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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
(a) The rotor turns through 4600 rad; (b) the angular acceleration is 200 rad/s^2.

Step by step solution

01

Identify Given Values

We are given the initial angular speed \( \omega_i = 420 \, \mathrm{rad/s} \), the final angular speed \( \omega_f = 1420 \, \mathrm{rad/s} \), and the time interval \( t = 5.00 \, \mathrm{s} \). We need to find the angle \( \theta \) through which the rotor turns and the angular acceleration \( \alpha \).
02

Calculate Angular Acceleration

Angular acceleration \( \alpha \) can be calculated using the formula: \( \alpha = \frac{\omega_f - \omega_i}{t} \). Substituting the given values: \( \alpha = \frac{1420 \, \mathrm{rad/s} - 420 \, \mathrm{rad/s}}{5.00 \, \mathrm{s}} \). Simplifying gives \( \alpha = \frac{1000}{5} = 200 \, \mathrm{rad/s^2} \).
03

Calculate the Angular Displacement

To find the angular displacement \( \theta \), we use the formula \( \theta = \omega_i t + \frac{1}{2} \alpha t^2 \). Substituting the known values: \( \theta = 420 \, \mathrm{rad/s} \times 5.00 \, \mathrm{s} + \frac{1}{2} \times 200 \, \mathrm{rad/s^2} \times (5.00 \, \mathrm{s})^2 \). Calculating gives \( \theta = 2100 + 2500 = 4600 \, \mathrm{rad} \).

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Key Concepts

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

Angular Speed
When we talk about angular speed, we're referring to how quickly an object rotates or revolves relative to an axis or point. It is a measure of the angle an object rotates through in a certain amount of time and is expressed in radians per second (rad/s). For instance, in our problem, the centrifuge rotor starts at an angular speed of 420 rad/s and speeds up to 1420 rad/s.
Angular speed is constant when there is no change in how fast the rotor is spinning. However, when the speed changes, that's when angular acceleration comes into play.
  • Initial Angular Speed: This is the speed at which rotational motion begins. In our example, it's 420 rad/s.
  • Final Angular Speed: This is the speed at the end of our observed time, here 1420 rad/s.
  • Constant Angular Speed: Occurs when the rotor moves at the same speed without speeding up or slowing down.
The change between the initial and final speeds is crucial for determining the angular acceleration and displacement.
Angular Acceleration
Angular acceleration is the rate of change of angular speed. It's measured in radians per second squared (rad/s²). In essence, it describes how quickly an object starts to spin faster or slower. To calculate angular acceleration, we subtract the initial angular speed from the final angular speed and divide it by the time over which this change happens.
Using our centrifuge rotor as an example, the calculation is as follows:\[\alpha = \frac{\omega_f - \omega_i}{t} = \frac{1420 \, \mathrm{rad/s} - 420 \, \mathrm{rad/s}}{5.00 \, \mathrm{s}} = 200 \, \mathrm{rad/s^2}\]
This shows that throughout the 5 second time span, the rotor's angular speed increased by 200 rad/s².
Here are some key points:
  • Positive Angular Acceleration: Indicates an increase in angular speed, like our problem where the rotor sped up.
  • Negative Angular Acceleration: Would indicate the rotor is slowing down.
By understanding the angular acceleration, we can gain insights into how forces influence rotational movement.
Centrifuge Rotor
A centrifuge rotor is a rotating part of the centrifuge device, often used in laboratories to separate substances of different densities. The nature of its operation relies heavily on rotational dynamics, with angular speed and acceleration being essential components.
The rotor's performance is crucial for accurately separating materials during processes such as:
  • Chemical analysis
  • Biological experiments, including DNA or cellular fractionation
  • Clinical laboratory work
The rotor must spin at high speeds to create sufficient centrifugal force to isolate these substances, and understanding its speed and acceleration can ensure efficiency in scientific procedures. Safe operation is also vital, so limitations on maximum angular speed are generally enforced through design considerations.
Angular Displacement
Angular displacement accounts for the total angle through which an object rotates over a certain time period. It's measured in radians and helps us understand how far around an axis something has rotated. Using angular displacement, we can gauge how much rotation was completed by the rotor in our example.
To find the angular displacement, we use:\[\theta = \omega_i t + \frac{1}{2} \alpha t^2\]
Plugging in our example values gives:\[\theta = 420 \, \mathrm{rad/s} \times 5.00 \, \mathrm{s} + \frac{1}{2} \times 200 \, \mathrm{rad/s^2} \times (5.00 \, \mathrm{s})^2 = 4600 \, \mathrm{rad}\]
This indicates that the centrifuge rotor turned through an angle of 4600 radians during the given time frame.
Key things to remember:
  • Angular displacement is cumulative and always considers the entirety of the motion.
  • It helps visualize the rotation extent relative to the initial position.
By understanding angular displacement, we get a clearer picture of the total rotational journey of an object.

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Most popular questions from this chapter

Energy of a Bullet Dissipated by Plywood. As part of a criminal investigation, you need to determine how much of a bullet's energy is dissipated by a 0.500 -inch piece of plywood. You construct a device that consists of three disks that are separated by a distance \(d=0.950 \mathrm{m}\) and rotate on a common axis. The bullet is fired through the first disk (a few inches above its center), which is composed of a light plastic that has a negligible effect on the speed of the bullet. The bullet then passes through the second disk, which is composed of 0.500-inch plywood. Finally, the bullet strikes the third disk, where it becomes embedded. The disks rotate with an angular velocity of \(\omega=92.0 \mathrm{rad} / \mathrm{s} .\) The angular displacement between holes in the first and second disks is \(\Delta \theta_{12}=0.255\) rad, and the angular displacement between the holes in the second and third disks is \(\Delta \theta_{23}=0.273\) rad. If the mass of the bullet is \(15.0 \mathrm{g}\), find \((\mathrm{a})\) the initial speed of the bullet and \((\mathrm{b})\) the energy dissipated by the 0.50-inch plywood.

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 pitcher throws a curveball that reaches the catcher in 0.60 s. The ball curves because it is spinning at an average angular velocity of 330 rev/min (assumed constant) on its way to the catcher’s mitt. What is the angular displacement of the baseball (in radians) as it travels from the pitcher to the catcher?

Conceptual Example 2 provides some relevant background for this problem. A jet is circling an airport control tower at a distance of \(18.0 \mathrm{km}\) An observer in the tower watches the jet cross in front of the moon. As seen from the tower, the moon subtends an angle of \(9.04 \times 10^{-3}\) radians. Find the distance traveled (in meters) by the jet as the observer watches the nose of the jet cross from one side of the moon to the other.

A racing car, starting from rest, travels around a circular turn of radius \(23.5 \mathrm{m}\). At a certain instant, the car is still accelerating, and its angular speed is \(0.571 \mathrm{rad} / \mathrm{s}\). At this time, the total acceleration (centripetal plus tangential) makes an angle of \(35.0^{\circ}\) with respect to the radius. (The situation is similar to that in Interactive Figure \(8.12 b .\) ) What is the magnitude of the total acceleration?

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