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Chapter 5: Problem 53

In a Beams ultracentrifuge, the rotor is suspended magnetically in a vacuum. Since there is no mechanical connection to the rotor, the only friction is the air resistance due to the few air molecules in the vacuum. If the rotor is spinning with an angular speed of $5.0 \times 10^{5} \mathrm{rad} / \mathrm{s}$ and the driving force is turned off, its spinning slows down at an angular rate of \(0.40 \mathrm{rad} / \mathrm{s}^{2} .\) (a) How long does the rotor spin before coming to rest? (b) During this time, through how many revolutions does the rotor spin?

Short Answer

Expert verified
Answer: It takes approximately \(1.25 \times 10^6 \, \mathrm{s}\) for the rotor to come to rest, and it goes through about \(4.98 \times 10^{10}\) revolutions during that time.

Step by step solution

01

Identify the relevant equations

From the problem, we have the initial angular velocity \(\omega_0 = 5.0 \times 10^{5} \, \mathrm{rad/s}\) and the angular acceleration \(\alpha = -0.40 \, \mathrm{rad/s^{2}}\) (negative because the rotor is slowing down). We need to find how long it takes to come to rest and how many revolutions it goes through during that time. For the first part, we can use the following equation relating angular velocity, angular acceleration, and time: \(\omega = \omega_0 + \alpha t\) Here, \(\omega\) is the final angular velocity, which is \(0 \, \mathrm{rad/s}\) since the rotor comes to rest.
02

Calculate the time taken to come to rest

We will now plug in the given values into the equation to find the time it takes to come to rest: \(0 = 5.0 \times 10^{5} - 0.40 t\) Now, we'll solve for \(t\): \(\Rightarrow t = \frac{5.0 \times 10^{5}}{0.40}\) \(\Rightarrow t = 1.25 \times 10^{6} \, \mathrm{s}\) So, it takes \(1.25 \times 10^{6} \, \mathrm{s}\) for the rotor to come to rest.
03

Find the total angular displacement

To find the number of revolutions the rotor goes through before coming to rest, we need to find the total angular displacement \(\theta\). We'll use another angular kinematic equation: \(\theta = \omega_0 t + \frac{1}{2} \alpha t^2\) We have the values for \(\omega_0\), \(t\), and \(\alpha\). Let's substitute these values and find \(\theta\): \(\theta = (5.0 \times 10^{5})(1.25 \times 10^{6}) - \frac{1}{2} (0.40)(1.25 \times 10^{6})^2\) \(\theta = 6.25 \times 10^{11} - 3.125 \times 10^{11}\) \(\theta = 3.125 \times 10^{11} \, \mathrm{rad}\) So, the total angular displacement during the slowing down process is \(3.125 \times 10^{11} \, \mathrm{rad}\).
04

Calculate the number of revolutions

Now, we'll convert the angular displacement from radians to revolutions. There are \(2\pi\) radians in one revolution, so we'll divide \(\theta\) by \(2\pi\) to get the number of revolutions: Revolutions \(=\frac{3.125 \times 10^{11}}{2\pi}\) Revolutions \(= 4.98 \times 10^{10}\) Therefore, the rotor spins through approximately \(4.98 \times 10^{10}\) revolutions before coming to rest.

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