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a c = v 2 r a c = 2 . size 12{a rSub { size 8{c} } = { {v rSup { size 8{2} } } over {r} } `; a rSub { size 8{c} } =rω rSup { size 8{2} } "."} {}

Recall that the direction of a c size 12{a rSub { size 8{c} } } {} is toward the center. You may use whichever expression is more convenient, as illustrated in examples below.

A centrifuge (see [link] b) is a rotating device used to separate specimens of different densities. High centripetal acceleration significantly decreases the time it takes for separation to occur, and makes separation possible with small samples. Centrifuges are used in a variety of applications in science and medicine, including the separation of single cell suspensions such as bacteria, viruses, and blood cells from a liquid medium and the separation of macromolecules, such as DNA and protein, from a solution. Centrifuges are often rated in terms of their centripetal acceleration relative to acceleration due to gravity ( g ) size 12{g} {} ; maximum centripetal acceleration of several hundred thousand g is possible in a vacuum. Human centrifuges, extremely large centrifuges, have been used to test the tolerance of astronauts to the effects of accelerations larger than that of Earth’s gravity.

How does the centripetal acceleration of a car around a curve compare with that due to gravity?

What is the magnitude of the centripetal acceleration of a car following a curve of radius 500 m at a speed of 25.0 m/s (about 90 km/h)? Compare the acceleration with that due to gravity for this fairly gentle curve taken at highway speed. See [link] (a).


Because v size 12{v} {} and r size 12{r} {} are given, the first expression in a c = v 2 r a c = 2 size 12{a rSub { size 8{c} } = { {v rSup { size 8{2} } } over {r} } `; a rSub { size 8{c} } =rω rSup { size 8{2} } } {} is the most convenient to use.


Entering the given values of v = 25 . 0 m/s size 12{v="25" "." 0`"m/s"} {} and r = 500 m size 12{r="500"} {} into the first expression for a c size 12{a rSub { size 8{c} } } {} gives

a c = v 2 r = ( 25 . 0 m/s ) 2 500 m = 1 . 25 m/s 2 . size 12{a rSub { size 8{c} } = { {v rSup { size 8{2} } } over {r} } = { { \( "25" "." 0" m/s" \) rSup { size 8{2} } } over {"500 m"} } =1 "." "25"" m/s" rSup { size 8{2} } "."} {}


To compare this with the acceleration due to gravity ( g = 9 . 80 m/s 2 ) size 12{g=9 "." 8`"m/s" rSup { size 8{2} } } {} , we take the ratio of a c / g = 1 . 25 m/s 2 / 9 . 80 m/s 2 = 0 . 128 size 12{a rSub { size 8{c} } /g= left (1 "." "25"`"m/s" rSup { size 8{2} } right )/ left (9 "." "80"`"m/s" rSup { size 8{2} } right )=0 "." "128"} {} . Thus, a c = 0 . 128 g size 12{a rSub { size 8{c} } =0 "." "128"} {} and is noticeable especially if you were not wearing a seat belt.

In figure a, a car shown from top is running on a circular road around a circular path. The center of the park is termed as the center of this circle and the distance from this point to the car is taken as radius r. The linear velocity is shown in perpendicular direction toward the front of the car, shown as v the centripetal acceleration is shown with an arrow pointed towards the center of rotation. In figure b, a centrifuge is shown an object of mass m is rotating in it at a constant speed. The object is at the distance equal to the radius, r, of the centrifuge. The centripetal acceleration is shown towards the center of rotation, and the velocity, v is shown perpendicular to the object in the clockwise direction.
(a) The car following a circular path at constant speed is accelerated perpendicular to its velocity, as shown. The magnitude of this centripetal acceleration is found in [link] . (b) A particle of mass in a centrifuge is rotating at constant angular velocity . It must be accelerated perpendicular to its velocity or it would continue in a straight line. The magnitude of the necessary acceleration is found in [link] .

How big is the centripetal acceleration in an ultracentrifuge?

Calculate the centripetal acceleration of a point 7.50 cm from the axis of an ultracentrifuge    spinning at 7.5 × 10 4 rev/min. Determine the ratio of this acceleration to that due to gravity. See [link] (b).


The term rev/min stands for revolutions per minute. By converting this to radians per second, we obtain the angular velocity ω size 12{ω} {} . Because r size 12{r} {} is given, we can use the second expression in the equation a c = v 2 r ; a c = 2 size 12{a rSub { size 8{c} } = { {v rSup { size 8{2} } } over {r} } `; a rSub { size 8{c} } =rω rSup { size 8{2} } } {} to calculate the centripetal acceleration.


To convert 7 . 50 × 10 4 rev / min size 12{7 "." "50" times "10" rSup { size 8{4} } {"rev"} slash {"min"} } {} to radians per second, we use the facts that one revolution is rad size 12{2π`"rad"} {} and one minute is 60.0 s. Thus,

ω = 7.50 × 10 4 rev min × rad 1 rev × 1 min 60 . 0 s = 7854  rad/s. size 12{ω="75","000" { {"rev"} over {"min"} } times { {2π" rad"} over {"1 rev"} } times { {1" min"} over {"60" "." "0 s"} } ="7850"" rad/s."} {}

Now the centripetal acceleration is given by the second expression in a c = v 2 r a c = 2 size 12{a rSub { size 8{c} } = { {v rSup { size 8{2} } } over {r} } `; a rSub { size 8{c} } =rω rSup { size 8{2} } } {} as

a c = 2 . size 12{a rSub { size 8{c} } =rω rSup { size 8{2} } "."} {}

Converting 7.50 cm to meters and substituting known values gives

a c = ( 0 . 0750 m ) ( 7854 rad/s ) 2 = 4 . 63 × 10 6 m/s 2 . size 12{a rSub { size 8{c} } = \( 0 "." "0750"" m" \) \( "7850"" rad/s" \) rSup { size 8{2} } =4 "." "62" times "10" rSup { size 8{6} } " m/s" rSup { size 8{2} } } {}

Note that the unitless radians are discarded in order to get the correct units for centripetal acceleration. Taking the ratio of a c size 12{a rSub { size 8{c} } } {} to g size 12{g} {} yields

a c g = 4 . 63 × 10 6 9 . 80 = 4 . 72 × 10 5 . size 12{ { {a rSub { size 8{c} } } over {g} } = { {4 "." "62" times "10" rSup { size 8{6} } } over {9 "." "80"} } 4 "." "71" times "10" rSup { size 8{5} } } {}


This last result means that the centripetal acceleration is 472,000 times as strong as g size 12{g} {} . It is no wonder that such high ω size 12{ω} {} centrifuges are called ultracentrifuges. The extremely large accelerations involved greatly decrease the time needed to cause the sedimentation of blood cells or other materials.

Questions & Answers

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
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I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
I'm not sure why it wrote it the other way
I got X =-6
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
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what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
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a perfect square v²+2v+_
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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
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or infinite solutions?
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
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rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
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Differences Between Laspeyres and Paasche Indices
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No. 7x -4y is simplified from 4x + (3y + 3x) -7y
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is it 3×y ?
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J, combine like terms 7x-4y
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how do you translate this in Algebraic Expressions
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Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
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Damian Reply
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
what is system testing
what is the application of nanotechnology?
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
anybody can imagine what will be happen after 100 years from now in nano tech world
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
silver nanoparticles could handle the job?
not now but maybe in future only AgNP maybe any other nanomaterials
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
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Source:  OpenStax, College physics arranged for cpslo phys141. OpenStax CNX. Dec 23, 2014 Download for free at http://legacy.cnx.org/content/col11718/1.4
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