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Chapter 6: Problem 64

During an Olympic bobsled run, the Jamaican team makes a turn of radius \(7.6 \mathrm{~m}\) at a speed of \(96.6 \mathrm{~km} / \mathrm{h}\). What is their acceleration in \(g\) -units? \((1 g\) -unit \(=\) \(\left.9.8 \mathrm{~m} / \mathrm{s}^{2} .\right)\)

Short Answer

Expert verified
The acceleration is 9.67 g-units.

Step by step solution

01

Convert Speed to Meters per Second

First, convert the speed from kilometers per hour to meters per second. Use the conversion factor: \[ 1 \text{ km/h} = \frac{1000 \text{ meters}}{3600 \text{ seconds}} = \frac{5}{18} \text{ m/s} \] Therefore, \[ 96.6 \text{ km/h} = 96.6 \times \frac{5}{18} \text{ m/s} = 26.83 \text{ m/s} \]
02

Calculate Centripetal Acceleration

Use the formula for centripetal acceleration, \[ a = \frac{v^2}{r} \] where \( v \) is the speed and \( r \) is the radius. Substituting the values: \[ a = \frac{(26.83 \text{ m/s})^2}{7.6 \text{ m}} = \frac{719.87}{7.6} \text{ m/s}^2 = 94.72 \text{ m/s}^2 \]
03

Convert Acceleration to g-units

To express the acceleration in g-units, divide by the value of one g-unit (9.8 \text{ m/s}^2). \[ a_{g} = \frac{94.72 \text{ m/s}^2}{9.8 \text{ m/s}^2} = 9.67 \text{ g-units} \]

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Key Concepts

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

Unit Conversion
Unit conversion is a crucial skill in physics. It allows you to switch between different measurement systems, making calculations consistent and straightforward.
For instance, in the given problem, we needed to convert the speed from kilometers per hour (km/h) to meters per second (m/s). We used the conversion factor where 1 km/h = \(\frac{1000 \text{ meters}}{3600 \text{ seconds}} = \frac{5}{18} \text{ m/s} \).
This conversion is essential as it standardizes the speed into a unit compatible with other measures in the problem, such as radius in meters.
Remember, performing accurate unit conversion is often the first step in solving many physics problems.
Physics Problems
Physics problems, like the one in the bobsled example, teach students how to apply formulas and theoretical knowledge to real-life scenarios. These problems often require:
  • Identifying known and unknown variables.
  • Making appropriate unit conversions.
  • Applying relevant physics principles and equations.
This problem involves finding the centripetal acceleration, which is an example of dynamic motion in a curved path. By breaking the problem into steps, such as converting units first, then calculating the desired quantity, and finally expressing it in the required format, students can systematically tackle even complex physics challenges.
g-Unit Conversion
Converting acceleration into g-units helps to relate the force experienced to Earth's gravitational force. This is particularly intuitive and useful in contexts like sports or aviation, where high accelerations are involved.
One g-unit is equivalent to the acceleration due to gravity on Earth, which is approximately \(9.8 \text{ m/s}^2 \).
In our problem, after computing the centripetal acceleration (\(94.72 \text{ m/s}^2 \)), we converted it into g-units by dividing the result by 9.8. This process made it easy to understand the intense force the bobsled team experienced during the turn, which was calculated to be about 9.67 g-units.
Centripetal Force
The concept of centripetal force is crucial for understanding motion in circular paths.
Centripetal force is the force that keeps an object moving in a curved path and is directed towards the center of rotation.
The formula for centripetal acceleration is: \[ a = \frac{v^2}{r} \]
where \(v\) is the velocity and \(r\) is the radius of the circular path.
In our example, the centripetal acceleration was determined using the bobsled's speed and the turn's radius. Calculating this force helps us understand the dynamics and stresses involved in high-speed turns, crucial for both safety and performance analysis.
Olympic Sports Physics
Physics plays a vital role in optimizing performance and ensuring safety in Olympic sports. Understanding concepts like centripetal force and acceleration is key.
In sports like bobsledding, athletes experience high accelerations and forces, especially during turns. By analyzing these forces, we can:
  • Improve sled design for better control and speed.
  • Enhance training regimens to condition athletes for high G-forces.
  • Ensure safety measures are in place to handle these intense conditions.
Studying the physics behind these movements provides valuable insights that can help push the boundaries of human performance and improve the overall spectacle and safety of athletic competitions.

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Most popular questions from this chapter

An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is \(5.0 \mathrm{kN}\), and the radius of the circle is \(10 \mathrm{~m}\). What are the magnitude and direction of the force of the boom on the car at the top of the circle if the car's speed there is (a) \(5.0 \mathrm{~m} / \mathrm{s}\) and (b) \(12 \mathrm{~m} / \mathrm{s}\) ?

A \(40 \mathrm{~kg}\) skier comes directly down a frictionless ski slope that is inclined at an angle of \(10^{\circ}\) with the horizontal while a strong wind blows parallel to the slope. Determine the magnitude and direction of the force of the wind on the skier if (a) the magnitude of the skier's velocity is constant, (b) the magnitude of the skier's velocity is increasing at a rate of \(1.0 \mathrm{~m} / \mathrm{s}^{2} .\) and \((\mathrm{c})\) the magnitude of the skier's velocity is increasing at a rate of \(2.0 \mathrm{~m} / \mathrm{s}^{2}\).

A block is projected up a frictionless inclined plane with initial speed \(v_{1}=3.50 \mathrm{~m} / \mathrm{s}\). The angle of incline is \(\theta=32.0^{\circ}\). (a) How far up the plane does it go? (b) How long does it take to get there? (c) What is its speed when it gets back to the bottom?

One day I was coming home late from work and stopped to pick up a pizza for dinner. I put the pizza box on the dashboard of my car and pushed it forward against the windshield and left against the steering wheel to prevent it from falling. (See Fig. 6-84.) Before I started driving, I realized that the box could still slide to the right or back toward the seat. When driving, do I have to worry more about it sliding when I turn left or when I turn right? Do I have to worry more when I speed up or when I slow down? Explain your answer in terms of the physics you have learned.

the terminal speed of a sky diver is \(160 \mathrm{~km} / \mathrm{h}\) in the spread-eagle position and \(310 \mathrm{~km} / \mathrm{h}\) in the nosedive position. Assuming that the diver's drag coefficient \(C\) does not change from one position to the other, find the ratio of the effective cross-sectional area \(A\) in the slower position to that in the faster position.
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