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The angular speed of the rotor in a centrifuge increases from 420 to \(1420 \mathrm{rad} / \mathrm{s}\) in a time of \(5.00 \mathrm{s}\). (a) Obtain the angle through which the rotor turns. (b) What is the magnitude of the angular acceleration?

Short Answer

Expert verified
The rotor turns through 4600 rad with an angular acceleration of 200 rad/s².

Step by step solution

01

Identify Given Information

The initial angular speed \( \omega_i \) is 420 rad/s, the final angular speed \( \omega_f \) is 1420 rad/s, and the time taken \( t \) is 5.00 seconds.
02

Calculate Angular Acceleration

Using the formula \( \alpha = \frac{\omega_f - \omega_i}{t} \), substitute the given values to find the angular acceleration \( \alpha \). \[ \alpha = \frac{1420 - 420}{5.00} \] \[ \alpha = 200 \text{ rad/s}^2 \]
03

Compute the Angle Turned

Use the kinematic equation for angular motion: \( \theta = \omega_i t + \frac{1}{2} \alpha t^2 \). Substitute the known values to find \( \theta \). \[ \theta = 420 \times 5.00 + \frac{1}{2} \times 200 \times (5.00)^2 \] \[ \theta = 2100 + 2500 \] \[ \theta = 4600 \text{ rad} \]

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Key Concepts

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

Understanding Angular Speed
Angular speed is a fundamental concept in rotational motion and helps us understand how fast an object is spinning. It's defined as the rate of change of angular displacement and is typically measured in radians per second (rad/s). In our problem, we first note the initial angular speed, \( \omega_i = 420 \) rad/s, and the final angular speed, \( \omega_f = 1420 \) rad/s. This gives us a clear indication that the centrifuge rotor is speeding up over time.

Angular speed is analogous to linear speed, but instead of measuring how fast something is traveling along a straight path, we measure how fast it is rotating. Understanding this helps us later when we need to determine the angle through which the rotor has turned.

**Key Points to Remember:**
  • Angular speed is uniform when the rate of rotation is constant.
  • In scenarios where the speed is changing, like our exercise, we must consider both initial and final angular speeds.
  • Changing angular speeds lead us to explore angular acceleration.
Delving into Angular Acceleration
Angular acceleration is the rate at which the angular speed of an object changes over time. Like linear acceleration in straight-line motion, angular acceleration tells us how quickly the rotation is speeding up or slowing down. In our exercise example, it is given by the formula:

\[ \alpha = \frac{\omega_f - \omega_i}{t} \]

Substituting the values from the problem, we find \( \alpha = 200 \text{ rad/s}^2 \). This tells us that every second, the angular speed of the centrifuge increases by 200 rad/s.

**Important Aspects of Angular Acceleration:**
  • Positive angular acceleration indicates an increase in angular speed, while negative acceleration indicates a decrease.
  • In uniform circular motion, angular acceleration is zero as the speed remains constant.
  • The unit of angular acceleration is rad/s squared.
To calculate angular displacement using angular acceleration, we utilize kinematic equations as used in the next section.
Kinematic Equations in Rotational Motion
Kinematic equations provide us with a powerful tool to relate different aspects of motion such as displacement, speed, and acceleration. In rotational motion, these include similar principles to those in linear motion.

For the exercise, we use the equation:

\[ \theta = \omega_i t + \frac{1}{2} \alpha t^2 \]

This formula helps us calculate the angle \(\theta\) through which the rotor turns. When we plug in our known values, \( \theta = 4600 \) rad.

**Using Kinematic Equations:**
  • Make sure to always identify the given values and plug them accurately into the equations.
  • Kinematic equations help predict one variable of motion when others are known.
  • These equations are applicable to both linear and rotational motion with slight differences in terminology and units.
  • Understanding which equation to apply based on given parameters is crucial.
These equations bridge the connection between angular displacement, angular speed, and angular acceleration, offering a complete picture of the motion of rotating objects.

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Most popular questions from this chapter

A stroboscope is a light that flashes on and off at a constant rate. It can be used to illuminate a rotating object, and if the flashing rate is adjusted properly, the object can be made to appear stationary. (a) What is the shortest time between flashes of light that will make a three-bladed propeller appear stationary when it is rotating with an angular speed of \(16.7 \mathrm{rev} / \mathrm{s} ?(\mathbf{b})\) What is the next shortest time?

In a large centrifuge used for training pilots and astronauts, a small chamber is fixed at the end of a rigid arm that rotates in a horizontal circle. A trainee riding in the chamber of a centrifuge rotating with a constant angular speed of 2.5 rad/s experiences a centripetal acceleration of 3.2 times the acceleration due to gravity. In a second training exercise, the centrifuge speeds up from rest with a constant angular acceleration. When the centrifuge reaches an angular speed of \(2.5 \mathrm{rad} / \mathrm{s},\) the trainee experiences a total acceleration equal to 4.8 times the acceleration due to gravity. (a) How long is the arm of the centrifuge? (b) What is the angular acceleration of the centrifuge in the second training exercise?

A baseball pitcher throws a baseball horizontally at a linear speed of \(42.5 \mathrm{m} / \mathrm{s}\) (about 95 \(\mathrm{mi} / \mathrm{h}\) ). Before being caught, the baseball travels a horizontal distance of \(16.5 \mathrm{m}\) and rotates through an angle of 49.0 rad. The baseball has a radius of \(3.67 \mathrm{cm}\) and is rotating about an axis as it travels, much like the earth does. What is the tangential speed of a point on the "equator" of the baseball?

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?

A compact disc (CD) contains music on a spiral track. Music is put onto a CD with the assumption that, during playback, the music will be detected at a constant tangential speed at any point. Since \(v_{\mathrm{T}}=r \omega, \mathrm{a}\) CD rotates at a smaller angular speed for music near the outer edge and a larger angular speed for music near the inner part of the disc. For music at the outer edge \((r=0.0568 \mathrm{m}),\) the angular speed is \(3.50 \mathrm{rev} / \mathrm{s} .\) Find (a) the constant tangential speed at which music is detected and (b) the angular speed (in rev/s) for music at a distance of \(0.0249 \mathrm{m}\) from the center of a CD.

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