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An advertisement claims that a centrifuge takes up only \(0.127 \mathrm{~m}\) of bench space but can produce a radial acceleration of \(3000 g\) at 5000 rev \(/\) min. Calculate the required radius of the centrifuge. Is the claim realistic?

Short Answer

Expert verified
After solving the formula for r, you'll find that r is much less than \(0.127\mathrm{~m}\). Therefore, the claim that a centrifuge takes up only \(0.127\mathrm{~m}\) of bench space but can produce a radial acceleration of 3000 g at 5000 rev/min is realistic.

Step by step solution

01

Understanding the problem

The key is to convert the given acceleration from g to m/s^2 and rev/min to rad/s by applying the provided conversion unit. Then apply the formula \(a = \omega^2r \) to calculate the radius.
02

Conversion of units

Before proceeding, convert the provided acceleration from g to m/s^2 and rotation speed from rev/min to rad/s. Since 1g is equal to \(9.81 \mathrm{ ~m/s^2}\), 3000g is \(3000 * 9.81 = 29430 \mathrm{~m/s^2}\). Next, let's convert the rotation speed: as you know, 1 rev corresponds to \(2\pi\) rad, and one minute contains 60 seconds. Thus , 5000 rev/min is \(5000 * \frac{2\pi}{60} = \frac{500\pi}{6} \mathrm{~rad/s}\)
03

Calculate the Radius

Now that you have the acceleration and angular velocity, it's time to calculate the radius using the formula \(a = \omega^2r \). Substitute the values into the formula which will look like this: \(29430 = (\frac{500\pi}{6})^2 * r\). Now, solve the equation for r.
04

Check whether the claim is realistic or not

Check if the calculated r is less than or equal to the given bench space of \(0.127\mathrm{~m}\). If the r value does not exceed \(0.127\mathrm{~m}\), then the claim is realistic, otherwise it is not.

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Key Concepts

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

Radial Acceleration
Radial acceleration is a crucial concept when analyzing the motion of objects moving in a circular path, such as in a centrifuge. It's the acceleration directed towards the center of the circle and is responsible for changing the direction of an object's velocity without altering its speed. In a centrifuge, this force is generated by its spinning motion and results in the separation of substances with different densities.

In our exercise, radial acceleration is given as a multiple of the Earth's gravitational acceleration, 'g', where 1g equals the acceleration due to gravity at the Earth's surface (9.81 \( \mathrm{m/s^2} \)). The centrifuge in the advertisement claims to achieve an acceleration of 3000g, which is 3000 times the standard gravity. What this means in practice is that the substances within the centrifuge are experiencing 3000 times the force they would feel if they were simply resting on the Earth's surface. This massive increase in acceleration is what enables the centrifuge to effectively separate its contents.

Understanding radial acceleration is key to assessing whether the centrifuge's claim is realistic. Higher radial acceleration requires a correspondingly designed centrifuge, which must be reinforced to withstand the increased forces without failure.
Unit Conversion
Unit conversion is a fundamental skill in physics problems, where we often need to convert between different units of measurement to make meaningful calculations. With the centrifuge problem, conversions from 'g' to \( \mathrm{m/s^2} \) and from rev/min to rad/s are necessary to use the radial acceleration formula correctly.

For acceleration, the conversion involves multiplying the value in 'g' by the acceleration due to gravity. Specifically, for 3000g, we calculate 3000 \( * \) 9.81 to convert to \( \mathrm{m/s^2} \). Similarly, for angular velocity, we convert rotations per minute (rev/min) to radians per second (rad/s) by utilizing the relationship that one revolution equals \( 2\flexibleSpace\pi\) radians and one minute equals 60 seconds.

Accurate unit conversions are critical for making precise calculations in physical applications. Mismatched or incorrect units can lead to substantial errors, potentially leading to the failure of the system being designed or analyzed, such as the centrifuge in our exercise.
Angular Velocity
Angular velocity represents how fast an object rotates or revolves in relation to a center point. In mathematical terms, it's the rate of change of the angle that the spinning object subtends at the center of rotation. This concept is vital when dealing with centrifuges, as their effectiveness is directly linked to the angular velocity.

In our centrifuge exercise, angular velocity is given in rev/min, but we convert it to the standard unit of rad/s for our calculations. With a higher angular velocity, the centrifuge can induce greater radial acceleration, leading to faster and more effective separation of mixtures.

The formula that relates radial acceleration to angular velocity is \( a = \flexibleSpace\omega^2r \), where 'a' is the radial acceleration, '\(\omega\)' is the angular velocity, and 'r' is the radius of the circular path. By understanding angular velocity, we can determine the centrifuge design's efficacy and safety, ensuring that it will function as intended under the specified conditions.

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Most popular questions from this chapter

An electric turntable \(0.750 \mathrm{~m}\) in diameter is rotating about a fixed axis with an initial angular velocity of \(0.250 \mathrm{rev} / \mathrm{s}\) and a constant angular acceleration of \(0.900 \mathrm{rev} / \mathrm{s}^{2}\). (a) Compute the angular velocity of the turntable after \(0.200 \mathrm{~s}\). (b) Through how many revolutions has the turntable spun in this time interval? (c) What is the tangential speed of a point on the rim of the turntable at \(t=0.200 \mathrm{~s} ?\) (d) What is the magnitude of the resultant acceleration of a point on the rim at \(t=0.200 \mathrm{~s} ?\)

You are designing a rotating metal flywheel that will be used to store energy. The flywheel is to be a uniform disk with radius \(25.0 \mathrm{~cm} .\) Starting from rest at \(t=0,\) the flywheel rotates with constant angular acceleration \(3.00 \mathrm{rad} / \mathrm{s}^{2}\) about an axis perpendicular to the flywheel at its center. If the flywheel has a density (mass per unit volume) of \(8600 \mathrm{~kg} / \mathrm{m}^{3},\) what thickness must it have to store \(800 \mathrm{~J}\) of kinetic energy at \(t=8.00 \mathrm{~s} ?\)

A thin, rectangular sheet of metal has mass \(M\) and sides of length \(a\) and \(b\). Use the parallel-axis theorem to calculate the moment of inertia of the sheet for an axis that is perpendicular to the plane of the sheet and that passes through one corner of the sheet.

A thin, light wire is wrapped around the rim of a wheel as shown in Fig. E9.49. The wheel rotates about a stationary horizontal axle that passes through the center of the wheel. The wheel has radius \(0.180 \mathrm{~m}\) and moment of inertia for rotation about the axle of \(I=0.480 \mathrm{~kg} \cdot \mathrm{m}^{2}\). A small block with mass \(0.340 \mathrm{~kg}\) is suspended from the free end of the wire. When the system is released from rest, the block descends with constant acceleration. The bearings in the wheel at the axle are rusty, so friction there does \(-9.00 \mathrm{~J}\) of work as the block descends \(3.00 \mathrm{~m}\). What is the magnitude of the angular velocity of the wheel after the block has descended \(3.00 \mathrm{~m} ?\)

A uniform disk with radius \(R=0.400 \mathrm{~m}\) and mass \(30.0 \mathrm{~kg}\) rotates in a horizontal plane on a frictionless vertical axle that passes through the center of the disk. The angle through which the disk has turned varies with time according to \(\theta(t)=(1.10 \mathrm{rad} / \mathrm{s}) t+\left(6.30 \mathrm{rad} / \mathrm{s}^{2}\right) t^{2}\) What is the resultant linear acceleration of a point on the rim of the disk at the instant when the disk has turned through 0.100 rev?

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