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Chapter 5: Problem 59

In a loop-the-loop roller coaster, show that a car moving too slowly would leave the track at an angle \(\phi\) given by \(\cos \phi=v^{2} / r g,\) where \(\phi\) is the angle made by a vertical line through the center of the circular track and a line from the center to the point where the car leaves the track.

Short Answer

Expert verified
The angle at which the roller coaster car would leave the track when it is not moving fast enough is given by \(\phi = \arccos \left( \frac{v^2}{rg} \right)\)

Step by step solution

01

Understanding the Problem

A key to solving this problem is the understanding of the motion of an object on a circular track. When the speed is too slow to keep the car on the loop, it will eventually leave the track. We can use the principle of conservation of energy to understand when this actually happens and where the car leaves the track, by equating gravitational and centripetal forces.
02

Setting Up the Equations

The force acting on the car as it goes around the loop is the centripetal force, and this is provided by the gravity when the car is just about to leave the track. Therefore, we have \(\cos \phi = \frac{v^2}{rg}\), where \(v\) is the velocity, \(r\) is the radius of the loop, and \(g\) is the acceleration due to gravity.
03

Solving the Equations

As we have a single equation, and we want to find the angle \(\phi\) at which the car loses contact with the track, we can simply rearrange the equation to solve for \(\phi\), resulting in \(\phi = \arccos \left( \frac{v^2}{rg} \right)\) .

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