} = 215 \times \frac{1000}{3600} \mathrm{m/s} \approx 59.72 \mathrm{m/s}\]
02

Convert Deceleration

Next, convert the deceleration from \(18 \mathrm{km/h/s}\) to \(\mathrm{m/s^2}\). Use the same conversion factor:\[18 \mathrm{km/h/s} = 18 \times \frac{1000}{3600} \mathrm{m/s^2} \approx 5 \mathrm{m/s^2}\]
03

Identify Forces Involved

At the bottommost position, the accelerometer records a total acceleration of \(3.5g\) where \(g = 9.81 \mathrm{m/s^2}\). Hence, the total acceleration \(a_t = 3.5 \times 9.81 \mathrm{m/s^2} = 34.335 \mathrm{m/s^2}\). Let the radial (centripetal) acceleration be \(a_r\) and tangential acceleration (deceleration) be \(a_t\), where \(a_t = 5 \mathrm{m/s^2}\).
04

Calculate Radial Acceleration

The total acceleration of \(34.335 \mathrm{m/s^2}\) is composed of the radial and tangential components. Since these are orthogonal, use:\[a_t^2 = a_r^2 + a_{ ext{tangential}}^2\]Substitute the values:\[34.335^2 = a_r^2 + 5^2\]Solving for \(a_r\):\[a_r^2 = 34.335^2 - 25\]\[a_r = \sqrt{34.335^2 - 25} \approx 33.96 \mathrm{m/s^2}\]
05

Solve for Radius of Curvature

The radial (centripetal) acceleration \(a_r\) is related to the velocity \(v\) and the radius of curvature \(\rho\) by the equation:\[a_r = \frac{v^2}{\rho}\]Solve for \(\rho\):\[\rho = \frac{v^2}{a_r}\]Substitute \(v = 59.72 \mathrm{m/s}\) and \(a_r = 33.96 \mathrm{m/s^2}\):\[\rho = \frac{59.72^2}{33.96} \approx 105.08 \mathrm{m}\]

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Key Concepts

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

Centripetal Acceleration
When an object moves in a circular path, it experiences acceleration directed towards the center of the circle. This is known as centripetal acceleration. Even if the speed of the object remains constant, the direction of its velocity vector changes. This change in velocity is what causes centripetal acceleration. Using the formula:
  • \( a_r = \frac{v^2}{\rho} \)
we can determine the magnitude of this acceleration, where \( v \) is the speed and \( \rho \) is the radius of curvature.

Centripetal acceleration is crucial in situations like cars moving around curves or roller-coasters maneuvering loops. This force enables objects to make turns. Without centripetal acceleration, objects would move off in a straight line rather than following a curved path. Understanding this concept is key for solving problems involving circular motion.
Radius of Curvature
The radius of curvature \( \rho \) describes the circular path's bending radius an object follows. In other words, it tells us how tight or loose a curve is at any particular point. A smaller radius indicates a more severe curve, while a larger radius indicates a gentler curve.Calculating the radius of curvature using centripetal acceleration involves knowing both the acceleration and the speed:
  • \( \rho = \frac{v^2}{a_r} \)
Here, \( v \) is the speed, and \( a_r \) is the radial (centripetal) acceleration.

In roller-coasters, the radius of curvature design impacts both safety and the thrill factor. A properly calculated radius ensures that the ride is exhilarating but still comfortable for passengers, avoiding excessive forces that could be unsafe or unpleasant.
Kinematics
Kinematics is the branch of mechanics concerned with objects in motion without considering the causes of motion. It involves studying the position, velocity, and acceleration of a moving object. In kinematics, we use different equations to predict the future position, velocity, or acceleration of an object:
  • \( v = u + at \)
  • \( s = ut + \frac{1}{2}at^2 \)
  • \( v^2 = u^2 + 2as \)
where \( u \) is initial velocity, \( v \) is final velocity, \( a \) is acceleration, \( s \) is displacement, and \( t \) is time.

In our roller-coaster example, kinematics helps us understand how it transitions through different sections of the track with varying speeds and accelerations. By applying these principles, we can determine and control the motion dynamics, ensuring the roller-coaster operates within safe and thrilling parameters.
Acceleration
Acceleration refers to the rate of change of velocity of an object. It can occur as changes in speed, direction, or both. There are different types of acceleration in motion, such as:
  • Linear Acceleration: A change in speed along a straight path, like a car speeding up on a highway.
  • Deceleration: A negative acceleration where the object slows down.
  • Centripetal Acceleration: Occurs when the direction changes but speed remains constant, as in circular motion.
In the exercise, the roller-coaster experiences both centripetal and tangential components of acceleration.

Understanding acceleration is crucial for calculating how quickly a system can respond to inputs, changes in conditions, and for designing systems that need precise control over speed and motion—like amusement park rides or automotive safety systems. Key equations help connect acceleration with forces, allowing engineering predictions and optimizations.

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Most popular questions from this chapter

A boy throws a ball upward with a speed \(v_{0}=12 \mathrm{m} / \mathrm{s}\) The wind imparts a horizontal acceleration of \(0.4 \mathrm{m} / \mathrm{s}^{2}\) to the left. At what angle \(\theta\) must the ball be thrown so that it returns to the point of release? Assume that the wind does not affect the vertical motion.

The car \(C\) increases its speed at the constant rate of \(1.5 \mathrm{m} / \mathrm{s}^{2}\) as it rounds the curve shown. If the magnitude of the total acceleration of the car is \(2.5 \mathrm{m} / \mathrm{s}^{2}\) at point \(A\) where the radius of curvature is \(200 \mathrm{m},\) compute the speed \(v\) of the car at this point.

The velocity of a particle is given by \(v=25 t^{2}-80 t-\) \(200,\) where \(v\) is in feet per second and \(t\) is in seconds. Plot the velocity \(v\) and acceleration \(a\) versus time for the first 6 seconds of motion and evaluate the velocity when \(a\) is zero.

The launching catapult of the aircraft carrier gives the jet fighter a constant acceleration of \(50 \mathrm{m} / \mathrm{s}^{2}\) from rest relative to the flight deck and launches the aircraft in a distance of \(100 \mathrm{m}\) measured along the angled takeoff ramp. If the carrier is moving at a steady 30 knots \((1 \mathrm{knot}=1.852 \mathrm{km} / \mathrm{h}),\) determine the magnitude \(v\) of the actual velocity of the fighter when it is launched.

The boom \(O A B\) pivots about point \(O,\) while section \(A B\) simultaneously extends from within section OA. Determine the velocity and acceleration of the center \(B\) of the pulley for the following conditions: \(\theta=20^{\circ}, \dot{\theta}=5\) deg/sec, \(\ddot{\theta}=2\) deg/sec \(^{2}, l=7 \mathrm{ft}\) \(i=1.5 \mathrm{ft} / \mathrm{sec}, \ddot{l}=-4 \mathrm{ft} / \mathrm{sec}^{2} .\) The quantities \(l\) and \(l\) are the first and second time derivatives, respectively, of the length \(l\) of section \(A B\).

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