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Chapter 5: Problem 63

An astronaut is training in an earthbound centrifuge that consists of a small chamber whirled horizontally at the end of a \(5.1-\mathrm{m}-\) long shaft. The astronaut places a notebook on the vertical wall of the chamber and it stays in place. If the coefficient of static friction is \(0.62,\) what's the minimum rate at which the centrifuge must be revolving?

Short Answer

Expert verified
The minimum rate at which the centrifuge must be revolving can be calculated from the aforementioned equations. The gravitational force must equal the frictional force, so setting those two equal to each other and solving for the velocity provides the solution.

Step by step solution

01

Identify Known Variables

Given in the problem is the length of the shaft (radius of the circular path) \( r = 5.1 \, m \), and the coefficient of static friction \( \mu_s = 0.62 \). The acceleration due to gravity \( g \) is known to be \( 9.8 \, m/s^2 \).
02

Apply the Condition for Static Equilibrium

The notebook does not move relative to the centrifuge. This means the frictional force \( F_f = \mu_s \cdot F_n \), should balance with the gravitational force \( F_g = m \cdot g \). Here, \( m \) is the mass of the notebook and \( F_n \) is the normal force. Considering that in a rotating frame the normal force is supplied by the centripetal (radial) force \( F_n = m \cdot v^2 / r \), where \( v \) is the centrifuge velocity.
03

Equate and Solve for Velocity

Equating the frictional force and gravitational force, we get \( m \cdot g = \mu_s \cdot m \cdot v^2 / r \). The mass \( m \) cancels out, leaving us with the equation for the velocity \( v \) which yields \( v = \sqrt{g \cdot r / \mu_s} \). Substituting the given and known values into this equation and solving for \( v \), we can find the minimum velocity required for the notebook to stay in place.

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