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Chapter 6: Problem 44

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 terms of \(g\) ?

Short Answer

Expert verified
The acceleration is approximately \(9.66g\).

Step by step solution

01

Convert Speed to Meters per Second

First, convert the given speed from kilometers per hour to meters per second. Since there are 1000 meters in a kilometer and 3600 seconds in an hour, we use the conversion factor: \[1 \text{ km/h} = \frac{1000}{3600} \text{ m/s} = \frac{5}{18} \text{ m/s}\]Therefore, the speed in meters per second is:\[v = 96.6 \cdot \frac{5}{18} \approx 26.833 \text{ m/s}\]
02

Calculate the Centripetal Acceleration

Use the formula for centripetal acceleration, which is given by:\[a_c = \frac{v^2}{r}\]Substitute the values for \(v\) and \(r\):\[a_c = \frac{(26.833)^2}{7.6} \approx 94.75 \text{ m/s}^2\]
03

Express Acceleration in terms of g

To express the acceleration in terms of \(g\), where \(g \approx 9.81 \text{ m/s}^2\), divide the centripetal acceleration by \(g\):\[a_g = \frac{94.75}{9.81} \approx 9.66g\]

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Key Concepts

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

Bobsled Dynamics
Bobsled dynamics is a fascinating topic that deals with the motion of a bobsled as it moves through the course, encompassing everything from speed to navigation.
  • A crucial element of bobsled dynamics is understanding how the bobsled interacts with the track, especially during turns.
  • The forces acting on the bobsled, such as gravity, friction, and especially centripetal force, play significant roles in maintaining its speed and direction.
In the context of a bobsled run, maintaining control while achieving high speeds is key. During turns, the centripetal force is essential to keep the sled on the curved path. This force is generated by the friction between the sled's runners and the track, pulling the sled towards the center of the circle of the turn. Evaluating these dynamics helps athletes and coaches optimize bobsled design and strategize about the best path to take on the course to maximize speed and performance.
Circular Motion
Circular motion involves any object moving along a circular path, and it is a fundamental principle in the study of dynamics.
  • Key to understanding circular motion is the concept of centripetal acceleration, which is required to change the direction of velocity as an object travels along the path.
  • This acceleration is always directed towards the center of the circle and is necessary to keep an object moving in a curve rather than in a straight line.
For the bobsled, centripetal acceleration (\[a_c\]) can be calculated using the formula:\[a_c = \frac{v^2}{r}\]where \(v\) is the velocity and \(r\) is the radius of the turn. In the given example, the bobsled's velocity and the radius enable us to determine the force needed to keep it on its path.Understanding this concept aids in designing bobsled runs and improving performance by calculating the speed that can be achieved without losing track of control.
Conversion of Units
The conversion of units is critical when dealing with any scientific calculations, as it ensures accuracy and consistency in measurements.
  • In the context of the bobsled problem, the initial speed is given in kilometers per hour (\[\text{km/h}\]), a common unit for measuring speed.
  • However, to use it in the formula for centripetal acceleration, it needs to be expressed in meters per second (\[\text{m/s}\]).
To convert from \[\text{km/h}\] to \[\text{m/s}\], we use the conversion factor:\[1 \text{ km/h} = \frac{5}{18} \text{ m/s}\]Applying this factor allows for an accurate computation of velocity in a unit system compatible with the SI units used in physics formulas. This conversion step is pivotal as it avoids potential errors in calculations and provides results that are standardized and widely accepted.

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Most popular questions from this chapter

Calculate the magnitude of the drag force on a missile \(53 \mathrm{~cm}\) in diameter cruising at \(250 \mathrm{~m} / \mathrm{s}\) at low altitude, where the density of air is \(1.2 \mathrm{~kg} / \mathrm{m}^{3}\). Assume \(C=0.75\).

A student wants to determine the coefficients of static friction and kinetic friction between a box and a plank. She places the box on the plank and gradually raises one end of the plank. When the angle of inclination with the horizontal reaches \(30^{\circ}\), the box starts to slip, and it then slides \(2.5 \mathrm{~m}\) down the plank in \(4.0 \mathrm{~s}\) at constant acceleration. What are (a) the coefficient of static friction and (b) the coefficient of kinetic friction between the box and the plank?

Suppose the coefficient of static friction between the road and the tires on a car is \(0.60\) and the car has no negative lift. What speed will put the car on the verge of sliding as it rounds a level curve of \(30.5 \mathrm{~m}\) radius?

SSM A warehouse worker exerts a constant horizontal force of magnitude \(85 \mathrm{~N}\) on a \(40 \mathrm{~kg}\) box that is initially at rest on the horizontal floor of the warehouse. When the box has moved a distance of \(1.4 \mathrm{~m}\), its speed is \(1.0 \mathrm{~m} / \mathrm{s}\). What is the coefficient of kinetic friction between the box and the fioor?

A cat dozes on a stationary merry-go-round, at a radius of \(5.4\) \(\mathrm{m}\) from the center of the ride. Then the operator turns on the ride and brings it up to its proper turning rate of one complete rotation every \(6.0 \mathrm{~s}\). What is the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding?
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