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Chapter 2: Problem 169

An amusement ride called the "corkscrew" takes the passengers through the upside-down curve of a horizontal cylindrical helix. The velocity of the cars as they pass position \(A\) is \(15 \mathrm{m} / \mathrm{s}\), and the component of their acceleration measured along the tangent to the path is \(g \cos \gamma\) at this point. The effective radius of the cylindrical helix is \(5 \mathrm{m},\) and the helix angle is \(\gamma=40^{\circ} .\) Compute the magnitude of the acceleration of the passengers as they pass position \(A\)

Short Answer

Expert verified
The magnitude of the acceleration is approximately \(45.6 \text{ m/s}^2\).

Step by step solution

01

Identify Known Variables

We are given the velocity at position \(A\), which is \(v = 15\, \text{m/s}\). The tangent component of acceleration is \(g \cos \gamma\), where \(g = 9.81\, \text{m/s}^2\) and \(\gamma = 40^\circ\). The effective radius of the helix is \(R = 5\, \text{m}\).
02

Calculate Tangential Acceleration

Using the formula for the tangential acceleration component, \(a_t = g \cos \gamma\), we substitute the known values: \(a_t = 9.81 \cos 40^\circ\).
03

Compute Normal Acceleration

Normal acceleration \(a_n\) is computed using the formula for centripetal acceleration: \(a_n = \frac{v^2}{R}\). Substituting the known values: \(a_n = \frac{(15)^2}{5}\).
04

Evaluate Tangential and Normal Components

Calculate \(a_t\) using a calculator: \(a_t = 9.81 \times 0.766\), which simplifies to \(a_t \approx 7.51\, \text{m/s}^2\). Calculate \(a_n\) to find \(a_n = \frac{225}{5} = 45\, \text{m/s}^2\).
05

Calculate Total Acceleration Magnitude

The total acceleration is the vector sum of tangential and normal components: \(a = \sqrt{a_t^2 + a_n^2}\). Substitute the values: \(a = \sqrt{(7.51)^2 + (45)^2}\).
06

Simplify and Conclude

Calculate \((7.51)^2 = 56.45\) and \((45)^2 = 2025\). So, \(a = \sqrt{56.45 + 2025} = \sqrt{2081.45}\). Finally, \(a \approx 45.6\, \text{m/s}^2\).

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Key Concepts

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

Tangential Acceleration
Tangential acceleration is a measure of how fast the speed of an object is changing as it moves along a path. Imagine you are driving a car and you press down on the accelerator pedal; your car speeds up, creating tangential acceleration. In this case, it doesn't change the direction of the car, only how fast it is going.
For the amusement ride called the "corkscrew," the tangential acceleration at position A is determined by the formula:
  • The tangential component is given by: \( a_t = g \cos \gamma \)
  • Here, \( g \) is the acceleration due to gravity, typically \( 9.81\, \text{m/s}^2 \).
  • The angle \( \gamma \) represents the inclination of the helix, and in this scenario, \( \cos 40^\circ \) yields a value approximately equal to 0.766.
  • Calculating the tangential acceleration, we have \( a_t \approx 7.51\, \text{m/s}^2 \).
Thus, the tangential acceleration is purely responsible for changing the speed of the passengers along the ride, but not the direction in which they are moving.
Normal Acceleration
Normal acceleration, also known as radial or centripetal acceleration, is all about changing the direction of an object. When you're on a merry-go-round holding onto a horse, the normal acceleration keeps you moving in a circular path.
For the corkscrew ride, normal acceleration can be calculated to understand the forces acting on the passengers, which keep them in the loop:
  • The formula to find normal acceleration is \( a_n = \frac{v^2}{R} \).
  • Here, \( v \) is the constant velocity, valued at 15 m/s, and \( R \) is the effective radius of the helix, given as 5 m.
  • Substituting these values, we compute: \( a_n = \frac{225}{5} = 45\, \text{m/s}^2 \).
Normal acceleration is responsible for changing the direction as the corkscrew spins, ensuring that passengers remain on the intended path.
Cylindrical Helix
A cylindrical helix is a fascinating geometric shape that closely resembles the spirals or coils of a spring. Understanding this shape helps in comprehending the path followed by the amusement ride at position A.
Characteristics of a cylindrical helix:
  • It is a 3D curve that winds around a cylinder, like a screw. The ride uses this layout to loop and curve through the air.
  • The helix angle \( \gamma \) determines the steepness of the spiral. In this problem, \( \gamma = 40^\circ \) determines how tightly or loosely the helix is wound.
  • The effective radius, provided as 5 m, is crucial for calculating both the tangential and normal components of acceleration.
Such configurations are commonly employed in roller coasters to give thrilling experiences by skillfully combining twists and loops.

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Most popular questions from this chapter

A train which is traveling at \(80 \mathrm{mi} / \mathrm{hr}\) applies its brakes as it reaches point \(A\) and slows down with a constant deceleration. Its decreased velocity is observed to be \(60 \mathrm{mi} / \mathrm{hr}\) as it passes a point \(1 / 2 \mathrm{mi}\) beyond \(A\). A car moving at \(50 \mathrm{mi} / \mathrm{hr}\) passes point \(B\) at the same instant that the train reaches point \(A\) In an unwise effort to beat the train to the crossing, the driver "steps on the gas." Calculate the constant acceleration \(a\) that the car must have in order to beat the train to the crossing by 4 seconds and find the velocity \(v\) of the car as it reaches the crossing.

A test projectile is fired horizontally into a viscous liquid with a velocity \(v_{0} .\) The retarding force is proportional to the square of the velocity, so that the acceleration becomes \(a=-k v^{2}\). Derive expressions for the distance \(D\) traveled in the liquid and the corresponding time \(t\) required to reduce the velocity to \(v_{0} / 2 .\) Neglect any vertical motion.

The acceleration of a particle is given by \(a=-k s^{2}\) where \(a\) is in meters per second squared, \(k\) is a constant, and \(s\) is in meters. Determine the velocity of the particle as a function of its position \(s\). Evaluate your expression for \(s=5 \mathrm{m}\) if \(k=0.1 \mathrm{m}^{-1} \mathrm{s}^{-2}\) and the initial conditions at time \(t=0\) are \(s_{0}=3 \mathrm{m}\) and \\[ v_{0}=10 \mathrm{m} / \mathrm{s} \\]

The rectangular coordinates of a particle which moves with curvilinear motion are given by \(x=\) \(10.25 t+1.75 t^{2}-0.45 t^{3}\) and \(y=6.32+14.65 t-\) \(2.48 t^{2},\) where \(x\) and \(y\) are in millimeters and the time \(t\) is in seconds, beginning from \(t=0 .\) Determine the velocity \(\mathbf{v}\) and acceleration a of the particle when \(t=5\) s. Also, determine the time when the velocity of the particle makes an angle of \(45^{\circ}\) with the \(x\) -axis.

For the baseball of Prob. \(2 / 45\) thrown upward with an initial speed of \(30 \mathrm{m} / \mathrm{s}\), determine the time \(t_{u}\) from ground to apex and the time \(t_{d}\) from apex to ground.
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