Question

Jack sits in the chair of a Ferris wheel that is rotating at a constant 0.100 rev/s. As Jack passes through the highest point of his circular path, the upward force that the chair exerts on him is equal to one-fourth of his weight. What is the radius of the circle in which Jack travels? Treat him as a point mass.

Short answer

The radius of the Ferris wheel is approximately 24.8 meters.

Step-by-step solution

Understand the Problem

Jack is on a Ferris wheel rotating at 0.100 revolutions per second. At the top of the wheel, the force exerted by the chair on Jack is one-fourth of his weight. We need to find the radius of the wheel.

Express the Force Condition Mathematically

At the highest point, the upward normal force exerted by the chair is given as one-fourth of the gravitational force acting on Jack. Thus, we can write: \( N = \frac{mg}{4} \), where \( N \) is the normal force and \( mg \) is Jack's weight.

Set Up the Circular Motion Equation

At the top of the circle, the centripetal force needed for Jack's circular motion is provided by the gravitational force minus the normal force: \( mg - N = m\frac{v^2}{r} \). Substitute \( N = \frac{mg}{4} \) into the equation: \( mg - \frac{mg}{4} = m\frac{v^2}{r} \).

Simplify the Equation

Simplify the equation: \( mg - \frac{mg}{4} = m\frac{v^2}{r} \) becomes \( \frac{3mg}{4} = m\frac{v^2}{r} \). Cancel mass \( m \) from both sides: \( \frac{3g}{4} = \frac{v^2}{r} \).

Calculate the Velocity

The speed \( v \) of Jack can be expressed in terms of the rotation rate: \( v = ωr \). Here, \( ω = 0.100 \times 2π \) radians per second. Thus, \( v = 0.2πr \).

Solve for the Radius

Substitute \( v = 0.2πr \) into \( \frac{3g}{4} = \frac{v^2}{r} \): \( \frac{3g}{4} = \frac{(0.2πr)^2}{r} \). This simplifies to \( \frac{3g}{4} = 0.04π^2r \). Rearrange to isolate \( r \): \( r = \frac{3g}{4 imes 0.04π^2} \).

Substitute Known Values and Solve

Substitute \( g = 9.8 \text{ m/s}^2 \) into the equation: \( r = \frac{3 imes 9.8}{4 imes 0.04π^2} \). Calculate the value of \( r \) to find the radius.

Key concepts

Centripetal Force

Centripetal force is what keeps an object moving in a circular path. It's the inward force required for circular motion and always points towards the circle's center. In Jack's case on the Ferris wheel, it's crucial to understand that the centripetal force at play is what keeps him moving in a circle at the consistent rotation speed of the Ferris wheel.

For an object to move in a circle, the centripetal force must be equal to the object's mass times the square of its velocity divided by the radius of the circle:

• Formula: \( F_c = \frac{mv^2}{r} \).

At the top of the Ferris wheel, this force is provided by the combination of Jack's weight (\( mg \), acting downward) and the normal force (\( N \), acting upward).

• Therefore, the centripetal force equation used in solving Jack’s exercise became \( mg - N = m\frac{v^2}{r} \).

This equation helps determine the radius of the Ferris wheel by linking it to Jack's speed and the forces acting on him.

Ferris Wheel Physics

Ferris wheels are an excellent example of uniform circular motion in physics. As they rotate, the physics involved illustrates how forces act on riders throughout the ride.

When considering a Ferris wheel, the key aspect is the continuous circular motion at a stable speed. The physics encourages students to think about:

• The forces at play at different points, especially at the top and bottom of the wheel.
• How the feeling of weightlessness or increased heaviness occurs due to changes in these forces.

In Jack's scenario, his experience at the highest point of the Ferris wheel involves the normal force being one-quarter of his weight, showing how gravitational force dominates at the top. This use of a Ferris wheel showcases physics concepts like acceleration, angular velocity, and centripetal force in a real-world context.

Normal Force

Normal force is the force perpendicular to the object's surface, counteracting forces like gravity. For instance, when you sit in a chair, the normal force from the chair holds you up against gravity.

On a Ferris wheel, normal force fluctuates based on position:

• At the top, it combines with gravity to govern the centripetal motion.
• At the bottom, it acts against gravity, influencing how heavy or light a rider feels.

In Jack's problem, we observe the normal force is only one-fourth of his weight at the top of the Ferris wheel. This differential in force distribution is crucial, because it impacts how Jack feels as he moves, and plays into the calculation of the Ferris wheel's radius by affecting the centripetal force equation. Understanding normal force helps clarify how objects interact with surfaces during motion.

Angular Velocity

Angular velocity is a measure of how quickly an object rotates around a circle. It is typically expressed in radians per second, which defines the speed of rotation as an angle covered per unit of time.

In Jack's case, the Ferris wheel has a constant angular velocity of 0.100 revolutions per second. To work with equations involving velocity, it can be converted to radians per second:

• Formula for conversion: \( \omega = 0.100 \times 2\pi \)
• This equals roughly 0.2\pi radians per second.

Angular velocity is key to converting linear to angular measurements, bridging the gap between the circular path and the forces involved. It directly relates to the linear velocity, using the relationship \( v = \omega r \), where \( v \) is linear velocity and \( r \) is the radius. With angular velocity, students see the link between circular paths and rotational speed.