why is the triangle ABC and triangle PQR similar?

the vector v1 (PR) form a right angle to AC and v2 (PQ) form a right angle to AB. Given this and a given angle between AC and AB you can draw up the lines and prove that the angle between PR and PQ must have the same angle. Thus the triangles are similar :)

how can deltaS equal deltaR?

That's a good question. Notice how the article says: when Δθ is _very small_, Δs = Δr. That's calculus at work - by _very small_ it means infinitesimally small. On a bigger scale, it's obvious that an arc and a chord are not equal, but the bigger scale is just an illustration of the ideas, an approximation. As θ gets smaller and smaller, Δs and Δr (the arc length and chord length) get closer and closer to being the same length. By using limits, you can actually prove that as Δθ approaches 0, Δs = Δr. It seems to defy "common sense", but it's true and it works. To get a little more intuitive picture of what's going on, start with a circle of radius 1 and a 90° arc, figure out the arc length and the chord length. Then figure out arc and chord for a 9° angle, then 0.9°, etc. See how close arc and chord get in just a few steps? Imagine you did that infinitely many times.

what is the real forces that provide centripetal acceleration

centripetal actually means - towards the center .So centripetal force is not a new type of force .Any force which is acting towards center can be called as centripetal force. To understand it better think of gravitational force , it acts in downwards direction so we call it downwards force because of its direction .There are only four real forces in nature i.e. Gravitational , Electromagnetic ,weak nuclear and strong nuclear . whenever any of these four forces acts towards center that force is called centripetal(towards center) . In case of earth and sun system ,real force acting is Gravitational but as it is always pointing towards center we call it centripetal . In case of an electron revolving around nucleus ,the real force between electron and positive nucleus is electromagnetic but because of its direction always pointing towards center ,it is centripetal force. Friction too is electromagnetic force.

If an object has a centripetal acceleration towards the center, why does it not go towards the center?

Technically they are. They are "falling", but also moving sideways at a large velocity, so they maintain a circular path.

Why does centripetal force does not work?

because the force is always perpendicular to the displacement.

what is meant by utlracentrifuge?

An ultracentrifuge is just a centrifuge that operates at very high angular velocity.

Can centrifugal force be thought of as the "equal and opposite force" to centripetal force? When turning in a car, it seems as if one tends away from the turn (away from the center). Maybe centrifugal force is just a vernacular term for Newton's first law when moving in a circle. The object is "trying" to maintain its fixed velocity, and when centripetal force acts on the object, it tends to stay in motion at its fixed velocity.

No these are not action reaction pairs, if they were then they would have acted on two different bodies but centripetal and centrifugal force act on same body.

why is centripetal acceleration equal to negative of v^2/r

As to why the sign of centripetal acceleration is negative, this is because we denote it to be in the radial direction. The radial direction is the direction that starts at the center of a circle and goes directly outwards. Since the centripetal acceleration points inwards, we give it a negative sign. As to why the magnitude is v^2/r, there are many derivations, but a simple one that uses a more geometric picture is this one: https://www.youtube.com/watch?v=TNX-Z6XR3gA or using calculus: https://www.youtube.com/watch?v=YRBRarbMCyE

How do you know that the 7.5×10^4 is in cm? it will be equal to 7500cm? in 1 meter has 100cm which you divide the 7500/100= 75meters and no 0.0750cm. I'm confused

The radius is 7.5 cm, which equals .075 m (that is, 7.5 / 100). The *angular velocity*, ω, is 7.5×10^4 *revolutions per minute*. The two numbers are totally separate. They use ω to calculate v (tangential velocity) by saying that 1 revolution is 2π*r = 2π*(.075) meters, and 1 minute = 60 seconds.

Main content

Course: Physics library > Unit 4

Lesson 1: Circular motion and centripetal acceleration

What is centripetal acceleration?

Google Classroom

Learn what centripetal acceleration means and how to calculate it.

What is centripetal acceleration?

Can an object accelerate if it's moving with constant speed? Yup! Many people find this counter-intuitive at first because they forget that changes in the direction of motion of an object—even if the object is maintaining a constant speed—still count as acceleration.

Acceleration is a change in velocity, either in its magnitude—i.e., speed—or in its direction, or both. In uniform circular motion, the direction of the velocity changes constantly, so there is always an associated acceleration, even though the speed might be constant. You experience this acceleration yourself when you turn a corner in your car—if you hold the wheel steady during a turn and move at constant speed, you are in uniform circular motion. What you notice is a sideways acceleration because you and the car are changing direction. The sharper the curve and the greater your speed, the more noticeable this acceleration will become. In this section we'll examine the direction and magnitude of that acceleration.

The figure below shows an object moving in a circular path at constant speed. The direction of the instantaneous velocity is shown at two points along the path. Acceleration is in the direction of the change in velocity, which points directly toward the center of rotation—the center of the circular path. This direction is shown with the vector diagram in the figure. We call the acceleration of an object moving in uniform circular motion—resulting from a net external force—the centripetal acceleration

a_{c}

; centripetal means “toward the center” or “center seeking”.

The directions of the velocity of an object at two different points, $B$ ‍ and $C$ ‍, are shown, and the change in velocity, $Δ v$ ‍, is seen to point roughly toward the center of curvature. To see what happens instantaneously, the points $B$ ‍ and $C$ ‍ must be very close and $Δ θ$ ‍ very small. We'll then find that $Δ v$ ‍ points directly toward the center of curvature.

Because $a_{c} = \frac{Δ v}{Δ t}$ ‍, the acceleration is also toward the center. Because $Δ θ$ ‍ is very small, the arc length $Δ s$ ‍ is equal to the chord length $Δ r$ ‍ for small time differences. Image credit: Openstax College Physics

The direction of centripetal acceleration is toward the center of the circle, but what is its magnitude? Note that the triangle formed by the velocity vectors and the triangle formed by the radii

r

and

Δ s

are similar. Both the triangles

A B C

and

P Q R

are isosceles triangles with two equal sides. The two equal sides of the velocity vector triangle are the speeds

v_{1} = v_{2} = v

. Using the properties of two similar triangles, we obtain

\frac{Δ v}{v} = \frac{Δ s}{r}

Acceleration is

\frac{Δ v}{Δ t}

, so we first solve the above expression for

Δ v

Δ v = \frac{v}{r} Δ s

If we divide both sides by

Δ t

we get the following:

\frac{Δ v}{Δ t} = \frac{v}{r} \times \frac{Δ s}{Δ t}

Finally, noting that

\frac{Δ v}{Δ t} = a_{c}

and that

\frac{Δ s}{Δ t} = v

, the linear or tangential speed, we see that the magnitude of the centripetal acceleration is

a_{c} = \frac{v^{2}}{r}

This is the acceleration of an object in a circle of radius

r

at a speed

v

. So, centripetal acceleration is greater at high speeds and in sharp curves—smaller radii—as you have noticed when driving a car. But it is a bit surprising that

a_{c}

is proportional to speed squared, implying, for example, that it is four times as hard to take a curve at 100 km/hr than at 50 km/hr. A sharp corner has a small radius, so

a_{c}

is greater for tighter turns, as you have probably noticed.

What is a centrifuge?

A centrifuge 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

; 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.

What do solved examples involving centripetal acceleration look like?

Example 1: Curving car

What is the magnitude of the centripetal acceleration of a car following a curve, see figure below, of radius 500 m at a speed of 25 m/s—about 90 km/hr? Compare the acceleration with that due to gravity for this fairly gentle curve taken at highway speed.

Example 2: Ultracentrifuge

Calculate the centripetal acceleration of a point 7.5 cm from the axis of an ultracentrifuge spinning at

7.5 \times 10^{4}

revolutions per minute.

The term

\frac{rev}{min}

stands for revolutions per minute. This is sometimes called the angular velocity,

ω

. We need to use this quantity to determine the velocity

v

in units of meters per second. To do this, we convert revolutions into meters and minutes into seconds.

Since during one revolution the centrifuge will move through the entire circumference of a circle, one revolution is equivalent to a path length of

2 π r

. Note: We will write the radius in units of meters, so

7.50 cm = 0.075 m

v = 7.5 \times 10^{4} \frac{rev}{min} \times (\frac{2 π (0.0750 m)}{1 rev}) \times (\frac{1 min}{60 sec}) = 589 \frac{m}{s}

Using the formula for centripetal acceleration we get the following:

a_{c} = \frac{v^{2}}{r} = \frac{(589 \frac{m}{s})^{2}}{0.075 m} = 4.63 \times 10^{6} \frac{m}{s^{2}}

This is 472,000 times as strong as the acceleration due to gravity on Earth. It is no wonder that such high speed centrifuges are called ultracentrifuges. The extremely large accelerations involved greatly decrease the time needed to cause the sedimentation of blood cells or other materials.

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Taha Anouar
Posted 8 years ago. Direct link to Taha Anouar's post “how can deltaS equal delt...”
how can deltaS equal deltaR?
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(13 votes)
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- qrrqtx
  Posted 8 years ago. Direct link to qrrqtx's post “That's a good question. N...”
  That's a good question. Notice how the article says: when Δθ is very small, Δs = Δr. That's calculus at work - by very small it means infinitesimally small. On a bigger scale, it's obvious that an arc and a chord are not equal, but the bigger scale is just an illustration of the ideas, an approximation. As θ gets smaller and smaller, Δs and Δr (the arc length and chord length) get closer and closer to being the same length. By using limits, you can actually prove that as Δθ approaches 0, Δs = Δr. It seems to defy "common sense", but it's true and it works. To get a little more intuitive picture of what's going on, start with a circle of radius 1 and a 90° arc, figure out the arc length and the chord length. Then figure out arc and chord for a 9° angle, then 0.9°, etc. See how close arc and chord get in just a few steps? Imagine you did that infinitely many times.
  Comment on qrrqtx's post “That's a good question. N...”
  (47 votes)
Archi130679
Posted 8 years ago. Direct link to Archi130679's post “what is the real forces...”
what is the real forces that provide centripetal acceleration
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- Rajeev Agarwal
  Posted 8 years ago. Direct link to Rajeev Agarwal's post “centripetal actually mean...”
  centripetal actually means - towards the center .So centripetal force is not a new type of force .Any force which is acting towards center can be called as centripetal force. To understand it better think of gravitational force , it acts in downwards direction so we call it downwards force because of its direction .There are only four real forces in nature i.e. Gravitational , Electromagnetic ,weak nuclear and strong nuclear .
  whenever any of these four forces acts towards center that force is called centripetal(towards center) . In case of earth and sun system ,real force acting is Gravitational but as it is always pointing towards center we call it centripetal . In case of an electron revolving around nucleus ,the real force between electron and positive nucleus is electromagnetic but because of its direction always pointing towards center ,it is centripetal force. Friction too is electromagnetic force.
  Comment on Rajeev Agarwal's post “centripetal actually mean...”
  (24 votes)
Riya Mahajan
Posted 7 years ago. Direct link to Riya Mahajan's post “If an object has a centri...”
If an object has a centripetal acceleration towards the center, why does it not go towards the center?
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Answer
- Nikolay
  Posted 4 years ago. Direct link to Nikolay's post “Technically they are. The...”
  Technically they are. They are "falling", but also moving sideways at a large velocity, so they maintain a circular path.
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  (9 votes)
Esha
Posted 8 years ago. Direct link to Esha's post “why is the triangle ABC a...”
why is the triangle ABC and triangle PQR similar?
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Answer
- Björn Strömberg
  Posted 8 years ago. Direct link to Björn Strömberg's post “the vector v1 (PR) form a...”
  the vector v1 (PR) form a right angle to AC and v2 (PQ) form a right angle to AB. Given this and a given angle between AC and AB you can draw up the lines and prove that the angle between PR and PQ must have the same angle. Thus the triangles are similar :)
  Comment on Björn Strömberg's post “the vector v1 (PR) form a...”
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zqiu
Posted 5 years ago. Direct link to zqiu's post “Why does centripetal forc...”
Why does centripetal force does not work?
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- Andrew M
  Posted 5 years ago. Direct link to Andrew M's post “because the force is alwa...”
  because the force is always perpendicular to the displacement.
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  (4 votes)
Surbhi Kavishwar
Posted 8 years ago. Direct link to Surbhi Kavishwar's post “what is meant by utlracen...”
what is meant by utlracentrifuge?
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- vinnv226
  Posted 8 years ago. Direct link to vinnv226's post “An ultracentrifuge is jus...”
  An ultracentrifuge is just a centrifuge that operates at very high angular velocity.
  Comment on vinnv226's post “An ultracentrifuge is jus...”
  (5 votes)
Mister Owl
Posted 8 years ago. Direct link to Mister Owl's post “Can centrifugal force be ...”
Can centrifugal force be thought of as the "equal and opposite force" to centripetal force? When turning in a car, it seems as if one tends away from the turn (away from the center). Maybe centrifugal force is just a vernacular term for Newton's first law when moving in a circle. The object is "trying" to maintain its fixed velocity, and when centripetal force acts on the object, it tends to stay in motion at its fixed velocity.
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- neeraj bhale
  Posted 8 years ago. Direct link to neeraj bhale's post “No these are not action r...”
  No these are not action reaction pairs, if they were then they would have acted on two different bodies but centripetal and centrifugal force act on same body.
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  (3 votes)
siddharth kashyap
Posted 8 years ago. Direct link to siddharth kashyap's post “why is centripetal accele...”
why is centripetal acceleration equal to negative of v^2/r
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- Robby358
  Posted 8 years ago. Direct link to Robby358's post “As to why the sign of cen...”
  As to why the sign of centripetal acceleration is negative, this is because we denote it to be in the radial direction. The radial direction is the direction that starts at the center of a circle and goes directly outwards. Since the centripetal acceleration points inwards, we give it a negative sign.
  
  As to why the magnitude is v^2/r, there are many derivations, but a simple one that uses a more geometric picture is this one: https://www.youtube.com/watch?v=TNX-Z6XR3gA
  
  or using calculus: https://www.youtube.com/watch?v=YRBRarbMCyE
  Comment on Robby358's post “As to why the sign of cen...”
  (4 votes)
Luiz Valverde
Posted 7 years ago. Direct link to Luiz Valverde's post “How do you know that the ...”
How do you know that the 7.5×10^4 is in cm? it will be equal to 7500cm? in 1 meter has 100cm which you divide the 7500/100= 75meters and no 0.0750cm. I'm confused
Button navigates to signup pageButton navigates to signup page
(2 votes)
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- robshowsides
  Posted 7 years ago. Direct link to robshowsides's post “The radius is 7.5 cm, whi...”
  The radius is 7.5 cm, which equals .075 m (that is, 7.5 / 100).
  
  The angular velocity, ω, is 7.5×10^4 revolutions per minute. The two numbers are totally separate. They use ω to calculate v (tangential velocity) by saying that 1 revolution is 2π*r = 2π*(.075) meters, and 1 minute = 60 seconds.
  Comment on robshowsides's post “The radius is 7.5 cm, whi...”
  (3 votes)
Ishan Saha
Posted 5 years ago. Direct link to Ishan Saha's post “How would you consider an...”
How would you consider an object with changing magnitude and direction for centripetal acceleration?

a=V^2/r, but doesn't address change in speed
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- caleyandrewj
  Posted 2 years ago. Direct link to caleyandrewj's post “Ishan, the direction is a...”
  Ishan, the direction is already changing because the acceleration is towards the center but the velocity is tangential, so it travels in a circle constantly changing direction as mentioned. If the speed, or magnitude, weren't constant and changed, in order to plug in for V in the formula, you would take the average. This can be done by finding the initial speed and final speed and dividing by 2. Assuming the acceleration is constant.
  Comment on caleyandrewj's post “Ishan, the direction is a...”
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