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The drawing shows a version of the loop-the-loop trick for a small car. If the car is given an initial speed of \(4.0 \mathrm{~m} / \mathrm{s}\), what is the largest value that the radius \(r\) can have if the car is to remain in contact with the circular track at all times?

Short Answer

Expert verified
The largest radius the loop can have is approximately 0.163 m.

Step by step solution

01

Identify the Concept

The problem involves circular motion, specifically the minimum speed required at the highest point of the loop to maintain contact. At the top of the loop, the gravitational force provides the centripetal force needed to keep the car on the track.
02

Apply the Condition for Contact

For the car to remain in contact at the top of the loop, the gravitational force must equal the centripetal force. Therefore, we can use the equation \(mg = \frac{mv^2}{r}\), where \(m\) is the mass of the car, \(g\) is the acceleration due to gravity (\(9.8\, \text{m/s}^2\)), \(v\) is the speed at the top of the loop, and \(r\) is the radius of the loop.
03

Calculate Minimum Speed at the Top

Solve the equation \(mg = \frac{mv^2}{r}\) for \(v\). This gives \(v^2 = rg\) or \(v = \sqrt{rg}\). This is the minimum speed the car needs to maintain contact.
04

Use Conservation of Energy

Apply the conservation of mechanical energy principle. The initial kinetic energy plus initial potential energy equals the kinetic energy plus potential energy at the top of the loop: \(\frac{1}{2}mv_0^2 = \frac{1}{2}mv^2 + 2mgr\), where \(v_0 = 4.0 \, \text{m/s}\) is the initial speed, and the car has risen a height of \(2r\) at the top.
05

Solve for Radius

Substitute the expression for \(v^2\) at the top of the loop into the energy conservation equation: \(\frac{1}{2}m(4.0)^2 = \frac{1}{2}m(rg) + 2mgr\). Simplify and solve for \(r\): \(8 = rg + 4gr\). Simplifying further, \(8 = 5gr\). Substituting \(g = 9.8\) gives \(r = \frac{8}{5 \times 9.8}\).
06

Calculate the Radius

Evaluating the expression, \(r \approx \frac{8}{49}\) gives \(r \approx 0.163 \text{ m}\).

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Key Concepts

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

Loop-the-Loop Physics
When it comes to understanding loop-the-loop physics, imagine a car racing around a circular track, just like a roller coaster in an amusement park. The critical factor is the car's speed, especially at the highest point of the loop. Here, the pull of gravity must be carefully balanced with the car's speed to ensure it doesn't fall off the track.

The car starts off with a kinetic energy defined by its initial speed. As it ascends the loop, kinetic energy is gradually converted into potential energy until its speed at the top reaches a critical minimum. At this point, the car must still be moving fast enough to counteract gravity pulling it downwards. Hence, loop-the-loop physics centers around achieving this balance to maintain continuous contact with the track. Understanding this principle is key to safely designing and navigating a loop-the-loop scenario.
Centripetal Force
In loop-the-loop physics, centripetal force is essential to keep the car moving in a circular path. This force acts towards the center of the loop, ensuring that the car follows the circular track without falling out. At the top of the loop, the gravitational force acting downwards serves as this centripetal force.

To maintain contact with the track, the inward-directed gravitational force must equal the centripetal force needed to keep the car moving in the loop. This relationship can be understood through the formula:
  • Gravitational Force: \( mg \)
  • Centripetal Force: \( \frac{mv^2}{r} \)
Thus, for coherence, we set these forces equal at the highest point: \[ mg = \frac{mv^2}{r} \]
Here, \(m\) is the mass of the car, \(g\) is the acceleration due to gravity, \(v\) is the velocity, and \(r\) is the radius of the loop. Solving this at the highest point gives the car's minimum speed \(v = \sqrt{rg}\) to stay on the track.
Conservation of Energy
The conservation of energy is a compelling concept in physics, dictating how energy transitions from one form to another without being lost. For the loop-the-loop challenge, we start by considering both kinetic and potential energy.

Initially, the car's energy is entirely kinetic due to its speed. As it climbs the loop, some of this energy is converted into potential energy. At the top, the energy balance must still satisfy that the total energy remains constant, as shown through this equation:

\[ \frac{1}{2}mv_0^2 = \frac{1}{2}mv^2 + 2mgr \]
  • \(v_0\) represents the initial speed,
  • \(v\) is the speed at the top,
  • and \(2mgr\) accounts for the height of \(2r\) (diameter of the loop) lifted.
This mathematical expression reflects energy conservation, stringently dictating that initial energy should match the sum of potential and kinetic energies at the loop's peak. By rearranging and solving, we determine maximum possible loop radius \(r\) compatible with given initial speed, ensuring the car doesn't lose contact.

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