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Chapter 4: Problem 92

An astronaut is rotated in a horizontal centrifuge at a radius of \(5.0 \mathrm{~m}\). (a) What is the astronaut's speed if the centripetal acceleration has a magnitude of \(7.0 \mathrm{~g} ?\) (b) How many revolutions per minute are required to produce this acceleration? (c) What is the period of the motion?

Short Answer

Expert verified
(a) 18.52 m/s, (b) 35.34 rpm, (c) 1.70 s

Step by step solution

01

Understanding Centripetal Acceleration

Centripetal acceleration is the acceleration that keeps an object moving in a circular path. It's given by the formula \( a_c = \frac{v^2}{r} \), where \( v \) is the linear speed and \( r \) is the radius of the circular path. Here, we are given \( a_c = 7.0g = 7.0 \times 9.8 \frac{m}{s^2} = 68.6 \frac{m}{s^2} \) and \( r = 5.0 \mathrm{~m} \).
02

Calculating Astronaut's Speed

Using the formula for centripetal acceleration, \( a_c = \frac{v^2}{r} \), we solve for \( v \):\[v^2 = a_c \cdot r = 68.6 \frac{m}{s^2} \cdot 5.0 \mathrm{~m} = 343.0 \mathrm{~m}^2/\mathrm{s}^2\]Thus, \( v = \sqrt{343.0} \approx 18.52 \mathrm{~m/s} \).
03

Converting Speed to Revolutions Per Minute

First, find the circumference of the path: \( C = 2\pi r = 2\pi \times 5.0 \mathrm{~m} \). The number of revolutions per second is \( \frac{v}{C} = \frac{18.52}{10\pi} \approx 0.589 \mathrm{~revs/s} \). Multiply by 60 to convert to revolutions per minute: \( 0.589 \times 60 \approx 35.34 \mathrm{~rpm} \).
04

Finding the Period of the Motion

The period \( T \) is the reciprocal of the frequency of revolutions per second. From \( 0.589 \mathrm{~revs/s} \), \( T = \frac{1}{0.589} \approx 1.70 \mathrm{~s} \). This is the time for one complete revolution.

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Key Concepts

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

Understanding Circular Motion
Circular motion occurs when an object moves along a circular path. The motion can be uniform, where the speed remains constant, or non-uniform, where the speed varies.
A key aspect of circular motion is the continuous change in direction of the velocity vector, which requires acceleration towards the center of the circle鈥攖his is known as centripetal acceleration.
The formula for centripetal acceleration is given by:
  • \( a_c = \frac{v^2}{r} \)
where \( v \) is the velocity of the object and \( r \) is the radius of the circular path.
This acceleration does not change the speed of the object, only its direction, ensuring it moves in a circle instead of off at a tangent. This is fundamental in many scenarios, such as satellites orbiting a planet, or an astronaut being trained in a centrifuge.
The concept of circular motion is critical when analyzing objects subjected to forces that keep them in curved paths, and it's the basis for understanding how centripetal forces enable objects to maintain circular trajectories.
Calculating Revolutions per Minute (RPM)
Revolutions per minute, or RPM, measure how many times an object completes a circular path in one minute.
This value helps in understanding the speed of rotation in real-world terms, bridging the gap between abstract physics calculations and observable motion.
To determine RPM from the object's velocity, we need to compute how many full circles it completes in a second, then convert that to a one-minute cycle.
Here's how you derive it:
  • Find the circumference of the circle: \( C = 2\pi r \)
  • Calculate the revolutions per second: \( \frac{v}{C} \)
  • Convert to revolutions per minute by multiplying by 60
This calculation provides insight into how fast the object is spinning or rotating, essential in mechanical systems, such as engines, where RPM is a crucial metric.
Exploring Centrifugal Force
Centrifugal force is often described as the apparent force that seems to push an object outward when it is in circular motion.
It's important to note that centrifugal force is not a real force acting on the object; rather, it is the result of inertia鈥攖he object's tendency to continue in a straight line.
When observing an object in circular motion from a rotating reference frame, such as inside a spinning machine, it feels like a force is pushing outward due to the body's resistance to change in its circular path direction.
  • The real force acting on the object to maintain its circular motion is the centripetal force, directed inward towards the center of the circle.
  • The feeling of being "pushed" outward is solely from the perspective within the rotating system.
Understanding centrifugal force is crucial in designing systems like washing machines, amusement park rides, and centrifuges, where high-speed rotation occurs.
This concept is crucial for ensuring safety and stability in systems that rely on rotation for function.

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Most popular questions from this chapter

A football player punts the football so that it will have a "hang time" (time of flight) of \(4.5 \mathrm{~s}\) and land \(46 \mathrm{~m}\) away. If the ball leaves the player's foot \(150 \mathrm{~cm}\) above the ground, what must be the (a) magnitude and (b) angle (relative to the horizontal) of the ball's initial velocity?

A particle \(P\) travels with constant speed on a circle of radius \(r=\) \(3.00 \mathrm{~m}\) (Fig. \(4-56)\) and completes one revolution in \(20.0 \mathrm{~s}\). The particle passes through \(O\) at time \(t=0 .\) State the following vectors in magnitudeangle notation (angle relative to the positive direction of \(x\) ). With respect to \(O\), find the particle's position vector at the times \(t\) of (a) \(5.00 \mathrm{~s}\), (b) \(7.50 \mathrm{~s}\), and \((\mathrm{c}) 10.0 \mathrm{~s}\). (d) For the \(5.00 \mathrm{~s}\) interval from the end of the fifth second to the end of the tenth second, find the particle's displacement. For that interval, find (e) its average velocity and its velocity at the (f) beginning and (g) end. Next, find the acceleration at the (h) beginning and (i) end of that interval.

An electron's position is given by \(\vec{r}=3.00 t \hat{\mathrm{i}}-4.00 t \hat{\mathrm{j}}+2.00 \hat{\mathrm{k}}\), with \(t\) in seconds and \(\vec{r}\) in meters. (a) In unit-vector notation, what js the electron's velocity \(\vec{v}(t) ?\) At \(t=2.00 \mathrm{~s}\), what is \(\vec{v}(\mathrm{~b})\) in unitvector notation and as (c) a magnitude and (d) an angle relative to the positive direction of the \(x\) axis?

The range of a projectile depends not only on \(v_{0}\) and \(\theta_{0}\) but also on the value \(g\) of the free-fall acceleration, which varies from place to place. In 1936 , Jesse Owens established a world's running broad jump record of \(8.09 \mathrm{~m}\) at the Olympic Games at Berlin (where \(g=9.8128 \mathrm{~m} / \mathrm{s}^{2}\) ). Assuming the same values of \(v_{0}\) and \(\theta_{0}\), by how much would his record have differed if he had competed instead in 1956 at Melbourne (where \(\left.g=9.7999 \mathrm{~m} / \mathrm{s}^{2}\right)\) ?

A projectile is fired horizontally from a gun that is \(45.0 \mathrm{~m}\) above flat ground, emerging from the gun with a speed of \(250 \mathrm{~m} / \mathrm{s}\). (a) How long does the projectile remain in the air? (b) At what horizontal distance from the firing point does it strike the ground? (c) What is the magnitude of the vertical component of its velocity as it strikes the ground?
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