/*! 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 57 To test the effects of high acce... [FREE SOLUTION] | 91Ó°ÊÓ

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To test the effects of high acceleration on the human body, the National Acronautics and Space Administration (NASA) has constructed a large centrifuge at the Manned Spacecraft Center in Houston. In this device, astronauts are placed in a capsule that moves in a circular path with a radius of \(15 \mathrm{m}\). If the astronauts in this centrifuge experience a centripetal acceleration 9.0 times that of gravity, what is the linear speed of the capsule?

Short Answer

Expert verified
The linear speed of the capsule is approximately 36.37 m/s.

Step by step solution

01

Understand the problem

The question is asking for the linear speed of a capsule moving in a circle with a given radius, experiencing centripetal acceleration that is 9 times the acceleration due to gravity. We know the radius of the circular path is 15 m.
02

Formula for centripetal acceleration

The formula for centripetal acceleration is \( a_c = \frac{v^2}{r} \), where \( v \) is the linear speed and \( r \) is the radius. We need to rearrange this formula to solve for \( v \), the linear speed.
03

Relate centripetal acceleration to gravitational acceleration

We are given that the centripetal acceleration is 9 times the gravitational acceleration \( g \). Therefore, \( a_c = 9g \), where \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.
04

Set up the equation for linear speed

Substitute \( 9g \) for \( a_c \) in the formula: \( 9g = \frac{v^2}{r} \). Now solve for \( v \) by multiplying both sides by \( r \): \( v^2 = 9gr \).
05

Solve for the linear speed

Substitute the known values: \( r = 15 \, \text{m} \) and \( g = 9.8 \, \text{m/s}^2 \). So, \( v^2 = 9 \times 9.8 \times 15 \). Calculate \( v \) by taking the square root of this product.
06

Calculate and conclude

Perform the calculations: \( 9 \times 9.8 \times 15 = 1323 \, \text{m}^2/\text{s}^2 \). Then, \( v = \sqrt{1323} \approx 36.37 \, \text{m/s} \). Therefore, the linear speed of the capsule is approximately 36.37 m/s.

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Key Concepts

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

Linear Speed
Linear speed is a key concept when discussing objects in motion, particularly in circular paths. In simple terms, linear speed refers to how fast an object is moving along its path. It's the distance traveled over time, similar to how we calculate speed in everyday situations. However, when an object is moving in a circle, like the NASA centrifuge capsule, linear speed becomes even more crucial as it determines how fast the capsule covers the circular path.

While in circular motion, linear speed is related to two main factors:
  • The circumference of the circle - the larger the circle, the more distance the object needs to cover.
  • The time it takes to complete one revolution - the faster the object circles, the higher the linear speed.
In physics, linear speed in circular motion is often given by the equation \(v = r \times \omega\), where \(r\) is the radius and \(\omega\) is the angular speed. In typical problems like the one from NASA, we derive linear speed by rearranging centripetal acceleration formulas to suit given conditions.
Gravitational Acceleration
Gravitational acceleration is a fundamental force experienced by all objects with mass. On Earth, this acceleration due to gravity, denoted by \( g \), is approximately \(9.8 \, \text{m/s}^2\). It represents how fast an object will increase in speed as it falls toward the Earth.

In problems involving circular motion, gravitational acceleration is often a reference point. For instance, when astronauts experience forces in a centrifuge, we might compare these forces to gravity to make them more relatable. This exercise entails a centripetal acceleration of 9 times the gravitational acceleration, showing that the force felt by astronauts in the centrifuge is considerably more intense than normal gravitational pull.

It is important to grasp that, while we typically think of gravitational acceleration as a downward force, in circular motion exercises, we expand our understanding to include horizontal and radial forces, using \(g\) as a pivotal comparison.
Circular Motion
Circular motion refers to the movement of an object around a central point in a path that forms a circle. Objects moving in circular motion are constantly changing direction, even if they maintain the same speed. Hence, they are undergoing acceleration due to the continuous change in direction of velocity.

Several forces are at play in circular motion, but the most significant one is centripetal force, which acts towards the center of the circle, keeping the object on its curved path. This force is directly tied to centripetal acceleration, described by the formula \( a_c = \frac{v^2}{r} \), guiding calculations involving speed and radius.

For circular motion problems, understanding the interplay between speed, radius, and centripetal acceleration is crucial. In our NASA example, these concepts are practically applied by calculating linear speed when the acceleration drastically exceeds everyday gravitational forces, simulating intense conditions astronauts might experience in space missions.

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Most popular questions from this chapter

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.)

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A tie of uniform width is laid out on a table, with a fraction of its length hanging over the edge. Initially, the tie is at rest. (a) If the fraction hanging from the table is increased, the tie eventually slides to the ground. Explain. (b) What is the coefficient of static friction between the tie and the table if the tie begins to slide when one-fourth of its length hangs over the edge?

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