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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).

Strategy

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.

Solution

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} } "."} {}

Discussion

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).

Strategy

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.

Solution

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} } } {}

Discussion

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

Preparation and Applications of Nanomaterial for Drug Delivery
Hafiz Reply
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
Jyoti Reply
I only see partial conversation and what's the question here!
Crow Reply
what about nanotechnology for water purification
RAW Reply
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
what is the stm
Brian Reply
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
LITNING Reply
what is a peer
LITNING Reply
What is meant by 'nano scale'?
LITNING Reply
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
Bob Reply
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
Damian Reply
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
Stoney Reply
why we need to study biomolecules, molecular biology in nanotechnology?
Adin Reply
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
Adin
why?
Adin
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
Damian Reply
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
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Source:  OpenStax, Introduction to applied math and physics. OpenStax CNX. Oct 04, 2012 Download for free at http://cnx.org/content/col11426/1.3
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