Trning a high-speed train. A high speed train holds a constant speed of \(200 \mathrm{~km} / \mathrm{h}\). Your job is to design a \(90^{\circ}\) turn. Let us assume that you design the turn as a part of a circle. (a) Find an expression for the acceleration of the train while turning. (b) How large must the radius of the circle be for the acceleration to be smaller than \(0.1 \mathrm{~g}\), where \(g=9.8 \mathrm{~m} / \mathrm{s}^{2} ?\) (c) How long time does the turn take?

Centripetal acceleration occurs when an object moves in a circular path with a constant speed. This acceleration is always directed towards the center of the circle. The formula for centripetal acceleration, denoted as \( a \), is given by:
\[ a = \frac{v^2}{r} \]
Where:

• \( v \) is the linear speed of the object
• \( r \) is the radius of the circular path

In essence, the faster an object moves or the tighter the curve (smaller radius), the greater the centripetal acceleration required to keep the object on its circular path. For example, for our high-speed train scenario, as the train navigates the circular turn at a speed of \( 55.56 \, \mathrm{m/s} \), we can plug this value into the formula to determine the acceleration.

Objects in circular motion follow a path defined by a circle. The key aspect of circular motion is that even if the speed (magnitude of velocity) remains constant, the direction changes continuously, leading to an acceleration called centripetal acceleration.
To maintain this path, a central force acts towards the center of the circle. In the context of our high-speed train, the force could be due to the tracks and friction that continuously change the train’s direction, allowing it to follow the circular segment laid out.
The circular segment described in the problem is a quarter of a full circle, which illustrates that a 90-degree turn is one-fourth of the circle's entire circumference.

Speed and velocity are often given in different units. In physics, we typically need to convert them into standard SI units for calculations. Here, the problem provides the train's speed as \( 200 \, \mathrm{km/h} \). To convert this into \( \mathrm{m/s} \):

• Convert kilometers to meters: \( 200 \, \mathrm{km} = 200 \,000 \, \mathrm{m} \)
• Convert hours to seconds: \( 1 \, \mathrm{hour} = 3600 \, \mathrm{seconds} \)

So, the calculation becomes:
\[ 200 \, \mathrm{km/h} = \frac{200 \,000 \, \mathrm{m}}{3600 \, \mathrm{s}} \approx 55.56 \, \mathrm{m/s} \]

Using this conversion helps in applying physics formulas correctly, such as the formula for centripetal acceleration.

The radius of the circular path greatly impacts the centripetal acceleration. Given an allowable acceleration limit, we can find the minimum radius required. For instance, if the limit is set to less than \( 0.1 \times 9.8 \, \mathrm{m/s}^2 \), we first express it, giving:
\[ 0.1g = 0.1 \times 9.8 = 0.98 \, \mathrm{m/s}^2 \]

We substitute into the centripetal acceleration formula to find the radius, resulting in:
\[ 0.98 > \frac{(55.56)^2}{r} \]
By solving this inequality, we find:
\[ r > \frac{(55.56)^2}{0.98} \approx 3150.2 \, \mathrm{m} \]

Thus, for the centripetal acceleration to be under the given limit, the radius must be greater than approximately 3150.2 meters.

Time calculation in circular motion involves understanding the path covered and the speed of the object. The train's quarter turn is a part of a full circle, whose length is determined by the radius:
The circumference of a full circle is:
\[ 2\pi r \]
For a quarter turn:
\[ \text{Turn Length} = \frac{1}{4} \times 2\pi r = \frac{1}{2} \pi r \]
Inserting the calculated radius:
\[ \text{Turn Length} = \frac{1}{2} \times 3.1416 \times 3150.2 \approx 4953.1 \, \mathrm{m} \]

Using the train’s speed to find the turn time:
\[ t = \frac{\text{Turn Length}}{v} = \frac{4953.1}{55.56} \approx 89.2 \, \mathrm{s} \]

So, the train takes approximately 89.2 seconds to complete the 90-degree turn.