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Chapter 4: Problem 63

An ultracentrifuge spins a small test tube in a circle of radios \(10 \mathrm{~cm}\) at 1000 nevolutionu per second. What in the centripetal accelcration of the tent tabe? How many standard gis does this amount to?

Short Answer

Expert verified
The centripetal acceleration is approximately \( 3947841.76 \text{ m/s}^2 \), which is about \( 402533.5 \) gs.

Step by step solution

01

Convert Radius to Meters

Given that the radius is 10 cm, we need to convert this into meters for standard SI units. There are 100 cm in a meter, so the radius in meters is \( r = \frac{10}{100} = 0.1 \text{ meters} \).
02

Calculate Angular Velocity

The test tube spins at 1000 revolutions per second. First, convert revolutions per second to radians per second since one revolution is \(2\pi\) radians. Thus, the angular velocity \( \omega \) is \(1000 \cdot 2\pi = 2000\pi \text{ radians per second} \).
03

Calculate Centripetal Acceleration

Use the formula for centripetal acceleration \(a_c = r\omega^2\). Substitute \(r = 0.1\) m and \(\omega = 2000\pi\) into the formula:\[a_c = 0.1 \cdot (2000\pi)^2\]Calculate the value:\[a_c = 0.1 \cdot 4000000\pi^2 \approx 0.1 \cdot 3.14159^2 \cdot 4000000 \approx 3947841.76 \text{ m/s}^2\]
04

Convert to Standard gs

One standard \(g\) is equivalent to \(9.81\; \text{m/s}^2\). To find how many gs the centripetal acceleration amounts to, divide by 9.81:\[\text{gs} = \frac{3947841.76}{9.81} \approx 402533.5\]

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Key Concepts

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

Angular Velocity
Angular velocity is a measure of how fast an object is rotating around a specific point or axis. It is represented by the symbol \(\omega\), and its unit is typically in radians per second. This concept is essential when calculating quantities related to circular motion, such as centripetal acceleration.
  • To calculate angular velocity when the frequency of rotation is given in revolutions per second (rev/s), multiply the number of revolutions by \(2\pi\), since one complete revolution equals \(2\pi\) radians.
  • For example, if a test tube spins at 1000 rev/s, like in our problem, its angular velocity \(\omega\) would be \(1000 \times 2\pi = 2000\pi\) radians per second.
  • Understanding this conversion is crucial as most rotational physics calculations require angular velocity in radians per second.
Thus, mastering the concept of angular velocity allows one to analyze and solve problems involving rotational or circular motions accurately.
Revolutions per Second
Revolutions per second (rev/s) express how many complete cycles an object makes in one second. It is a unit of frequency specifically used for rotational motion and provides a straightforward way to describe how fast an object is spinning.
  • In problems involving circular motion, revolutions per second often need to be converted into radians per second, which is the standard unit for angular velocity.
  • To perform this conversion, recognize that one revolution comprises \(2\pi\) radians. Therefore, the conversion factor for revolutions to radians is \(2\pi\).
  • In our example, converting 1000 revolutions per second, this means the test tube completes 1000 cycles each second, which translates into \(1000 \times 2\pi\) radians per second for further calculations involving rotational dynamics.
Understanding revolutions per second is essential for moving seamlessly from qualitative to quantitative analysis of rotational systems.
Standard g
The term "standard g" refers to the standard acceleration due to gravity at the Earth's surface, which is approximately \(9.81 \; \text{m/s}^2\). It is a key reference value in physics to express gravitational forces relative to Earth's gravity.
  • Standard g is often used to compare accelerations with gravitational acceleration, making it easier to comprehend the magnitude of forces involved.
  • In problems involving centripetal acceleration, like the ultracentrifuge problem, expressing the acceleration in terms of standard g gives an intuitive understanding of the forces at play compared to gravity.
  • To convert an acceleration in \(\text{m/s}^2\) to g's, divide by 9.81. For instance, the resulting centripetal acceleration of \(3947841.76 \text{ m/s}^2\) is approximately \(402533.5\) g's, illustrating the immense force involved compared to Earth's gravity.
This concept helps students and professionals alike bridge the gap between common everyday experiences of gravity and the often huge accelerations encountered in various technological processes.
Conversion to SI Units
Converting measurements to Standard International (SI) units is a fundamental practice in science and engineering. It ensures consistency, ease of communication, and accuracy in calculations.
  • In the context of the ultracentrifuge problem, the radius was initially given in centimeters. To adhere to SI units, this needed to be converted to meters.
  • Since one meter equals 100 centimeters, convert the given radius of 10 cm to meters by dividing by 100, resulting in \(0.1\) meters.
  • Performing conversions to SI units allows for direct application of standard equations and avoids potential calculations errors derived from improper unit conversion or usage.
Being proficient in converting units to the SI system is vital for solving physics problems accurately and efficiently. It also ensures the results can be universally understood and applied.

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Most popular questions from this chapter

The gun of a coautal battery in empleced on a hill \(50 \mathrm{~m}\) abowe the water level. It fires a shot with a mauzle speed of \(600 \mathrm{~m} / \mathrm{s}\). at a ship at a honizontal distance of \(12000 \mathrm{~m}\). What clevabion angle must the gun have if the shot is to hit the ship? Pretend there in no air resistance.

According to the Gwinness Book of World Recerds, during a catastrophic explosion in Halifax on December 6,1917, William Becher wae thrown throuph the air for nome \(1500 \mathrm{~m}\) and was found, itill alive, in a tree. Aseume that Becluer left the ground and rctumed to the gacund (ignore the height of the trcc) at an anple of \(45^{\circ}\). With what sipeed did he leave the pround? How high did he rise? How long did he stay in flight?

A hattelebip steming at \(45 \mathrm{~km} / \mathrm{h}\) fires u gun at right angles to the longitudinal axis of the ship. The cicvation angle of the gun is \(30^{\prime}\), and the muzde velncity of the shot is \(720 \mathrm{~m} / \mathrm{s}\) the gravitational acccieration in \(9.8 \mathrm{~m} / \mathrm{s}^{2}\), What is the range of thin abot in therefcrence frame of the grocnd? Prctend that there is no air retistaroc.

At the cntrance to Ambrone Channel at New York hubor, thc maximum tidal carrent has a velocity of \(4.2 \mathrm{~km} / \mathrm{h}\) in a dircetion \(20^{\circ}\) south of eact. What is the component of this velocity] in the east direction? In the north direction?

Two football players are initially \(20 \mathrm{~m}\) apert. The fint player (a roeciver) runs perpendicularly to the initial line joining the two players at a constant ipeed of \(7.0 \mathrm{~m} / \mathrm{s}\). After two seconds, the scoond player (the quarterback) throws the ball at a horizontal fpeed of \(15 \mathrm{~m} / \mathrm{n}\) (ignofe any vertical motion). In what herizontal difcetion should the quarterback aim wo that the hall reaches the name spot the recciver will be? At what tinc will the ball be cadpht?
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