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A centrifuge rotor rotating at 10,300 rpm is shut off and is eventually brought uniformly to rest by a frictional torque of \(1.20 \mathrm{~m} \cdot \mathrm{N}\). If the mass of the rotor is \(3.80 \mathrm{~kg}\) and it can be approximated as a solid cylinder of radius \(0.0710 \mathrm{~m},\) through how many revolutions will the rotor turn before coming to rest, and how long will it take?

Short Answer

Expert verified
The rotor turns approximately 741 revolutions and takes 8.644 seconds to stop.

Step by step solution

01

Calculate the Moment of Inertia

The moment of inertia \( I \) for a solid cylinder is given by the formula: \[ I = \frac{1}{2} m r^2 \]Substitute the given values: \( m = 3.80 \text{ kg} \), \( r = 0.0710 \text{ m} \):\[ I = \frac{1}{2} \times 3.80 \times (0.0710)^2 = 0.0096239 \text{ kg} \cdot \text{m}^2 \]
02

Calculate the Angular Velocity in Radians per Second

The initial angular velocity \( \omega_0 \) is given in revolutions per minute (rpm). Convert this to radians per second:\[ \omega_0 = 10,300 \times \frac{2\pi}{60} = 1,077.566 \text{ rad/s} \]
03

Calculate the Angular Deceleration

The angular deceleration \( \alpha \) can be found using \( \tau = I \alpha \), where \( \tau = 1.20 \text{ m} \cdot \text{N} \) is the torque:\[ \alpha = \frac{\tau}{I} = \frac{1.20}{0.0096239} = 124.679 \text{ rad/s}^2 \]Since the rotor is slowing down, \( \alpha \) is negative: \( \alpha = -124.679 \text{ rad/s}^2 \).
04

Calculate the Time to Come to Rest

Use the formula \( \omega = \omega_0 + \alpha t \) for calculating time:Set \( \omega = 0 \) (since it comes to rest):\[ 0 = 1,077.566 - 124.679 t \]Solving for \( t \):\[ t = \frac{1,077.566}{124.679} = 8.644 \text{ seconds} \]
05

Calculate Total Revolutions Before Stopping

The total angular displacement \( \theta \) in radians is found using \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \):\[ \theta = 1,077.566 \times 8.644 + \frac{1}{2} (-124.679) \times (8.644)^2 \]\[ \theta = 9,308.852 - 4,654.426 = 4,654.426 \text{ rad} \]Convert \( \theta \) from radians to revolutions by dividing by \( 2\pi \):\[ \text{Revolutions} = \frac{4,654.426}{2\pi} \approx 741 \text{ revolutions} \]

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Key Concepts

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

Moment of Inertia
The concept of moment of inertia is akin to mass in linear dynamics. It measures an object's resistance to changes in its rotational motion. Imagine pushing a heavy door. The door's mass and the way it's distributed about its hinge determine how hard it is to open or close.
For a cylinder, like the centrifuge rotor in our exercise, the moment of inertia (\( I \) for short) is calculated using the formula:
  • \[ I = \frac{1}{2} m r^2 \]
Where \( m \) is the mass and \( r \) is the radius.Plugging these values in from our exercise, with \( m = 3.80 \text{ kg} \) and \( r = 0.0710 \text{ m} \), our calculation becomes:
  • \[ I = \frac{1}{2} \times 3.80 \times (0.0710)^2 = 0.0096239 \text{ kg m}^2 \]
This value tells us how much effort is needed to change the rotor's spinning state. The bigger this number, the harder it is to speed up or slow down the rotation.
Angular Velocity
Angular velocity describes how fast something spins. Unlike linear velocity, which we measure in meters per second, angular velocity is usually given in revolutions per minute (rpm) or radians per second (rad/s). All spinning objects have angular velocity.
To convert angular velocity from rpm to radians per second, multiply by \( 2\pi \) (since one revolution equals \( 2\pi \) radians) and divide by 60 (since there are 60 seconds in a minute):
  • \[ \omega_0 = 10,300 \times \frac{2\pi}{60} = 1,077.566 \text{ rad/s} \]
This high number tells us that the rotor was spinning incredibly fast, which makes perfect sense for something designed to separate substances through centrifugal force. Understanding angular velocity is crucial for determining how long it will take the rotor to stop spinning when a brake force is applied.
Angular Deceleration
As soon as the centrifuge rotor is switched off, it begins to slow down. This change in spin rate is quantified by angular deceleration, which measures how quickly angular velocity decreases over time.
Using the formula \( \tau = I \alpha \), where \( \tau \) is the torque applied, you can solve for angular deceleration (\( \alpha \)). Given \( \tau = 1.20 \text{ m} \cdot \text{N} \):
  • \[ \alpha = \frac{\tau}{I} = \frac{1.20}{0.0096239} = 124.679 \text{ rad/s}^2 \]
This deceleration is negative because it's slowing down the rotor, reflecting the torque's direction. When analyzing motion, remember that negative signs indicate a decrease in speed.
Moreover, angular deceleration indicates how long it takes for the rotor to come to a complete stop. Knowing this helps in calculating the deceleration time through the equation \( \omega = \omega_0 + \alpha t \). Thus, understanding these concepts gives a complete picture of how forces influence rotational systems.

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Most popular questions from this chapter

A rotating merry-go-round makes one complete revolution in \(4.0 \mathrm{~s}\) (Fig. \(10-45)\). (a) What is the linear speed of a child seated \(1.2 \mathrm{~m}\) from the center? \((b)\) What is her acceleration (give components)?

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(II) A centrifuge rotor rotating at \(10,300\) rpm is shut off and is eventually brought uniformly to rest by a frictional torque of 1.20 \(\mathrm{m} \cdot \mathrm{N}\) . If the mass of the rotor is 3.80 \(\mathrm{kg}\) and it can be approximated as a solid cylinder of radius \(0.0710 \mathrm{m},\) through how many revolutions will the rotor turn before coming to rest, and how long will it take?

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A merry-go-round has a mass of \(1640 \mathrm{~kg}\) and a radius of \(7.50 \mathrm{~m} .\) How much net work is required to accelerate it from rest to a rotation rate of 1.00 revolution per 8.00 s? Assume it is a solid cylinder.

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