Question

The rotor is an amusement park ride where people stand against the inside of a cylinder. Once the cylinder is spinning fast enough, the floor drops out. (a) What force keeps the people from falling out the bottom of the cylinder? (b) If the coefficient of friction is 0.40 and the cylinder has a radius of \(2.5 \mathrm{m},\) what is the minimum angular speed of the cylinder so that the people don't fall out? (Normally the operator runs it considerably faster as a safety measure.) (IMAGE CAN'T COPY)

Short answer

The friction between the person and the wall keeps them from falling, with a minimum angular speed of approximately 3.13 rad/s.

Step-by-step solution

Understand the Force Involved

To prevent people from falling when the floor drops, a force should provide the necessary counteraction. In this case, the frictional force between the person and the wall of the spinning cylinder prevents them from falling. The cylindrical motion generates a normal force pointing perpendicular to the wall, and this normal force produces the friction needed.

Determine the Force of Friction

The frictional force required to keep individuals from falling is equal to their weight when the ride is spinning at minimum speed. This means that \( f_{friction} = mg \), where \( m \) is the mass of a person and \( g \) is the gravitational acceleration \( (9.8 \text{ m/s}^2) \).

Frictional Force Formula

The frictional force \( f_{friction} \) is given by \( \mu N \), where \( \mu \) is the coefficient of friction and \( N \) is the normal force. Set this equal to the weight: \( \mu N = mg \).

Relate Normal Force to Centripetal Force

Circular motion creates a centripetal force which equals \( N = m r \omega^2 \), where \( r \) is the radius and \( \omega \) is the angular speed in radians per second. Thus, replacing \( N \), we have \( \mu m r \omega^2 = mg \).

Solve for Angular Speed

Divide both sides by \( m \) and rearrange to solve for \( \omega \): \( \mu r \omega^2 = g \). Therefore, \( \omega = \sqrt{\frac{g}{\mu r}} \).

Plug in Values and Calculate

Substitute the given values: \( \mu = 0.40 \), \( r = 2.5 \text{ m} \), and \( g = 9.8 \text{ m/s}^2 \) into the formula: \( \omega = \sqrt{\frac{9.8}{0.40 \times 2.5}} \). Calculate this to find \( \omega \approx 3.13 \text{ rad/s} \).

Key concepts

Centripetal Force

Centripetal force is a fundamental concept in the physics of amusement rides, acting as the "center-seeking" force that keeps riders moving in a circular path. It is essential for maintaining circular motion, preventing objects from flying outwards. In the Rotor ride, centripetal force is generated by the rotation of the cylinder. This force pushes the riders against the wall of the cylinder, forming a balance with the frictional force that prevents them from falling. The normal force, which is crucial for friction, actually arises from this inward centripetal force. Without it, the necessary friction to keep riders from slipping down would be impossible to maintain.

In mathematical terms, centripetal force can be expressed as \( F_{c} = m r \omega^2 \), where \( m \) is the mass of the rider, \( r \) is the radius of the circular path, and \( \omega \) is the angular speed. This equation highlights how the speed of rotation and radius directly affect the force experienced by the rider.

Frictional Force

Frictional force plays a vital role in amusement rides like the Rotor. It is the force that resists the sliding of riders down the wall when the floor drops out. The frictional force is directly proportional to the normal force (which, as we mentioned, comes from centripetal force) and is calculated with the formula \( f_{friction} = \mu N \). Here, \( \mu \) represents the coefficient of friction, a measure of the stickiness between surfaces. This coefficient is crucial to ensuring safety on the ride.

For the Rotor, we require that the frictional force equals the gravitational pull on the riders. That is, \( f_{friction} = mg \), balancing the downward gravitational force with the upward frictional resistance. When designing such rides, operators choose materials and speeds that ensure \( \mu N \) is always greater than or equal to weight. This guarantees that riders cling safely to the wall.

Angular Speed

Angular speed is a measure of how quickly something spins around a central point. For the Rotor ride, angular speed determines how fast the cylinder rotates—and ultimately, how strong the centripetal force will be. Measured in radians per second (rad/s), angular speed connects with other aspects like radius and mass to define the dynamics of the ride.

The key relation involving angular speed is rooted in balancing forces. The frictional force must equal the gravitational force for riders to stay put when the floor disappears. By substituting forces with measurable quantities, we arrive at \( \mu r \omega^2 = g \), ultimately solving for angular speed: \( \omega = \sqrt{\frac{g}{\mu r}} \). This formula shows how angular speed decreases with increasing friction coefficient and radius, ensuring rides can operate safely at the minimum speed necessary.

Circular Motion

Circular motion is the underlying principle of many amusement park rides, defining the path through which riders travel. In the context of the Rotor, circular motion underpins the entire operation, with riders experiencing forces continually directed towards the center of the cylinder. This is what allows them to stick to the wall and not fall when the floor drops out.

Understanding circular motion involves knowing that even at constant speed, velocity is not constant—because the direction changes continually. This direction change requires a continuous inward force, provided by centripetal force, which in practical terms is exerted by the wall of the rotor. These dynamics explain how riders experience what feels like a gravitational pull outward as they are pushed into the wall, even though it is the wall pushing back on them to maintain their circular path.