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In a popular amusement park ride, a rotating cylin- der of radius \(3.00 \mathrm{~m}\) is set in rotation at an angular speed of \(5.00 \mathrm{rad} / \mathrm{s}\), as in Figure P7.75. The floor then drops away, leav- ing the riders suspended against the wall in a verti- cal position. What mini- mum coefficient of friction between a rider's clothing and the wall is needed to keep the rider from slipping?

Short Answer

Expert verified
The minimum coefficient of friction needed to prevent the rider from slipping is approximately 7.65.

Step by step solution

01

Consider the forces in play

In this scenario, the forces acting on a rider are the gravitational force (the weight of the person) and the frictional force which prevents them from slipping down. The normal force exerted by the wall provides the necessary centripetal force for circular motion. So, first write down the equations representing these forces.
02

Set up the equations

Since the frictional force must counteract gravity to prevent the rider from falling, we can set up the equation \(f = mg\), where \(m\) is the mass of the rider, \(g\) is the acceleration due to gravity, and \(f\) is the frictional force. Additionally, we know that the normal force from the wall provides the centripetal force necessary for circular motion: \(N = mR\omega^2\), where \(R\) is the radius of the cylinder and \(\omega\) is the angular speed. Here, \(N\) is also the normal force, and is the product of the coefficient of friction (\(\mu\)) and the frictional force (\(f\)). So, we can revise the previous equation to \(N = \mu mg\). Therefore, our system of two equations is \(f = mg\) and \(N = \mu mg = mR\omega^2\).
03

Solve for the coefficient of friction

We need to find \(\mu\), the coefficient of friction. Combining above two equations, we get \(\mu mg = mR\omega^2\). Solving for \(\mu\), we have: \(\mu = R\omega^2 / g\). Now we can substitute the given values: \(R = 3.00 \mathrm{~m}\), \(\omega = 5.00 \mathrm{rad} / \mathrm{s}\), and \(g = 9.8 \mathrm{~m/s^2}\).
04

Calculate the minimum coefficient of friction

Substituting the known values into the equation \(\mu = R\omega^2 / g\) gives: \(\mu = (3.00 \mathrm{~m} * (5.00 \mathrm{~rad/s})^2) / 9.8 \mathrm{~m/s^2}\). After calculating this, we get the minimum coefficient of friction needed to prevent the rider from slipping.

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Key Concepts

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

Centripetal Force
Centripetal force plays a vital role in the physics of amusement park rides. When a rider is inside a rotating cylinder, they experience circular motion. This motion requires a continuous inward force to keep the rider moving along the curved path.

This inward force is what we call the centripetal force. In the case of a rotating amusement park ride, the wall of the cylinder provides the necessary centripetal force. The normal force from the wall acts towards the center of the circular path, ensuring the rider maintains their circular trajectory.

Understanding centripetal force is key to solving any problem involving objects in circular motion, especially when considering the balance of forces that keep a rider safely against the wall.
Friction Coefficient
The friction coefficient (\(\mu\)) is a measure of how much frictional force can be exerted between two surfaces. In our amusement park scenario, the coefficient of friction between a rider's clothing and the wall determines whether the rider will remain safely suspended when the floor drops away.

Frictional force, calculated as the product of the coefficient of friction and the normal force (\(f = \mu N\)), is crucial to counteract the force of gravity acting on the rider. The higher this coefficient, the greater the frictional force, and the better the rider is held in place.

Calculating the minimum friction coefficient involves balancing the gravitational force downwards with the frictional force upwards, ensuring the rider won't slip.
Angular Motion
Angular motion describes the rotation of an object around a point or axis. In the context of our rotating cylinder ride, this motion is characterized by its angular speed. Angular speed (\(\omega\)) is the rate at which the object rotates and is usually measured in radians per second (rad/s).

As the cylinder spins, each rider moves along a circular path at a consistent angular speed. This uniform rotation implies that the centripetal force required to maintain this motion is directly proportional to the square of angular speed.

Thus, the angular motion not only defines how fast the ride spins but is also essential in calculating the forces involved, including those needed to keep the rider safely against the wall.
Cylindrical Motion
Cylindrical motion refers to the movement involved when an object travels along the inner surface of a cylindrical structure. In our amusement park ride, riders experience cylindrical motion as they cling to the walls of the rotating cylinder.

Much like angular motion, cylindrical motion is maintained by the meticulous balance of forces. The radius of the cylinder and the angular speed combine to create the conditions necessary for safe cylindrical motion.

As the cylinder rotates, the diameter and speed determine the centripetal force required. The normal force from the wall, coupled with the friction between the rider and the wall, keeps the rider from slipping downward, thereby achieving stable cylindrical motion.
Amusement Park Physics
Amusement park rides exemplify various physics principles, providing a real-world context for learning. Understanding how these principles apply helps one appreciate the forces at play and the safety measures employed.

In the rotating cylinder ride, for instance, we see the application of concepts like centripetal force, friction, and motion dynamics. Each element of the ride is carefully designed to maintain rider safety while providing thrills.

Learning about these principles is crucial for both understanding physics in a practical sense and for engineering designs that ensure safety and enjoyment in amusement park attractions.

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Most popular questions from this chapter

A \(45.0\)-cm diameter disk rotates with a constant angular acceleration of \(2.50 \mathrm{rad} / \mathrm{s}^{2}\). It starts from rest at \(t=\) 0 , and a line drawn from the center of the disk to a point \(P\) on the rim of the disk makes an angle of \(57.3^{\circ}\) with the positive \(x\)-axis at this time. At \(t=2.30 \mathrm{~s}\), find (a) the angular speed of the wheel, (b) the linear velocity and tangential acceleration of \(P\), and (c) the position of \(P\) (in degrees, with respect to the positive \(x\)-axis).

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A \(50.0\)-kg child stands at the rim of a merry-go-round of radius \(2.00 \mathrm{~m}\), rotating with an angular speed of \(3.00 \mathrm{rad} / \mathrm{s}\). (a) What is the child's centripetal acceleration? (b) What is the minimum force between her feet and the floor of the carousel that is required to keep her in the circular path? (c) What minimum coefficient of static friction is required? Is the answer you found reasonable? In other words, is she likely to stay on the merry-go-round?

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