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A horizontal plywood disk with mass 7.00 \(\mathrm{kg}\) and diameter 1.00 \(\mathrm{m}\) pivots on frictionless bearings about a vertical axis through its center. You attach a circular model-railroad track of negligible mass and average diameter 0.95 m to the disk. A 1.20 -kg, battery-driven model train rests on the tracks. To demonstrate conservation of angular momentum, you switch on the train's engine. The train moves counterclockwise, soon attaining a constant speed of 0.600 \(\mathrm{m} / \mathrm{s}\) relative to the tracks. Find the magnitude and direction of the angular velocity of the disk relative to the earth.

Short Answer

Expert verified
The disk rotates clockwise with an angular velocity of 0.387 rad/s.

Step by step solution

01

Understand the Initial Conditions

We start with a horizontal plywood disk of mass \(7.00 \, \text{kg}\) and diameter \(1.00 \, \text{m}\), thus leading to a radius \(R_d = 0.50 \, \text{m}\). A train of mass \(1.20 \, \text{kg}\) is on a track with average diameter \(0.95 \, \text{m}\), so its radius \(R_t = 0.475 \, \text{m}\). Initially, the system is at rest.
02

Calculate the Moment of Inertia

The moment of inertia of the disk \(I_d\) is given by the formula for a circular disk: \[ I_d = \frac{1}{2} M R_d^2 \]Plugging in the values: \[ I_d = \frac{1}{2} \times 7.00 \, \text{kg} \times (0.50 \, \text{m})^2 = 0.875 \, \text{kg} \, \text{m}^2 \]
03

Angular Momentum of the Train

When the train reaches a speed of \(0.600 \, \text{m/s}\), its linear velocity \(v_t\) is related to its angular velocity \(\omega_t\) by the relation \(v_t = R_t \omega_t\). Solving for \(\omega_t\):\[ \omega_t = \frac{v_t}{R_t} = \frac{0.600 \, \text{m/s}}{0.475 \, \text{m}} = 1.263 \, \text{rad/s} \]The angular momentum of the train \(L_t\) is given by:\[ L_t = m_t R_t^2 \omega_t = 1.20 \, \text{kg} \times (0.475 \, \text{m})^2 \times 1.263 \, \text{rad/s} = 0.339 \, \text{kg} \, \text{m}^2/\text{s} \]
04

Apply Conservation of Angular Momentum

Since the system was initially at rest and is closed, the conservation of angular momentum states that the sum of angular momentum is zero:\[ L_d + L_t = 0 \]Where \(L_d\) is the angular momentum of the disk with angular velocity \(\omega_d\). This gives us:\[ I_d \omega_d + L_t = 0 \]Solving for \(\omega_d\):\[ \omega_d = -\frac{L_t}{I_d} = -\frac{0.339 \, \text{kg} \, \text{m}^2/\text{s}}{0.875 \, \text{kg} \, \text{m}^2} = -0.387 \, \text{rad/s} \]
05

Determine the Direction

The negative sign indicates that the direction of the disk's angular velocity is opposite to that of the train's motion. Thus, if the train moves counterclockwise, the disk must rotate clockwise relative to the Earth.

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Key Concepts

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

Moment of Inertia
The moment of inertia is an essential concept in understanding rotation. It tells us how difficult it is to change an object's rotation rate. Think of it as the rotational equivalent of mass in linear motion.
It depends on how the mass is distributed in relation to the axis of rotation. For a circular disk, like our plywood disk, the formula is:
  • \( I_d = \frac{1}{2} M R_d^2 \)
Here, \(M\) is the mass and \(R_d\) is the radius of the disk.
In our example, substituting the mass and radius gives \(I_d = 0.875 \, \text{kg m}^2\). This tells us how the disk would resist changes to its rotational speed.
When multiple objects interact, like our disk and the train on a circular track, each has its moment of inertia. This value is critical for calculating angular momentum and understanding how they influence each other in rotational motion.
Angular Velocity
Angular velocity is a measure of how fast an object rotates or revolves relative to another point. It's the rotational equivalent of linear velocity. In our exercise, it's used to describe how quickly the train and the disk rotate.
The train's angular velocity, \( \omega_t \), is determined from its linear speed \( v_t \) and track radius \( R_t \):
  • \( v_t = R_t \omega_t \)
  • \( \omega_t = \frac{v_t}{R_t} \)
For the train moving at \(0.600 \, \text{m/s}\) on a \(0.475 \, \text{m}\) radius track, \(\omega_t\) evaluated to \(1.263 \, \text{rad/s}\).
The disk's angular velocity, \( \omega_d \), was found using conservation laws, calculated as \(-0.387 \, \text{rad/s}\), indicating it rotates opposite to the train. Angular velocities are crucial as they help understand how different parts of a system move relative to each other.
Angular Momentum
Angular momentum is a pivotal concept reflecting the rotational motion's quantity, similar to how linear momentum reflects straight motion. It's the product of an object's moment of inertia and its angular velocity.
In rotational systems, like our disk-and-train setup, it's essential for understanding equilibrium and dynamics. The angular momentum of the train was:
  • \( L_t = m_t R_t^2 \omega_t \)
Solving gives \(0.339 \, \text{kg} \, \text{m}^2/\text{s}\) for the train. According to the conservation of angular momentum:
  • \( L_d + L_t = 0 \)
This means no external torques act, so the initial and final angular momentum is balanced. As the train moves counterclockwise, it gains positive angular momentum, making the disk rotate clockwise to maintain the balance of the system. This exchange and conservation highlight how connected components of a system react to each other's motions. Understanding this can help solve complex problems involving rotational motion in physics.

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Most popular questions from this chapter

In a lab experiment you let a uniform ball roll down a curved track. The ball starts from rest and rolls without slipping.While on the track, the ball descends a vertical distance \(h .\) The lower end of the track is horizontal and extends over the edge of the lab table; the ball leaves the track traveling horizontally. While free-falling after leaving the track, the ball moves a horizontal distance \(x\) and a vertical distance \(y\) . (a) Calculate \(x\) in terms of \(h\) and \(y\) ignoring the work done by friction. (b) Would the answer to part (a) be any different on the moon? (c) Although you do the experiment very carefully, your measured value of \(x\) is consistently a bit smaller than the value calculated in part (a). Why? (d) What would \(x\) be for the same \(h\) and \(y\) as in part (a) if you let a silver dollar roll down the track? You can ignore the work done by friction.

A target in a shooting gallery consists of a vertical square wooden board, 0.250 \(\mathrm{m}\) on a side and with mass 0.750 \(\mathrm{kg}\) , that pivots on a horizontal axis along its top edge. The board is struck face-on at its center by a bullet with mass 1.90 \(\mathrm{g}\) that is traveling at 360 \(\mathrm{m} / \mathrm{s}\) and that remains embedded in the board. (a) What is the angular speed of the board just after the bullet's impact? (b) What maximum height above the equilibrium position does the center of the board reach before starting to swing down again? (c) What minimum bullet speed would be required for the board to swing all the way over after impact?

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A 2.00 -kg textbook rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is \(0.150 \mathrm{m},\) to a hanging book with mass 3.00 \(\mathrm{kg}\) . The system is released from rest, and the books are observed to move 1.20 \(\mathrm{m}\) in 0.800 s. (a) What is the tension in each part of the cord? (b) What is the moment of inertia of the pulley about its rotation axis?

A uniform hollow disk has two pieces of thin, light wire wrapped around its outer rim and is supported from the ceiling (Fig. P10.62). Suddenly one of the wires breaks, and the remaining wire does not slip as the disk rolls down. Use energy conservation to find the speed of the center of this disk after it has fallen a distance of 2.20 \(\mathrm{m} .\)

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