/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 47 A track is mounted on a large wh... [FREE SOLUTION] | 91Ó°ÊÓ

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A track is mounted on a large wheel that is free to turn with negligible friction about a vertical axis (Fig. \(11-48\) ). A toy train of mass \(m\) is placed on the track and, with the system initially at rest, the train's electrical power is turned on. The train reaches speed \(0.15\) \(\mathrm{m} / \mathrm{s}\) with respect to the track. What is the angular speed of the wheel if its mass is \(1.1 \mathrm{~m}\) and its radius is \(0.43 \mathrm{~m}\) ? (Treat the wheel as a hoop, and neglect the mass of the spokes and hub.)

Short Answer

Expert verified
The angular speed of the wheel is approximately 0.317 rad/s.

Step by step solution

01

Identify Given Variables and Formula

The problem states that the mass of the toy train is \( m \), the speed of the train is \( 0.15 \text{ m/s} \), the mass of the wheel (considered as a hoop) is \( 1.1m \), and the radius of the wheel is \( 0.43 \text{ m} \). We need to find the angular speed \( \omega \) of the wheel. To solve this, we use the conservation of angular momentum: \( L_{initial} = L_{final} \).
02

Calculate the Moment of Inertia of the Wheel

The wheel is treated as a hoop, so its moment of inertia \( I \) is given by \( I = M R^2 \), where \( M \) is the mass of the wheel and \( R \) is the radius. Substitute in the given values: \( I = 1.1m \times (0.43 \text{ m})^2 \).
03

Apply the Conservation of Angular Momentum

Initially, the system is at rest, so the initial angular momentum is zero. For the final angular momentum, we have contributions from both the train and the wheel. The angular momentum of the train is \( L_{train} = m v R \), and that of the wheel is \( L_{wheel} = I \omega \). By conservation of angular momentum, \( m v R = I \omega \).
04

Substitute Known Values and Solve for \(\omega\)

Using the equation from step 3, substitute \( v = 0.15 \text{ m/s} \), \( R = 0.43 \text{ m} \), and the expression for \( I \) from step 2. The equation becomes \( m \times 0.15 \times 0.43 = (1.1m \times 0.1849) \omega \). Simplifying, \( 0.0645m = 0.20339m \omega \). Thus, \( \omega = \frac{0.0645}{0.20339} \).
05

Calculate Angular Speed \(\omega\)

Solve the equation from step 4 to find \( \omega \): \( \omega \approx 0.317 \text{ rad/s} \).

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Key Concepts

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

Conservation of Angular Momentum
The conservation of angular momentum is a fundamental principle in physics. It states that if no external torque acts on a system, the total angular momentum remains constant. In simpler terms, this means: "When something is spinning and nothing is affecting it from the outside, its spin stays the same."
Angular momentum, often denoted by the symbol \( L \), is the product of an object's moment of inertia and its angular velocity.
In this exercise, initially both the toy train and wheel are stationary, so their total angular momentum is zero. As the train begins to move, the system's total angular momentum must remain zero because no external forces are acting on it. The train's movement generates momentum in one direction, while the wheel must move in the opposite direction to balance the momentum.
Applying this concept helps us find the angular speed of the wheel.
Moment of Inertia
The moment of inertia, symbolized by \( I \), is a measure of how difficult it is to change an object's rotational motion. It depends on the mass distribution relative to the axis of rotation. Simply put, it's like mass for rotation.
When a wheel is considered as a hoop, its mass is concentrated at a certain radius from the center of rotation. For a hoop, calculation of the moment of inertia is straightforward: \[ I = M R^2 \]where \( M \) is the mass of the wheel, and \( R \) is its radius.
In our example, calculating the wheel's moment of inertia is a crucial step. It helps understand how much resistance the wheel offers against changes to its motion and is pivotal in determining the resultant angular speed.
Angular Speed
Angular speed \( \omega \) defines how quickly something is spinning around an axis. It's similar to linear speed but for rotation. Typically measured in radians per second, angular speed can tell us how fast lines extending from the center of a rotating object are sweeping through space.
In our exercise, the train on the wheel leads to a situation where the wheel itself needs to spin to conserve angular momentum. Solving for \( \omega \) involved balancing the angular momentum contributions from the train and the wheel, derived through conservation laws. This provides the eventual angular speed of the wheel in radians per second, showcasing the tight interplay between linear motion and rotation in physics.
Physics Problem Solving
Solving physics problems often requires a structured, step-by-step approach:
  • First, identify given data and what you need to find. It's important to underline all known variables and the end goal.
  • Next, understand the involved concepts and formulas. For instance, recognize when to apply conservation laws.
  • Break the problem into smaller parts. Like calculating the moment of inertia separately before finding the angular speed.
  • Finally, insert known values into your formulas and solve step by step.
Using these strategies ensures that even complex physics problems become manageable and less intimidating.

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Most popular questions from this chapter

A solid brass ball of mass \(0.280 \mathrm{~g}\) will roll smoothly along a loop-the-loop track when released from rest along the straight section. The circular loop has radius \(R=14.0 \mathrm{~cm}\), and the ball has radius \(r \ll R\). (a) What is \(h\) if the ball is on the verge of leaving the track when it reaches the top of the loop? If the ball is released at height \(h=\) \(6.00 R\), what are the (b) magnitude and (c) direction of the horizontal force component acting on the ball at point \(Q\) ?

A wheel rotates clockwise about its central axis with an angular momentum of \(600 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\). At time \(t=0\), a torque of magnitude \(50 \mathrm{~N} \cdot \mathrm{m}\) is applied to the wheel to reverse the rotation. At what time \(t\) is the angular speed zero?

A uniform thin rod of length \(0.500 \mathrm{~m}\) and mass \(4.00 \mathrm{~kg}\) can rotate in a horizontal plane about a vertical axis through its center. The rod is at rest when a \(3.00 \mathrm{~g}\) bullet traveling in the rotation plane is fired into one end of the rod. As viewed from above, the bullet's path makes angle \(\theta=60.0^{\circ}\) with the rod (Fig. 11-50). If the bullet lodges in the rod and the angular velocity of the rod is \(10 \mathrm{rad} / \mathrm{s}\) immediately after the collision, what is the bullet's speed just before impact?

A constant horizontal force \(\vec{F}_{\text {app }}\) of magnitude \(12 \mathrm{~N}\) is applied to a uniform solid cylinder by fishing line wrapped around the cylinder. The mass of the cylinder is \(10 \mathrm{~kg}\), its radius is \(0.10 \mathrm{~m}\), and the cylinder rolls smoothly on the horizontal surface. (a) What is the magnitude of the acceleration of the center of mass of the cylinder? (b) What is the magnitude of the angular acceleration of the cylinder about the center of mass? (c) In unit-vector notation, what is the frictional force acting on the cylinder?

Is an overhead view of a thin uniform rod of length \(0.800 \mathrm{~m}\) and mass \(M\) rotating horizontally at angular speed \(20.0 \mathrm{rad} / \mathrm{s}\) about an axis through its center. A particle of mass \(M / 3.00\) initially attached to one end is ejected from the rod and travels along a path that is perpendicular to the rod at the instant of ejection. If the particle's speed \(v_{p}\) is \(6.00 \mathrm{~m} / \mathrm{s}\) greater than the speed of the rod end just after ejection, what is the value of \(v_{p}\) ?

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