Question

During an Olympic bobsled run, a European team takes a turn of radius \(25 \mathrm{ft}\) at a speed of \(60 \mathrm{mi} / \mathrm{h}\). What acceleration do the riders experience \((a)\) in \(\mathrm{ft} / \mathrm{s}^{2}\) and \((b)\) in units of \(g\) ?

Short answer

The riders experience an acceleration of \(a = 309.76 ft/s^2\) or equivalently \(9.62 g\).

Step-by-step solution

Convert speed from miles/hour to feet/second

First, convert the speed from miles per hour to feet per second. There are 5280 feet in one mile and 3600 seconds in one hour. So, \(v = 60 mi/hr * (5280 ft / 1 mi) * (1 hr / 3600 s) = 88 ft/s\).

Substitute values into the acceleration formula

Next, use the equation \(a = v^2/r\) to find the acceleration. Substitute \(v = 88 ft/s\) and \(r = 25 ft\) into the equation to get \(a = (88 ft/s)^2 / 25 ft = 309.76 ft/s^2\) .

Convert acceleration from \(ft/s^2\) to \(g\)

Finally, convert the acceleration from \(ft/s^2\) to \(g\). Since \(1 g = 32.2 ft/s^2\), divide 309.76 by 32.2 to get the acceleration in terms of \(g\). So, \(a_g = 309.76 ft/s^2 / 32.2 ft/s^2/g = 9.62 g\).

Key concepts

Centripetal Acceleration

When an object moves in a circular path, it experiences an acceleration towards the center of the circle. This is called *centripetal acceleration* and is essential for maintaining circular motion. Centripetal acceleration is calculated using the formula \( a = \frac{v^2}{r} \), where \( v \) is the velocity of the object and \( r \) is the radius of the circular path.

In the context of the bobsled problem, this acceleration ensures that the bobsled stays on its curved path as it goes around the turn. The formula tells us that the faster the bobsled goes or the sharper the turn (smaller radius), the greater this centripetal acceleration will be. This acceleration pulls the riders towards the inside of the turn, creating the sensation of being pushed against the outer wall of the bobsled.

Understanding centripetal acceleration is key to interpreting scenarios involving circular motion, such as the motions of vehicles on curved tracks or satellites orbiting planets.

Unit Conversion

Unit conversion is a critical skill in physics, enabling us to express measurements in different units without altering the underlying value. In the bobsled problem, we begin by converting the speed from miles per hour to feet per second. Understanding how to convert units is essential, as it ensures all expressions are consistent and compatible for calculations.

Here's a quick breakdown of the steps for converting miles per hour to feet per second:

• Recognize that 1 mile equals 5280 feet.
• Understand there are 3600 seconds in an hour.
• Use this information to convert 60 miles per hour to feet per second: \( 60 \text{ mi/hr} = 60 \times \frac{5280 \text{ ft}}{1 \text{ mi}} \times \frac{1 \text{ hr}}{3600 \text{ s}} = 88 \text{ ft/s} \).

Unit conversions like these are invaluable tools in physics, providing a foundation for working through problems accurately without being tripped up by mismatched units.

Radius of Curvature

The radius of curvature is a measure that describes how sharply a curve is turning. In the context of circular motion, it defines the radius of the circular path. This concept is crucial because it directly affects the centripetal acceleration an object experiences.

In the bobsled problem, the turn the bobsled takes has a radius of 25 feet. This radius of curvature implies how tightly the turn is. A smaller radius would mean a sharper turn, requiring a greater centripetal acceleration to keep the bobsled on its path. Conversely, a larger radius would mean a gentler curve with a reduced need for acceleration.

• Radius is directly inversely proportional to centripetal acceleration when speed is constant.
• In practical scenarios, managing the radius of curvature is important for safety and comfort.

Understanding radius of curvature helps in planning paths for racing tracks, roadways, and in designing fairground rides, ensuring both excitement and safety.