/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 56 Find the linear speed of the bot... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 56

Find the linear speed of the bottom of a test tube in a centrifuge if the centripetal acceleration there is 52,000 times the acceleration of gravity. The distance from the axis of rotation to the bottom of the test tube is \(7.5 \mathrm{cm}\).

Short Answer

Expert verified
The linear speed of the bottom of the test tube is approximately 195.46 m/s.

Step by step solution

01

Convert Distance Units

First, convert the distance from centimeters to meters. The distance given is 7.5 cm. By converting this into meters, we get:\[ 7.5 \text{ cm} = 0.075 \text{ m} \]
02

Define Centripetal Acceleration

The centripetal acceleration at the bottom of the test tube is given as 52,000 times the acceleration due to gravity. The acceleration due to gravity \( g \) is approximately 9.81 m/s虏. So, the centripetal acceleration \( a_c \) is:\[ a_c = 52,000 \times 9.81 \text{ m/s}^2 \]
03

Calculate Centripetal Acceleration

Now, calculate the centripetal acceleration using the equation from the previous step:\[ a_c = 52,000 \times 9.81 \approx 509,320 \text{ m/s}^2 \]
04

Use the Formula for Linear Speed

The formula for centripetal acceleration \( a_c \) in terms of linear speed \( v \) and radius \( r \) is:\[ a_c = \frac{v^2}{r} \]We need to solve for \( v \):\[ v = \sqrt{a_c \times r} \]
05

Substitute Values into Formula

Substitute the values of \( a_c \) and \( r \) into the formula \[ v = \sqrt{509,320 \times 0.075} \]
06

Calculate Linear Speed

Now, compute the linear speed by solving the equation:\[ v = \sqrt{509,320 \times 0.075} \]\[ v = \sqrt{38,199} \]\[ v \approx 195.46 \text{ m/s} \]

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Key Concepts

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

Understanding Linear Speed
Linear speed is a way to describe how fast an object is traveling in a straight path. Even when objects are rotating or revolving in a circle, we often want to know their speed in a straight line, which is what linear speed tells us.
In the case of a rotating object like a centrifuge, the linear speed refers to how fast a point on the edge, or at a specific radius, is traveling. Here, in our exercise, we're interested in the speed at the bottom of the test tube.
Simply put, linear speed describes how quickly an object would be moving if you were to "unroll" its circular motion into a straight path. It helps connect rotational motion to familiar straight-line motion.
  • Linear speed is determined by both the radius from the axis of rotation and the angular speed of the rotating object.
  • In this exercise, we use the centripetal acceleration formula to find linear speed.
  • Units for linear speed are often meters per second (m/s), which is similar to other speeds we encounter in daily life.
Centrifuge Mechanics
A centrifuge is a machine that spins objects around a central point. This process forces objects outwards, which is ideal for separating substances.
In a centrifuge, samples like test tubes are placed at a certain distance from the axis of rotation. When the machine spins, it exerts a force that makes denser substances move outward.
  • The key element of a centrifuge is its capability to create extremely high centripetal accelerations, much greater than gravity.
  • In our exercise, the centripetal acceleration is 52,000 times the acceleration due to gravity, which results in the substantial outward force.
  • Understanding how a centrifuge works gives insights into how different components are separated based on density.
The Role of the Rotation Axis
The rotation axis in any rotating object or system is the line about which the object spins. In our exercise with the centrifuge, this axis is crucial because it helps determine the radius to the point where linear speed and centripetal acceleration are measured.
  • The distance from the rotation axis to a particular point on the object, like our test tube's bottom, is referred to as the radius.
  • The greater the radius, the higher the linear speed for the same angular speed, because the outer points need to cover more distance in one revolution.
  • Understanding the rotation axis helps us determine the right parameters to use when calculating speed and forces in rotating systems.

The axis serves as a fixed point that aids in calculating how fast different parts of a rotating object are moving relative to one another.

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Most popular questions from this chapter

When a block is placed on top of a vertical spring, the spring compresses \(3.15 \mathrm{cm}\). Find the mass of the block, given that the force constant of the spring is \(1750 \mathrm{N} / \mathrm{m}\).

As plants grow, they tend to align their stems and roots along the direction of the gravitational field. This tendency, which is related to differential concentrations of plant hormones known as auxins, is referred to as gravitropism. As an illustration of gravitropism, experiments show that seedlings placed in pots on the rim of a rotating turntable do not grow in the vertical direction. Do you expect their stems to tilt inward-toward the axis of rotation -or outward-away from the axis of rotation?

Force Times Distance I At the local hockey rink, a puck with a mass of \(0.12 \mathrm{kg}\) is given an initial speed of \(v=5.3 \mathrm{m} / \mathrm{s}\) (a) If the coefficient of kinetic friction between the ice and the puck is 0.11 , what distance \(d\) does the puck slide before coming to rest? (b) If the mass of the puck is doubled, does the frictional force \(F\) exerted on the puck increase, decrease, or stay the same? Explain. (c) Does the stopping distance of the puck increase, decrease, or stay the same when its mass is doubled? Explain. (d) For the situation considered in part (a), show that \(F d=\frac{1}{2} m v^{2}\). The significance of this result will be discussed in Chapter 7 , where we will see that \(\frac{1}{2} m v^{2}\) is the kinetic energy of an object.)

A force of \(9.4 \mathrm{N}\) pulls horizontally on a \(1.1-\mathrm{kg}\) block that slides on a rough, horizontal surface. This block is connected by a horizontal string to a second block of mass \(m_{2}=1.92 \mathrm{kg}\) on the same surface. The coefficient of kinetic friction is \(\mu_{k}=0.24\) for both blocks. (a) What is the acceleration of the blocks? (b) What is the tension in the string?

The equilibrium length of a certain spring with a force constant of \(k=250 \mathrm{N} / \mathrm{m}\) is \(0.18 \mathrm{m}\). (a) What is the magnitude of the force that is required to hold this spring at twice its equilibrium length? (b) Is the magnitude of the force required to keep the spring compressed to half its equilibrium length greater than, less than, or equal to the force found in part (a)? Explain.
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