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A potter's wheel is spinning with an initial angular velocity of \(12 \mathrm{rad} / \mathrm{s} .\) It rotates through an angle of \(60 \mathrm{rad}\) in the process of coming to rest. (a) What is the angular acceleration of the wheel? (b) How long does it take for it to come to rest?

Short Answer

Expert verified
Angular acceleration is \(-1.2 \text{ rad/s}^2\), and time to rest is 10 seconds.

Step by step solution

01

Identify Known Values and Required Equations

We are given the initial angular velocity \( \omega_0 = 12 \text{ rad/s} \), the final angular velocity \( \omega = 0 \text{ rad/s} \) since it comes to rest, and the angular displacement \( \theta = 60 \text{ rad} \). We need to find the angular acceleration \( \alpha \) using the equation \( \omega^2 = \omega_0^2 + 2\alpha\theta \).
02

Solve for Angular Acceleration

Using the equation \( \omega^2 = \omega_0^2 + 2\alpha\theta \), substitute the known values: \( 0 = (12)^2 + 2\alpha(60) \). Simplifying gives \( 0 = 144 + 120\alpha \). Rearrange to find \( \alpha = -\frac{144}{120} = -1.2 \text{ rad/s}^2 \).
03

Solve for Time to Come to Rest

Use the equation for angular velocity: \( \omega = \omega_0 + \alpha t \). Substitute the known values: \( 0 = 12 + (-1.2)t \). Solve for \( t \) to get \( 1.2t = 12 \) yielding \( t = 10 \text{ s} \).
04

Interpret the Results

The negative sign of \( \alpha \) indicates deceleration. The potter's wheel takes 10 seconds to slow down to rest due to this deceleration.

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Key Concepts

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

Understanding Angular Velocity
Angular velocity is a fundamental concept in angular kinematics. It refers to how quickly an object rotates or spins around an axis. Just as we use linear velocity to describe how fast something moves in a straight path, angular velocity describes motion in a circular path.
  • Angular velocity is denoted by the symbol \( \omega \).
  • Measured in radians per second (rad/s).
  • Represents the rate of change of angular displacement over time.
When thinking about angular velocity, remember it is similar to the concept of speed but in a circular motion. For example, the wheels of a potter's wheel spinning around its center have an angular velocity of 12 rad/s, which tells us how many radians it covers in one second.
The formula \( \omega = \frac{\Delta \theta}{\Delta t} \) can be used when constant angular velocity is involved, where \( \Delta \theta \) is the change in angular displacement, and \( \Delta t \) is the time period. This formula helps to connect how far an object has turned with how long it has taken.
Exploring Angular Acceleration
Angular acceleration is about how quickly the angular velocity of an object changes with time. If a spinning object speeds up or slows down, it experiences angular acceleration.
  • Represented by the symbol \( \alpha \).
  • Measured in rad/s².
  • Describes the rate of change of angular velocity.
In the potter's wheel exercise, the angular acceleration can be calculated using the equation \( \omega^2 = \omega_0^2 + 2\alpha\theta \). This equation is derived from kinematics and connects angular velocity, angular displacement, and angular acceleration.
The negative sign in the solution, \( -1.2 \text{ rad/s}^2 \), indicates a deceleration, meaning the wheel is slowing down. This value tells us that for every second, the angular velocity decreases by 1.2 rad/s until the wheel stops.
Understanding Angular Displacement
Angular displacement refers to the angle through which an object moves on a circular path. Unlike linear displacement, which measures straight-line paths, angular displacement focuses on rotational movement.
  • Denoted by \( \theta \).
  • Measured in radians.
  • Represents the angle covered by the rotating object.
In the context of the potter's wheel problem, the wheel rotates through an angular displacement of 60 radians before coming to a stop. This value shows how much the wheel has turned from its initial position.
Understanding angular displacement is crucial as it links directly with angular variables like velocity and acceleration, providing a complete picture of the rotational motion of objects. By knowing the displacement and initial velocity, one can determine vital information about the system's acceleration and time taken to reach a certain state, just as demonstrated in the exercise.

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Most popular questions from this chapter

A flywheel having constant angular acceleration requires \(4.00 \mathrm{~s}\) to rotate through 162 rad. Its angular velocity at the end of this time is \(108 \mathrm{rad} / \mathrm{s}\). Find (a) the angular velocity at the beginning of the 4.00 s interval; (b) the angular acceleration of the flywheel.

An airplane propeller is \(2.08 \mathrm{~m}\) in length (from tip to tip) with mass \(117 \mathrm{~kg}\) and is rotating at \(2400 \mathrm{rpm}\) (rev/min) about an axis through its center. You can model the propeller as a slender rod. (a) What is its rotational kinetic energy? (b) Suppose that, due to weight constraints, you had to reduce the propeller's mass to \(75.0 \%\) of its original mass, but you still needed to keep the same size and kinetic energy. What would its angular speed have to be, in rpm?

It has been suggested that we should use our power plants to generate energy in the off-hours (such as late at night) and store it for use during the day. One idea put forward is to store the energy in large flywheels. Suppose we want to build such a flywheel in the shape of a hollow cylinder of inner radius \(0.500 \mathrm{~m}\) and outer radius \(1.50 \mathrm{~m},\) using concrete of density \(2.20 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\). (a) If, for stability, such a heavy flywheel is limited to 1.75 seconds for each revolution and has negligible friction at its axle, what must be its length to store \(2.5 \mathrm{MJ}\) of energy in its rotational motion? (b) Suppose that by strengthening the frame you could safely double the flywheel's rate of spin. What length of flywheel would you need in that case? (Solve this part without reworking the entire problem!)

A wheel rotates with a constant angular velocity of \(6.00 \mathrm{rad} / \mathrm{s}\) (a) Compute the radial acceleration of a point \(0.500 \mathrm{~m}\) from the axis, using the relation \(a_{\mathrm{rad}}=\omega^{2} r .(\mathrm{b})\) Find the tangential speed of the point, and compute its radial acceleration from the relation \(a_{\mathrm{rad}}=v^{2} / r\)

A laser beam aimed from the earth is swept across the face of the moon. (a) If the beam is rotated at an angular velocity of \(1.50 \times 10^{-3} \mathrm{rad} / \mathrm{s},\) at what speed does the laser light move across the moon's surface? (See Appendix E for the moon's orbital radius.) (b) If the diameter of the laser spot on the moon is \(6.00 \mathrm{~km},\) what is the angle of divergence of the laser beam?

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