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A wheel, starting from rest, rotates with a constant angular acceleration of \(2.00 \mathrm{rad} / \mathrm{s}^{2} .\) During a certain \(3.00 \mathrm{~s}\) interval, it turns through 90.0 rad. (a) What is the angular velocity of the wheel at the start of the \(3.00 \mathrm{~s}\) interval? (b) How long has the wheel been turning before the start of the 3.00 s interval?

Short Answer

Expert verified
(a) The angular velocity is 27.0 rad/s at the start of the interval. (b) The wheel has been turning for 13.5 s.

Step by step solution

01

Write the Angular Motion Equation

The equation for angular motion that relates angular displacement \( \theta \), initial angular velocity \( \omega_0 \), time \( t \), and angular acceleration \( \alpha \) is given by:\[ \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \]For the time interval of interest, \( \theta = 90.0 \text{ rad} \), \( \alpha = 2.00 \text{ rad/s}^2 \), and \( t = 3.00 \text{ s} \). Substituting these values will allow us to solve for the initial angular velocity \( \omega_0 \) at the start of the interval.
02

Substitute the Known Values

Substituting the known values into the equation:\[ 90.0 = \omega_0 (3.00) + \frac{1}{2} (2.00) (3.00)^2 \]Simplify the equation to:\[ 90.0 = 3\omega_0 + 9 \]
03

Solve for Initial Angular Velocity \(\omega_0\)

Rearrange the equation to solve for \( \omega_0 \):\[ 90.0 - 9 = 3\omega_0 \]\[ 81 = 3\omega_0 \]\[ \omega_0 = \frac{81}{3} \]\[ \omega_0 = 27.0 \text{ rad/s} \]Thus, the angular velocity of the wheel at the start of the \(3.00 \text{ s}\) interval is \(27.0 \text{ rad/s} \).
04

Determine the Total Time the Wheel Has Been Turning

We know that the wheel started from rest, so its initial angular velocity \( \omega_i = 0 \text{ rad/s} \). Using the relationship between angular velocity, time, and angular acceleration:\[ \omega = \omega_i + \alpha t \]Rearrange to find \( t \) (the total time the wheel has been turning):\[ t = \frac{\omega - \omega_i}{\alpha} \]Given that \( \omega = 27.0 \text{ rad/s} \) and \( \alpha = 2.00 \text{ rad/s}^2 \),
05

Solve for Total Time

Substitute the known values into the time equation:\[ t = \frac{27.0 \text{ rad/s} - 0}{2.00 \text{ rad/s}^2} \]\[ t = \frac{27.0}{2.00} \]\[ t = 13.5 \text{ s} \]Therefore, the wheel has been rotating for a total of \(13.5 \text{ s}\) before the start of the \(3.00 \text{ s}\) interval.

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Key Concepts

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

Angular Velocity
Angular velocity is a concept used to describe how fast something is rotating. It tells us how many radians an object sweeps through per second as it spins around an axis. Think of it as the rotational counterpart to linear velocity, which tells us how fast something is moving forward.

In this exercise, we calculated the angular velocity at the beginning of a time interval using the known values of angular displacement, time, and angular acceleration. The formula to find angular velocity \[ \omega = \omega_0 + \alpha t \]relates the final angular velocity \( \omega \), initial angular velocity \( \omega_0 \), angular acceleration \( \alpha \), and time \( t \).

### Practical Uses- Understanding of how gears and wheels work in machinery and vehicles.- Assisting engineers in designing rotating parts in engines.
  • Useful in sports to determine the speed of rotating objects, like a skater or a turning bicycle wheel.
  • Helps astronomers calculate the spin of planets and stars.
Angular Acceleration
Angular acceleration is how much the angular velocity of an object changes over time. It is like the pedal in a car, which controls how fast the car speeds up or slows down. When you apply angular acceleration to a wheel, it affects how quickly the wheel spins faster or slower.

For a wheel that starts from rest and reaches certain value of angular velocity, the angular acceleration\( \alpha \) can be calculated using:\[ \omega = \omega_i + \alpha t \]Here, \( \omega \) is the final angular velocity, \( \omega_i \) is the initial angular velocity (often zero if starting from rest), and \( t \) is the time over which acceleration occurs.

### Applications- Used in robotics where precise control of motor speeds is required.- Important in designing roller coasters, where the rotation rates of rides are calculated for safety and thrill.

Understanding angular acceleration helps in designing systems that require controlled rotational movements, ensuring smooth and optimized operations.
Angular Displacement
Angular displacement is about the angle through which an object has rotated over a certain period. It tells you how much an object has turned or rotated, which is crucial in both engineering and physics problems.

In our exercise, we worked with an angular displacement of 90 radians. This value was used to determine the initial angular velocity and how long the wheel had been spinning before a specific time interval. The equation relating angular displacement \( \theta \) with angular velocity and acceleration is:\[ \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \]

### Key Uses- Helps in navigating systems like gyroscopes, which rely on precise angular measurements.
- Fundamental in calculating movements in gears and mechanical systems.
  • Valuable in measuring rotations in satellites and spacecraft.
  • Utilized in animation and graphics to calculate how objects move on screen.
Understanding angular displacement is key in any task that involves turning, from a simple door hinge to complex airplane maneuvers.

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Most popular questions from this chapter

A \(32.0 \mathrm{~kg}\) wheel, essentially a thin hoop with radius \(1.20 \mathrm{~m}\) is rotating at 280 rev/min. It must be brought to a stop in \(15.0 \mathrm{~s}\). (a) How much work must be done to stop it? (b) What is the required average power?

The angular position of a point on the rim of a rotating wheel is given by \(\theta=4.0 t-3.0 t^{2}+t^{3},\) where \(\theta\) is in radians and \(t\) is in seconds. What are the angular velocities at (a) \(t=2.0 \mathrm{~s}\) and \((\mathrm{b}) t=4.0 \mathrm{~s} ?\) (c) What is the average angular acceleration for the time interval that begins at \(t=2.0 \mathrm{~s}\) and ends at \(t=4.0 \mathrm{~s} ?\) What are the instantaneous angular accelerations at (d) the beginning and (e) the end of this time interval?

Cheetahs running at top speed have been reported at an astounding \(114 \mathrm{~km} / \mathrm{h}\) (about \(71 \mathrm{mi} / \mathrm{h})\) by observers driving alongside the animals. Imagine trying to measure a cheetah's speed by keeping your vehicle abreast of the animal while also glancing at your speedometer, which is registering \(114 \mathrm{~km} / \mathrm{h}\). You keep the vehicle a constant \(8.0 \mathrm{~m}\) from the cheetah, but the noise of the vehicle causes the cheetah to continuously veer away from you along a circular path of radius \(92 \mathrm{~m}\). Thus, you travel along a circular path of radius \(100 \mathrm{~m} .\) (a) What is the angular speed of you and the cheetah around the circular paths? (b) What is the linear speed of the cheetah along its path? (If you did not account for the circular motion, you would conclude erroneously that the cheetah's speed is \(114 \mathrm{~km} / \mathrm{h},\) and that type of error was apparently made in the published reports.)

If a \(32.0 \mathrm{~N} \cdot \mathrm{m}\) torque on a wheel causes angular acceleration \(25.0 \mathrm{rad} / \mathrm{s}^{2}\), what is the wheel's rotational inertia?

An object rotates about a fixed axis, and the angular position of a reference line on the object is given by \(\theta=0.40 e^{2 t},\) where \(\theta\) is in radians and \(t\) is in seconds. Consider a point on the object that is \(4.0 \mathrm{~cm}\) from the axis of rotation. At \(t=0,\) what are the magnitudes of the point's (a) tangential component of acceleration and (b) radial component of acceleration?

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