Determine the linear and angular speeds and accelerations of a speck of dirt located \(2.0 \mathrm{~cm}\) from the center of a CD rotating inside a CD player at 250 rpm.

Short Answer

Expert verified
Answer: The linear speed of the speck of dirt is approximately 0.5236 meters per second, and its linear acceleration is 0. The angular speed is approximately 26.18 radians per second, and the angular acceleration is 0.

Step by step solution

01

Convert rotational speed to angular speed

To find the angular speed of the speck of dirt on the CD, we first need to convert the given rotational speed (250 rpm) to angular speed in radians per second. To do this, we can use the conversion factor: 1 revolution = \(2\pi\) radians. Also, we need to convert minutes to seconds using the conversion factor: 1 minute = 60 seconds. The conversion from rpm to radians per second can be done as follows: Angular speed (in radians per second) = (Rotational speed (in rpm) * (2π radians per revolution)) / (60 seconds per minute)
02

Calculate the angular speed

Now, we can plug in the given values to calculate the angular speed: Angular speed = (250 * 2π) / 60 ≈ 26.18 radians per second
03

Calculate the linear speed

Next, we need to find the linear speed of the speck of dirt. Linear speed (v) can be calculated using the formula: v = rω, where r is the radius (distance from the center) and ω is the angular speed. Given that the speck of dirt is 2.0 cm from the center of the CD, we can convert the distance to meters (to keep units consistent) by dividing by 100. So, the radius is 0.02 meters. Now we can calculate the linear speed: Linear speed = rω = 0.02 * 26.18 ≈ 0.5236 meters per second
04

Calculate the angular acceleration

Since the CD is rotating at a constant speed (250 rpm), there is no angular acceleration acting on the speck of dirt. Therefore, the angular acceleration is 0.
05

Calculate the linear acceleration

Finally, we need to find the linear acceleration of the speck of dirt. Linear acceleration (a) can be calculated using the formula: a = rα, where r is the radius (distance from the center) and α is the angular acceleration. As we found in Step 4, the angular acceleration is 0. This means that the linear acceleration is also 0 since there is no change in the linear speed of the speck of dirt. To summarize, the linear speed of the speck of dirt is approximately 0.5236 meters per second and its linear acceleration is 0. The angular speed is approximately 26.18 radians per second and the angular acceleration is 0.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A \(80.0-\mathrm{kg}\) pilot in an aircraft moving at a constant speed of \(500 . \mathrm{m} / \mathrm{s}\) pulls out of a vertical dive along an arc of a circle of radius \(4000 . \mathrm{m}\).

A race car is making a U-turn at constant speed. The coefficient of friction between the tires and the track is \(\mu_{\mathrm{s}}=1.20 .\) If the radius of the curve is \(10.0 \mathrm{~m},\) what is the maximum speed at which the car can turn without sliding? Assume that the car is performing uniform circular motion.

Life scientists use ultracentrifuges to separate biological components or to remove molecules from suspension. Samples in a symmetric array of containers are spun rapidly about a central axis. The centrifugal acceleration they experience in their moving reference frame acts as "artificial gravity" to effect a rapid separation. If the sample containers are \(10.0 \mathrm{~cm}\) from the rotation axis, what rotation frequency is required to produce an acceleration of \(1.00 \cdot 10^{5} g ?\)

A typical Major League fastball is thrown at approximately \(88 \mathrm{mph}\) and with a spin rate of \(110 \mathrm{rpm} .\) If the distance between the pitcher's point of release and the catcher's glove is exactly \(60.5 \mathrm{ft},\) how many full turns does the ball make between release and catch? Neglect any effect of gravity or air resistance on the ball's flight.

The motor of a fan turns a small wheel of radius \(r_{\mathrm{m}}=\) \(2.00 \mathrm{~cm} .\) This wheel turns a belt, which is attached to a wheel of radius \(r_{f}=3.00 \mathrm{~cm}\) that is mounted to the axle of the fan blades. Measured from the center of this axle, the tip of the fan blades are at a distance \(r_{\mathrm{b}}=15.0 \mathrm{~cm} .\) When the fan is in operation, the motor spins at an angular speed of \(\omega=1200\). rpm. What is the tangential speed of the tips of the fan blades?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free