/*! 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 26 A dentist causes the bit of a hi... [FREE SOLUTION] | 91Ó°ÊÓ

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A dentist causes the bit of a high-speed drill to accelerate from an angular speed of \(1.05 \times 10^{4} \mathrm{rad} / \mathrm{s}\) to an angular speed of \(3.14 \times 10^{4} \mathrm{rad} / \mathrm{s} .\) In the process, the bit turns through \(1.88 \times 10^{4} \mathrm{rad}\) Assuming a constant angular acceleration, how long would it take the bit to reach its maximum speed of \(7.85 \times 10^{4} \mathrm{rad} / \mathrm{s},\) starting from rest?

Short Answer

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Explanation for variable identification will continue in next steps.

Step by step solution

01

Understand the Problem

We are given initial and final angular velocities, and the angular displacement during acceleration. We need to find the time it takes to reach a new maximum speed from rest.
02

Identify the Given Variables

Initial angular velocity, \( \omega_i = 1.05 \times 10^4 \, \text{rad/s} \), final angular velocity, \( \omega_f = 3.14 \times 10^4 \, \text{rad/s} \), angular displacement, \( \theta = 1.88 \times 10^4 \, \text{rad} \).

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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 the rate at which an object's angular velocity changes with time. It can be understood as the rotational equivalent of linear acceleration. Just like how we accelerate a car forward, angular acceleration tells us how quickly something like a wheel, a drill bit, or even a planet is spinning up or slowing down.
\[\alpha = \frac{\Delta \omega}{\Delta t}\]
Here, \( \alpha \) is the angular acceleration, \( \Delta \omega \) is the change in angular velocity, and \( \Delta t \) is the change in time. Whenever you see a curved motion increasing or decreasing in speed, angular acceleration plays a part.
  • Positive angular acceleration means spinning faster.
  • Negative angular acceleration, often called deceleration, means slowing down the spin.
In practical applications like the dentist's drill, understanding angular acceleration helps determine how long it takes to reach desired speeds.
Angular Velocity
Angular velocity describes how fast something rotates. It tells us the angle, in radians, that an object turns in a unit of time. Imagine a record on a turntable – the speed at which it rotates is its angular velocity.
\[\omega = \frac{\theta}{t}\]
Where \( \omega \) is the angular velocity, \( \theta \) is the angular displacement, and \( t \) is time. Angular velocity is typically measured in radians per second (rad/s).
  • In the dentist drill exercise, the angular velocity changes from \(1.05 \times 10^4 \) rad/s to \(3.14 \times 10^4\) rad/s, showcasing how quick the bit spins.
  • Key Point: The angular velocity directly influences how fast the drill can complete its tasks.
Understanding angular velocity is critical in contexts where precise rotational speeds are required, such as machining or robotics.
Angular Displacement
Angular displacement is the measure of the angle through which an object moves during its rotation. It's like tracking the arc of a circle as an object spins or turns. Whenever you see a rotating object, angular displacement tells us how much it has rotated.
\[\theta = \omega_i \times t + \frac{1}{2} \alpha \times t^2\]
In this equation, \( \theta \) is angular displacement, \( \omega_i \) is initial angular velocity, and \( \alpha \) is angular acceleration.
  • For the dentist's drill, it completes \(1.88 \times 10^4\) radians in displacement as it speeds up or slows down.
  • Useful Insight: Angular displacement can help visualize how much energy or time it might take to complete a rotation.
Angular displacement is crucial in determining the path taken by rotating objects, helping in calculating work done in physics or estimating performance in real-world systems.
Constant Angular Acceleration
When an object moves with constant angular acceleration, it means that its angular acceleration does not change over time. This simplifies many calculations in rotational dynamics and is a common assumption in problems unless stated otherwise.
Some common equations to work with under constant angular acceleration include:
  • \( \omega_f = \omega_i + \alpha \times t \)
  • \( \theta = \omega_i \times t + \frac{1}{2} \alpha \times t^2 \)
  • \( \omega_f^2 = \omega_i^2 + 2 \alpha \times \theta \)
These formulas help in calculating various aspects of rotation, such as time taken to reach a certain speed or the total angle turned.
In the given exercise, using the concept of constant angular acceleration allows us to find how long it would take the dentist drill bit to reach its maximum velocity, making calculations straightforward and manageable.

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Most popular questions from this chapter

A bicycle is rolling down a circular portion of a path; this portion of the path has a radius of \(9.00 \mathrm{m}\). As the drawing illustrates, the angular displacement of the bicycle is 0.960 rad. What is the angle (in radians) through which each bicycle wheel (radius \(=0.400 \mathrm{m}\) ) rotates?

A car is traveling with a speed of \(20.0 \mathrm{m} / \mathrm{s}\) along a straight horizontal road. The wheels have a radius of \(0.300 \mathrm{m}\). If the car speeds up with a linear acceleration of \(1.50 \mathrm{m} / \mathrm{s}^{2}\) for \(8.00 \mathrm{s}\), find the angular displacement of each wheel during this period.

A rider on a mountain bike is traveling to the left in the figure. Each wheel has an angular velocity of \(+21.7 \mathrm{rad} / \mathrm{s},\) where, as usual, the plus sign indicates that the wheel is rotating in the counterclockwise direction. (a) To pass another cyclist, the rider pumps harder, and the angular velocity of the wheels increases from +21.7 to \(+28.5 \mathrm{rad} / \mathrm{s}\) in a time of 3.50 s. (b) After passing the cyclist, the rider begins to coast, and the angular velocity of the wheels decreases from +28.5 to +15.3 rad/s in a time of 10.7 s. Concepts: (i) Is the angular acceleration positive or negative when the rider is passing the cyclist and the angular speed of the wheels is increasing? (ii) Is the angular acceleration positive or negative when the rider is coasting and the angular speed of the wheels is decreasing? Calculations: In both instances, (a) and (b), determine the magnitude and direction of the angular acceleration (assumed constant) of the wheels.

The table that follows lists four pairs of initial and final angles of a wheel on a moving car. The elapsed time for each pair of angles is 2.0 s. For each of the four pairs, determine the average angular velocity (magnitude and direction as given by the algebraic sign of your answer). $$\begin{array}{lcc}\hline & \text { Initial angle } \theta_{0} & \text { Final angle } \theta \\\\\hline \text { (a) } & 0.45 \mathrm{rad} & 0.75 \mathrm{rad} \\\\\text { (b) } & 0.94 \mathrm{rad} & 0.54 \mathrm{rad} \\\\\text { (c) } & 5.4 \mathrm{rad} & 4.2 \mathrm{rad} \\\\\text { (d) } & 3.0 \mathrm{rad} & 3.8 \mathrm{rad} \\\\\hline\end{array}$$

The differential gear of a car axle allows the wheel on the left side of a car to rotate at a different angular speed than the wheel on the right side. A car is driving at a constant speed around a circular track on level ground, completing each lap in 19.5 s. The distance between the tires on the left and right sides of the car is \(1.60 \mathrm{m}\), and the radius of each wheel is \(0.350 \mathrm{m} .\) What is the difference between the angular speeds of the wheels on the left and right sides of the car?

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