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A merry-go-round is stationary. A dog is running on the ground just outside its circumference, moving with a constant angular speed of \(0.750 \mathrm{rad} / \mathrm{s} .\) The dog does not change his pace when he sees what he has been looking for: a bone resting on the edge of the merry-go-round one third of a revolution in front of him. At the instant the dog sees the bone \((t=0),\) the merry-go-round begins to move in the direction the dog is running, with a constant angular acceleration of \(0.0150 \mathrm{rad} / \mathrm{s}^{2},\) (a) At what time will the dog reach the bone? (b) The confused dog keeps running and passes the bone. How long after the merry-go-round starts to turn do the dog and the bone draw even with each other for the second time?

Step by step solution

01

Determine the equation of motion for the dog

The dog is moving at a constant angular speed. The angular displacement of the dog relative to its initial position is given by \(\omega t\), where \(\omega\) is the angular speed and \(t\) is the time. With given \(\omega = 0.750 rad/s\), the equation of motion for the dog is \(\theta_{d}=0.750t\).

02

Determine the equation of motion for the merry-go-round

The merry-go-round is accelerating from rest, so its angular displacement is given by \(0.5\alpha t^{2}\), where \(\alpha\) is the angular acceleration and \(t\) is the time. With \(\alpha=0.0150 rad/s^{2}\), the equation of motion is \(\theta_{m}=0.5 * 0.0150 * t^{2}\).

03

Solve for the time at which the dog reaches the bone

The point where the bone is located is 1/3 of the revolution or \(2\pi / 3\) rad ahead of the dog at \(t=0\). The dog will reach the bone when its angular position equals to the angular position of the bone. Equating the two equations of motion, \(0.750t = 2\pi/3 + 0.5 * 0.0150 * t^{2}\) and solving for \(t\), we have approximately \(t = 9.17s\).

04

Solve for the time at which the dog and the bone are even for the second time

The dog and the bone draw even with each other for the second time when they both completed one full revolution since their first meeting. For the dog, it will take \((2\pi)/0.750 \approx 8.38s\) to run a full circle. For the merry-go-round which is accelerating, it completes one full revolution in a time \(t'\) that can be solved from the equation \(0.5 * 0.0150 * t'^{2} = 2\pi\). Adding the time it takes for them to meet the first time, the total time when the dog and the bone are even for the second time is approximately \(t' + 9.17s\).

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Key Concepts

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

Angular Speed

Angular speed is a measure of how fast an object rotates or revolves relative to another point. It tells us how quickly the angular position of an object is changing. In the context of rotational motion, it is denoted by the Greek letter omega, \( \omega \), and is usually measured in radians per second. In our example, the dog's constant angular speed is given as \(0.750 \ \text{rad/s}\). This means that every second, the dog covers an angular path of \(0.750 \ \text{radians}\).

• Constant angular speed: The dog runs with a steady speed, making calculations simpler.
• Angular speed equation: \(\theta = \omega t\) describes a direct relationship between angular displacement (\(\theta\)), angular speed (\(\omega\)), and time (\(t\)).
• Understanding the unit: Radians helps differentiate angular speed from linear speed, emphasizing rotational motion.

Remember that angular speed is crucial for determining how quickly an object will complete a revolution or a portion thereof.

Angular Acceleration

Angular acceleration is the rate of change of angular speed. This concept is similar to linear acceleration, but it applies to objects rotating around a point instead of moving in a straight line.Given by the Greek letter alpha, \( \alpha \), it measures how quickly an object speeds up or slows down its rotation, and is commonly expressed in \( \text{rad/s}^2 \).In our merry-go-round scenario, the merry-go-round begins to move with an angular acceleration of \(0.0150 \ \text{rad/s}^2\), which means every second, its angular speed increases by \(0.0150 \text{rad/s}\).

• Starts from rest: Initially, the merry-go-round is not moving, so it begins from zero angular speed.
• Constant acceleration: The merry-go-round accelerates at a steady rate, making it easy to apply physics equations.
• Equation of motion relies on \(\alpha\): The angular displacement grows quadratically with time, as described by \(\theta = \frac{1}{2} \alpha t^2 \).

Understanding angular acceleration helps predict how rotational speed changes over time. It is essential for determining when the dog catches up with the bone.

Equations of Motion

Equations of motion for rotational systems work similarly to linear motion, allowing us to solve for various parameters, like time, angular displacement, or angular speed. These equations are derived from the basic principles of kinematics, adapted for rotational scenarios.

• Constant angular speed: \(\theta = \omega t\) for objects moving in a circle at a fixed speed, like the running dog.
• Constant angular acceleration: \(\theta = \frac{1}{2} \alpha t^2\) for accelerating objects, like the merry-go-round.
• Combining motions: For complex problems, you often equate different equations to find when two moving objects meet.

In our problem, the dog's constant speed is modeled by one equation, while the merry-go-round's accelerating motion uses another. Solving the problem involves equating these equations to find the time when the dog catches up with or overtakes the bone. This involves solving for time \(t\) using the condition that the angular positions must equate.

Angular Displacement

Angular displacement refers to the angle through which a point or line has been rotated in a specified sense about a specified axis. Unlike linear displacement, it does not depend on the path taken and is measured in radians.In the problem, angular displacement of both the dog and the merry-go-round determines when they meet the bone. It can be calculated using different approaches depending on whether the object is moving with a constant speed or is accelerating:

• Linear function for dogs: Since the dog runs at constant speed, its angular displacement is \(\theta_d = 0.750t\), showing a linear relationship with time.
• Quadratic function for merry-go-round: The equation \(\theta_m = \frac{1}{2} * 0.0150 * t^2\) illustrates how displacement increases faster when accelerating.
• Meeting point displacement: For the dog to reach the bone, we set displacements equal, adjusted for initial bone offset.

Understanding angular displacement helps solve rotational movement problems, as it integrates the effects of both speed and acceleration.