Centripetal Force Note Taking Guide
Centrifugal force is ubiquitous in our daily lives. We experience it when we round a corner in a car or when an airplane banks into a turn. We see it in the spin cycle of a washing machine or when children ride on a merry-go-round. One day it may even provide artificial gravity for space ships and space stations. Some people confuse centrifugal force with its counterpart, centripetal force, because they are so closely related. One might say they are two sides of the same coin.
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Is defined as, “The component of force acting on a body in curvilinear motion that is directed toward the center of curvature or axis of rotation,” while is defined as, “The apparent force, equal and opposite to the centripetal force, drawing a rotating body away from the center of rotation, caused by the inertia of the body,” according to the American Heritage Dictionary. Note that while centripetal force is an actual force, centrifugal force is defined as an apparent force. In other words, when twirling a mass on a string, the string exerts an inward centripetal force on the mass, while mass appears to exert an outward force on the string. “The difference between centripetal and centrifugal force has to do with different ‘frames of reference,’ that is, different viewpoints from which you measure something,” according to Andrew A.
Ganse, a research physicist at the University of Washington. If you are observing a rotating system from the outside, you see an inward centripetal force acting to constrain the rotating body to a circular path. However, if you are part of the rotating system, you experience an apparent centrifugal force pushing you away from the center of the circle, even though what you are actually feeling is the inward centripetal force that is keeping you from literally going off on a tangent. This apparent outward force is described. States that “A body at rest will remain at rest, and a body in motion will remain in motion unless it is acted upon by an external force.” If a massive body is moving through space in a straight line, its inertia will cause it to continue in a straight line unless an outside force causes it to speed up, slow down or change direction. In order for it to follow a circular path without changing speed, a continuous centripetal force must be continuously applied at a right angle to its path.
The radius r of this circle is equal to the mass m times the square of the velocity v divided by the centripetal force F, or r = mv 2/ F. The force can be calculated by simply rearranging the equation, F= mv 2/ r.
Note Taking Guide – Centripetal Force Notes on centripetal force: • • Physics Challenge: When a car turns to the left, why do passengers slide to the right? Circular motion is covered in almost every physics class. This article steps you through the algebra-based derivation of the centripetal force equation.
Centripetal force is the inward-seeking pull of an object. Centrifugal force is the opposite; it appears to pull objects away from the center or axis of rotation. Credit: Fouad A. Saad Shutterstock There are many applications that exploit centripetal force. One is to simulate the acceleration of a space launch for astronaut training.
When a rocket is first launched, it is so laden with fuel and oxidizer that it can barely move. However, as it ascends, it burns fuel at a tremendous rate, continuously losing mass. States that force equals mass times acceleration, or F = ma. In most situations, mass remains constant. With a rocket, though, its mass changes drastically, while the force, in this case the thrust of the rocket motors, remains nearly constant. This causes the acceleration toward the end of the boost phase to increase to several times that of normal gravity.
To test astronauts and prepare them for this extreme acceleration. In this application, the centripetal force is provided by the seat back pushing inward on the astronaut. Another application for centripetal force is the, which is used to accelerate the precipitation of particles suspended in liquid. One common use of this technology is for fractionating blood samples.
According to, “The unique structure of blood makes it very easy to separate red blood cells from plasma and the other formed elements by differential centrifugation.” Under the normal force of gravity, thermal motion causes continuous mixing which prevents blood cells from settling out of a whole blood sample. However, a typical centrifuge can achieve accelerations that are 600 to 2,000 times that of normal gravity.
This forces the heavy red blood cells to settle at the bottom and stratifies the other various components into layers according to their density. Can also be used to separate the lighter and rarer uranium-235 isotope from the heavier and much more common uranium-238. A uranium oxide powder called yellow cake (U 3O 8) is first converted to uranium hexafluoride gas (UF 6) and piped into a spinning cylinder. The heavier U-238 is forced to the outside while the lighter U-235 moves to the inside, where the cylinder is tapped and the gas is piped to the next centrifuge in the array for further enrichment. It takes multiple steps to enrich the gas to just 3 to 5 percent U-235, which is required to fuel nuclear power plants, and many more steps are required to produce highly enriched weapons-grade uranium which is over 90 percent pure U-235. According to the Georgia State University website, this apparent centrifugal force depends on the mass of the object being rotated, times its square of its angular velocity, i.e., its rate of rotation, divided by its distance from the center of rotation.
This is expressed as F = mv 2/ r. This gives rise to an interesting phenomenon. When a container filled with a liquid is rotated at a constant rate, the liquid moves toward the outside of the container against the force of gravity which is pulling down on the liquid trying to flatten it out. The angular velocity increases linearly the farther you get from the center. However, because the effective outward force depends on the square of the angular velocity, the height of the liquid is proportional to the square of the distance from the center.
For this reason, the surface takes on the shape of a, which happens to be the ideal shape for a telescope mirror. The focal length of the mirror is determined by the rotational rate and can be calculated as f = g/ ω 2, where f is the focal length, g is the acceleration of gravity and ω is the rate of rotation in radians per second (2π radians equals one complete circle). This is the principle behind a (LMT). LMTs use a reflecting liquid metal (usually mercury) that is rotated so as to form a parabolic reflecting surface. In order to save weight and cost, the bowl containing the mercury is close to the desired parabolic shape so only a thin layer of the liquid metal is needed. The advantage of this system is that it avoids the cost of casting, grinding, polishing, and coating a solid mirror.
The disadvantage is that large LMTs can only look straight up and scan the sky as the earth turns. Conversely, smaller LMTs with mirror diameters of a few inches can use flat mirrors in the optical path to look in any direction. According to Ganse, “Centripetal force and centrifugal force are really the exact same force, just in opposite directions because they're experienced from different frames of reference.” This brings us to, which states, “For every action, there is an equal and opposite reaction.” Just as gravity causes you to exert a force on the ground, the ground appears to exert an equal and opposite force on your feet. When you are in an accelerating car, the seat exerts a forward force on you just as you appear to exert a backward force on the seat. In the case of a rotating system, the centripetal force pulls the mass inward to follow a curved path, while the mass appears to push outward due to its inertia. In each of these cases, though, there is only one real force being applied, while the other is only an apparent force. Additional resources.
Jim Lucas, Live Science Contributor Jim Lucas is a contributing writer for Live Science. He covers physics, astronomy and engineering.
Jim graduated from Missouri State University, where he earned a bachelor of science degree in physics with minors in astronomy and technical writing. After graduation he worked at Los Alamos National Laboratory as a network systems administrator, a technical writer-editor and a nuclear security specialist. In addition to writing, he edits scientific journal articles in a variety of topical areas.
Explanation: We are simply asked to find the centripetal acceleration, which is given by: We were given in the problem statement (radius will be equal to the length of the string), so we only need to find the velocity of the ball. We are told that it travels in a circle with radius 1.5m and completes two full rotations per second. The length of each rotation is just the circumference of the circle: The velocity can be found by multiplying that distance by the frequency: Now we have all of our variables and can plug into our first equation. Explanation: There is a lot going on in this problem and it will take several steps to get to the answer. However, when you boil the question down, we are pretty much asked how fast must the ride spin so that the centripetal force on the student provides enough static friction to keep the student from falling.
Let's work through this problem one step at a time. First, let's figure out what minimal static frictional force is required. This force will be equal to the weight of the student; the student's weight will pull downward, while the friction of the wall pushes upward. Now we can calculate the normal force required to reach that magnitude of frictional force. Note that the vector for the normal force will be perpendicular to the wal, directed toward the center of the circlel.
This normal force is the minimum centripital force required to keep the student pinned to the wall. We can convert this to centripital acceleration: We can now convert centripital acceleration to a translational velocity using the equation: Rearranging for velocity, we get: This is the velocity that the outer wall of the ride must be spinning.
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Since we know the radius of the ride, we can convert this velocity into a maximum period, the final answer: If you weren't sure how to come about this equation, just think about your units. You know you need to get to units of seconds, and you have a value with units of m/s. Therefore, you need to cancel out the meter and get the second on top.
Explanation: We can use the expression for conservation of energy to solve this problem: Substituting in our expressions for each variable and removing initial kinetic energy and final potential energy (which will each be zero), we get: Rearranging for final velocity: The only variable we don't have at this point is the distance the gumball travels. However, we can calculate it knowing the height of the track and its slope. We can imagine that the spiral track is unwound, creating a right triangle with an angle of 10 degrees and a height of 1.5m.
For this triangle, the hypotenuse will be the total distance of the track. Now that we have all of our variables, we can solve for the final velocity: We can then use this to calculate the centripetal force on the gumball. Explanation: First, we need to identify at which point in the circle the string is experiencing the most tension. There are two total forces in the system: gravity and tension. It is important to note that the tension isn't only resulting from gravity; it also includes the centripetal force required to keep the ball in circular motion.
Thinking practically, we can say that the greatest tension will be when the ball is at its lowest point (gravity and tension are in opposite directions). At this point we can write: Expand our terms for force: We know the acceleration due to gravity, but we need to determine the centripetal accerleration. The formula for that is: We know the radius (length of the string), so we need to develop an expression for velocity. We can use the period and circumference of circle: Here, we use to denote period. Substituting this into the expression for centripetal acceleration: Substituting this back into the equation for tension, we get: We have all of these values, allowing us to solve. Explanation: To start off with, it is useful to consider the energy of the system.
Initially, the block is at a certain unknown height, and will thus have gravitational potential energy. Since we are assuming that there is no friction in this case, then we know that total mechanical energy will be conserved. Therefore, all of the gravitational potential energy contained in the block will become kinetic energy when it slides down to the bottom of the loop. But, once the block begins to slide up the loop, it will lose kinetic energy and will regain some gravitational potential energy. Therefore, at the top of the loop, the block will have a combination of kinetic and gravitational potential energy whose sum is equal to the initial energy of the system. Due to conservation of energy, we can equate the two.
In addition to considering energy, it is also necessary to consider the forces acting on the block in this scenario. When at the top of the loop, the block will experience a downward force due to its weight, and another downward force due to the normal force of the loop on the block. Furthermore, because the block is traveling along a circular path while in the loop, it will experience a centripetal force. At the top of the loop, the centripetal force will be due to a combination of the weight of the block as well as the normal force. We're looking for a starting height that will just allow the block to travel around the loop. The minimum height will be the height such that the block will just start to fall off.
When falling off, the block will no longer be touching the loop and therefore, the normal force will be equal to zero. This is the situation we are looking for, and since the normal force is zero, only the block's weight will contribute to the centripetal force at the top of the loop. The preceding expression gives us the value of velocity that will allow the block to have enough kinetic energy while at the top of the loop to not fall off.
We can plug this expression into the previous energy equation. Explanation: Like all force problems, this one starts with a clear free body diagram: The tension points along the vine (tensions can only pull), so it goes straight up. The force of gravity points straight down, as it always does. The two do not add to zero, however, since the person is undergoing circular motion. Instead, they add to a net force pointing towards the center of the circle that the person is making, which is up at the place where the vine is attached to the tree. Draw a vector diagram: Then write the equation about the lengths of the vectors: the length of the gravity vector plus the length of the net force vector equals the length of the tension vector: The net force for an object undergoing circular motion is mass times speed squared divided by the radius of the circle. Is the gravity force constant.
Some use, but the AP physics 1 test allows you to use, which makes it a lot easier. Plug in the numbers, and solve for the tension: This answer is reasonable since the vine has to both hold the person up and provide a centripetal force; that is why the tension is is greater than his weight alone. If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.
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