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Chapter 9: Problem 44

A bicycle racer is going downhill at \(11.0 \mathrm{~m} / \mathrm{s}\) when, to his horror, one of his \(2.25 \mathrm{~kg}\) wheels comes off when he is \(75.0 \mathrm{~m}\) above the foot

Short Answer

Expert verified
The wheel's final speed at the base is approximately \(28.2 \text{ m/s}\).

Step by step solution

01

Understand the Problem

We need to find out what happens to the wheel after it comes loose from the bicycle. We'll consider its motion given its initial speed and height.
02

Identify the Known Variables

The initial speed of the wheel is given as \(11.0 \text{ m/s}\), the mass of the wheel is \(2.25 \text{ kg}\), and it loses contact at a height of \(75.0 \text{ m}\).
03

Determine the Total Mechanical Energy Initially

The wheel has both kinetic and potential energy. Kinetic energy \( KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 2.25 \times (11.0)^2\) and potential energy \( PE = mgh = 2.25 \times 9.8 \times 75\).
04

Establish the Conservation of Mechanical Energy Principle

As the wheel goes downhill, its total mechanical energy will remain constant (ignoring air resistance and friction). Thus, \( E_{initial} = E_{final} \).
05

Solve for the Final Speed at the Foot of the Hill

At the foot of the hill, the height will be zero, so all initial potential energy is converted to kinetic energy. Thus, \( KE_{final} = KE_{initial} + PE_{initial} \). \( \frac{1}{2}mv^2_{final} = \frac{1}{2}mv^2_{initial} + mgh \). Solve for \(v_{final}\).
06

Calculate the Initial Kinetic Energy

Calculate \( KE_{initial} = \frac{1}{2} \times 2.25 \times (11.0)^2 = 136.125 \text{ J}\).
07

Calculate the Potential Energy Before the Descent

Calculate \( PE_{initial} = 2.25 \times 9.8 \times 75 = 1653.75 \text{ J}\).
08

Calculate the Final Speed

Using the equation from Step 5, solve for \(v_{final}\): \( \frac{1}{2} \times 2.25 \times v^2_{final} = 136.125 + 1653.75 \). Simplify and solve: \( v^2_{final} = \frac{1789.875}{2.25} \). \( v_{final} \approx \sqrt{795.5} \approx 28.2 \text{ m/s}\).

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Key Concepts

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

Kinetic Energy
Kinetic Energy is all about the energy that a body possesses due to its motion. Whenever you see an object moving, think kinetic energy. The faster the object moves, the more kinetic energy it has. In our bicycle wheel example, when the wheel was moving downhill initially at 11.0 m/s, it had a certain amount of kinetic energy.To calculate this energy, we use the formula:
  • \( KE = \frac{1}{2}mv^2 \)
Where \( m \) is the mass of the object and \( v \) is its velocity.
In our case:
  • Mass \( m = 2.25 \) kg
  • Velocity \( v = 11.0 \) m/s
  • Thus, \( KE = \frac{1}{2} \times 2.25 \times (11.0)^2 = 136.125 \text{ J} \) Joules
Understanding kinetic energy helps us grasp how the motion of objects contributes to their overall energy state.
Potential Energy
Potential Energy is the energy that an object holds due to its position relative to other objects. Think of it as stored energy. A wheel perched high up has potential to fall due to gravity, and this gives it gravitational potential energy.For our wheel example, it started from a height of 75.0 meters above the ground. The formula to calculate gravitational potential energy is:
  • \( PE = mgh \)
Where \( m \) is mass, \( g \) is the gravity acceleration (around \( 9.8 \text{ m/s}^2 \) on Earth), and \( h \) is the height.
For the wheel:
  • Mass \( m = 2.25 \) kg
  • Height \( h = 75.0 \) m
  • So, \( PE = 2.25 \times 9.8 \times 75 = 1653.75 \text{ J} \) Joules
When the wheel rolls down, this potential energy gradually converts into kinetic energy.
Speed Calculation
Speed Calculation in this exercise revolves around using the conservation of energy principles. As the wheel descends the hill, its potential energy is converted into kinetic energy. Total mechanical energy remains the same (ignoring external forces like friction).Initially, combined energies (kinetic + potential) are needed:
  • Initial total energy: \( 136.125 \text{ J} + 1653.75 \text{ J} = 1789.875 \text{ J} \)
By conserving mechanical energy:
  • At the foot of the hill: kinetic energy should be same as this total \( KE_{final} = 1789.875 \text{ J} \)
Using kinetic energy formula again, solve for final speed \( v_{final} \):
  • \( \frac{1}{2} 2.25 v_{final}^2 = 1789.875 \)
  • Simplifies to \( v_{final}^2 = \frac{1789.875}{2.25} = 795.5 \)
  • \( v_{final} = \sqrt{795.5} \approx 28.2 \text{ m/s} \)
This shows how all potential energy converted, illustrating the maximum speed the wheel reaches as it touches down.

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Most popular questions from this chapter

What fraction of the total kinetic energy is rotational for the following objects rolling without slipping on a horizontal surface? (a) a uniform solid cylinder; (b) a uniform sphere; (c) a thin-walled, hollow sphere; (d) a hollow cylinder with outer radius \(R\) and inner radius \(R / 2\).

A solid uniform spherical stone starts moving from rest at the top of a hill. At the bottom of the hill, the ground curves upward, launching the stone vertically a distance \(H\) below its start. How high will the stone go (a) if there is no friction on the hill and (b) if there is enough friction on the hill for the stone to roll without slipping? (c) Why do you get two different answers even though the stone starts with the same gravitational potential energy in both cases?

A wheel turns with a constant angular acceleration of \(0.640 \mathrm{rad} / \mathrm{s}^{2}\). (a) How much time does it take for the wheel to reach an angular velocity of \(8.00 \mathrm{rad} / \mathrm{s},\) starting from rest? (b) Through how many revolutions does the wheel turn in this interval?

A solid uniform sphere and a uniform spherical shell, both having the same mass and radius, roll without slipping along a horizontal surface at a speed \(v\). They then encounter a hill that rises at an angle \(\theta\) above the horizontal. To what height \(h\) does each sphere roll before coming to rest?

II A flywheel with a radius of \(0.300 \mathrm{~m}\) starts from rest and accelerates with a constant angular acceleration of \(0.600 \mathrm{rad} / \mathrm{s}^{2}\). Compute the magnitude of the tangential acceleration, the radial acceleration, and the resultant acceleration of a point on its rim (a) at the start, (b) after it has turned through \(60.0^{\circ},\) and \((\mathrm{c})\) after it has turned through \(120.0^{\circ} .\)
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