Problem 92
Question
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?
Step-by-Step Solution
Verified Answer
(a) 18.52 m/s, (b) 35.34 rpm, (c) 1.70 s
1Step 1: 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 \( r = 5.0 \mathrm{~m} \).
2Step 2: 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, \( v = \sqrt{343.0} \approx 18.52 \mathrm{~m/s} \).
3Step 3: 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: \( 0.589 \times 60 \approx 35.34 \mathrm{~rpm} \).
4Step 4: 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.
Key Concepts
Understanding Circular MotionCalculating Revolutions per Minute (RPM)Exploring Centrifugal Force
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—this is known as centripetal acceleration.
The formula for centripetal acceleration is given by:
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.
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—this is known as centripetal acceleration.
The formula for centripetal acceleration is given by:
- \( a_c = \frac{v^2}{r} \)
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:
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
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—the 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.
This concept is crucial for ensuring safety and stability in systems that rely on rotation for function.
It's important to note that centrifugal force is not a real force acting on the object; rather, it is the result of inertia—the 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.
This concept is crucial for ensuring safety and stability in systems that rely on rotation for function.
Other exercises in this chapter
Problem 88
Long flights at midlatitudes in the Northern Hemisphere encounter the jet stream, an eastward airflow that can affect a plane's speed relative to Earth's surfac
View solution Problem 89
A particle starts from the origin at \(t=0\) with a velocity of \(8.0 \hat{\mathrm{j}} \mathrm{m} / \mathrm{s}\) and moves in the \(x y\) plane with constant ac
View solution Problem 93
\(A\) is \(90 \mathrm{~km}\) due west of oasis \(B\). A desert camel leaves \(A\) and takes \(50 \mathrm{~h}\) to walk \(75 \mathrm{~km}\) at \(37^{\circ}\) nor
View solution Problem 96
For women's volleyball the top of the net is \(2.24 \mathrm{~m}\) above the floor and the court measures \(9.0 \mathrm{~m}\) by \(9.0 \mathrm{~m}\) on each side
View solution