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A 2000 -kg Ferris wheel accelerates from rest to an angular speed of \(20 \mathrm{rad} / \mathrm{s}\) in \(12 \mathrm{~s}\). Approximate the Ferris wheel as a circular disk with a radius of \(30 \mathrm{~m}\). What is the net torque on the wheel?

Short Answer

Expert verified
The net torque on the Ferris wheel is \( 1,500,000 \text{ N} \cdot \text{m} \).

Step by step solution

01

Identify Known Values

First, let's list out the known values from the problem statement:- Mass of the Ferris wheel, \( m = 2000 \text{ kg} \).- Radius of the Ferris wheel, \( r = 30 \text{ m} \).- Final angular speed, \( \omega_f = 20 \text{ rad/s} \).- Time to reach final angular speed, \( t = 12 \text{ s} \).- Initial angular speed, \( \omega_i = 0 \text{ rad/s} \).
02

Calculate Angular Acceleration

To find the angular acceleration, we use the formula: \[ \alpha = \frac{\omega_f - \omega_i}{t} \]Substitute the known values into the formula:\[ \alpha = \frac{20 \text{ rad/s} - 0}{12 \text{ s}} = \frac{20}{12} \text{ rad/s}^2 = \frac{5}{3} \text{ rad/s}^2 \].
03

Compute the Moment of Inertia

The moment of inertia \( I \) for a circular disk is calculated using:\[ I = \frac{1}{2} m r^2 \]Substitute the given values:\[ I = \frac{1}{2} \times 2000 \text{ kg} \times (30 \text{ m})^2 \]\[ I = 900,000 \text{ kg} \cdot \text{m}^2 \].
04

Find the Net Torque Using Angular Acceleration

Net torque \( \tau \) is given by the formula:\[ \tau = I \alpha \]Using the previously calculated values:\[ \tau = 900,000 \text{ kg} \cdot \text{m}^2 \times \frac{5}{3} \text{ rad/s}^2 \]\[ \tau = 1,500,000 \text{ N} \cdot \text{m} \].

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Key Concepts

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

Angular Acceleration
Angular acceleration is an important concept in circular motion. It tells us how quickly an object is speeding up or slowing down its rotation. Imagine you are spinning a toy top. At first, you spin it slow and then a bit faster. Angular acceleration measures how fast this change in spinning speed happens.
To find angular acceleration, we use the formula \( \alpha = \frac{\omega_f - \omega_i}{t} \). Here, \( \omega_f \) is the final angular speed, \( \omega_i \) is the initial angular speed, and \( t \) is the time it takes for this change.
This helps us understand the dynamics involved in rotational systems like Ferris wheels or car tires. When a Ferris wheel starts from a standstill and reaches a certain speed, it's angular acceleration that tells us how quickly this transition occurs.
Moment of Inertia
The moment of inertia is like the mass of objects in linear motion but for rotation. It represents how much torque, or rotational force, is needed to get an object spinning around an axis. Think of it as an indicator of rotational stubbornness.
For a Ferris wheel, which we often approximate as a solid disc, the moment of inertia is calculated using \( I = \frac{1}{2} m r^2 \). Here, \( m \) is the mass and \( r \) is the radius.
- A larger mass or a bigger radius means a larger moment of inertia, making it harder to start or stop the spinning. - Just like how heavier objects are harder to push in a straight line, larger or heavier wheels take more effort to spin.
Circular Motion
Circular motion is when something moves along a circular path. Anything moving in a loop, like a Ferris wheel, is experiencing circular motion. Typically, such objects are subject to forces that keep them on their path.
In circular motion, several unique dynamics come into play:
  • Centripetal Force: This keeps the object moving in the circle and prevents it from flying off the path.
  • Angular Speed: The rate at which the object travels around the circle.
Learning how these forces work together helps to understand what keeps structures like Ferris wheels safe and functioning.
Ferris Wheel Dynamics
Ferris wheel dynamics combine all principles of physics for circular motion into one structured experience. As the Ferris wheel turns, it offers more than just a thrilling ride. Understanding its dynamics provides insights on how it's engineered for both safety and fun.
Several key principles include:
  • Structural Design: Designing a Ferris wheel involves knowing the right materials and design to handle the immense forces.
  • Stability: Ensuring rides remain stable, even in strong winds or when fully loaded with people.
  • Torque and Acceleration: Calculating how powerful motors need to be to start, stop, and maintain a consistent speed.
By understanding these dynamics, engineers can design Ferris wheels that are both thrilling and reliable.

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Most popular questions from this chapter

A stationary ice skater with a mass of \(80.0 \mathrm{~kg}\) and a moment of inertia (about her central vertical axis) of \(3.00 \mathrm{~kg} \cdot \mathrm{m}^{2}\) catches a baseball with her outstretched arm. The catch is made at a distance of \(1.00 \mathrm{~m}\) from the central axis. The ball has a mass of \(145 \mathrm{~g}\) and is traveling at \(20.0 \mathrm{~m} / \mathrm{s}\) before the catch. (a) What linear speed does the system (skater \(+\) ball) have after the catch? (b) What is the angular speed of the system (skater \(+\) ball) after the catch? (c) What percentage of the ball's initial kinetic energy is lost during the catch? Neglect friction with the ice.

A wheel rolls uniformly on level ground without slipping. A piece of mud on the wheel flies off when it is at the 9 o'clock position (rear of wheel). Describe the subsequent motion of the mud.

An ice skater has a moment of inertia of \(100 \mathrm{~kg} \cdot \mathrm{m}^{2}\) when his arms are outstretched and a moment of inertia of \(75 \mathrm{~kg} \cdot \mathrm{m}^{2}\) when his arms are tucked in close to his chest. If he starts to spin at an angular speed of 2.0 rps (revolutions per second) with his arms outstretched, what will his angular speed be when they are tucked in?

A cylinder with a diameter of \(20 \mathrm{~cm}\) rolls with an angular speed of \(0.050 \mathrm{rad} / \mathrm{s}\) on a level surface. If the cylinder experiences a uniform tangential acceleration of \(0.018 \mathrm{~m} / \mathrm{s}^{2}\) without slipping until its angular speed is \(1.2 \mathrm{rad} / \mathrm{s},\) through how many complete revolutions does the cylinder rotate during the time it accelerates?

In a circus act, a uniform board (length \(3.00 \mathrm{~m}\), mass \(35.0 \mathrm{~kg}\) ) is suspended from a bungie-type rope at one end, and the other end rests on a concrete pillar. When a clown (mass \(75.0 \mathrm{~kg}\) ) steps out halfway onto the board, the board tilts so the rope end is \(30^{\circ}\) from the horizontal and the rope stays vertical. (a) In which situation will the rope tension be larger: (1) the board without the clown on it, (2) the board with the clown on it, or (3) you can't tell from the data given? (b) Calculate the force exerted by the rope in both situations.

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