What is the formula for centripetal acceleration? Circular motion

centripetal acceleration- point acceleration component, which characterizes the rate of change in the direction of the velocity vector for a trajectory with curvature (the second component, tangential acceleration, characterizes the change in the velocity modulus). Directed towards the center of curvature of the trajectory, which is the reason for the term. The magnitude is equal to the square of the speed divided by the radius of curvature. The term " centripetal acceleration" is equivalent to the term " normal acceleration". That component of the sum of forces that causes this acceleration is called the centripetal force.

Most simple example centripetal acceleration is the acceleration vector for uniform circular motion (directed towards the center of the circle).

Rapid acceleration projected onto a plane perpendicular to the axis, it appears as a centripetal.

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A n = v 2 R (\displaystyle a_(n)=(\frac (v^(2))(R))\ ) a n = ω 2 R , (\displaystyle a_(n)=\omega ^(2)R\ ,)
where a n (\displaystyle a_(n)\ )- normal (centripetal) acceleration, v (\displaystyle v\ )- (instantaneous) linear speed of movement along the trajectory, ω (\displaystyle \omega \ )- (instantaneous) angular velocity of this movement relative to the center of curvature of the trajectory, R (\displaystyle R\ )- radius of curvature of the trajectory at a given point. (The connection between the first formula and the second is obvious, given v = ω R (\displaystyle v=\omega R\ )).
The expressions above include absolute values. They can be easily written in vector form by multiplying by e R (\displaystyle \mathbf (e) _(R))- unit vector from the center of curvature of the trajectory to its given point:
a n = v 2 R e R = v 2 R 2 R (\displaystyle \mathbf (a) _(n)=(\frac (v^(2))(R))\mathbf (e) _(R)= (\frac (v^(2))(R^(2)))\mathbf (R) ) a n = ω 2 R . (\displaystyle \mathbf (a) _(n)=\omega ^(2)\mathbf (R) .)
These formulas are equally applicable to the case of motion with a constant (in absolute value) speed, and to an arbitrary case. However, in the second it must be borne in mind that centripetal acceleration is not complete vector acceleration, but only its component perpendicular to the trajectory (or, what is the same, perpendicular to the instantaneous velocity vector); the total acceleration vector then also includes the tangential component ( tangential acceleration) a τ = d v / d t (\displaystyle a_(\tau )=dv/dt\ ), coinciding in direction with the tangent to the trajectory (or, which is the same, with the instantaneous speed) .

Motivation and conclusion

That the decomposition of the acceleration vector into components - one along the vector tangent to the trajectory (tangential acceleration) and another orthogonal to it (normal acceleration) - can be convenient and useful is pretty obvious in itself. When moving with a constant modulo speed, the tangential component becomes equal to zero, that is, in this important particular case, it remains only normal component. In addition, as can be seen below, each of these components has pronounced properties and structure of its own, and the normal acceleration contains a rather important and non-trivial geometric content in the structure of its formula. Not to mention the important special case of motion in a circle.

Formal derivation

The expansion of the acceleration into tangential and normal components (the second of which is the centripetal or normal acceleration) can be found by differentiating with respect to time the velocity vector represented as v = v e τ (\displaystyle \mathbf (v) =v\,\mathbf (e) _(\tau )) through the unit tangent vector e τ (\displaystyle \mathbf (e) _(\tau )):
a = d v d t = d (v e τ) d t = d v d t e τ + v d e τ d t = d v d t e τ + v d e τ d l d l d t = d v d t e τ + v 2 R e n , (\displaystyle \mathbf (a) =(\frac (d\mathbf ( v) )(dt))=(\frac (d(v\mathbf (e) _(\tau )))(dt))=(\frac (\mathrm (d) v)(\mathrm (d) t ))\mathbf (e) _(\tau )+v(\frac (d\mathbf (e) _(\tau ))(dt))=(\frac (\mathrm (d) v)(\mathrm ( d) t))\mathbf (e) _(\tau )+v(\frac (d\mathbf (e) _(\tau ))(dl))(\frac (dl)(dt))=(\ frac (\mathrm (d) v)(\mathrm (d) t))\mathbf (e) _(\tau )+(\frac (v^(2))(R))\mathbf (e) _( n)\ ,)
Here we use the notation for the unit normal vector to the trajectory and l (\displaystyle l\ )- for the current length of the trajectory ( l = l (t) (\displaystyle l=l(t)\ )); the last transition also uses the obvious d l / d t = v (\displaystyle dl/dt=v\ ).
v 2 R e n (\displaystyle (\frac (v^(2))(R))\mathbf (e) _(n)\ )
Normal (centripetal) acceleration. At the same time, its meaning, the meaning of the objects included in it, as well as the proof of the fact that it is indeed orthogonal to the tangent vector (that is, that e n (\displaystyle \mathbf (e) _(n)\ )- indeed a normal vector) - will follow from geometric considerations (however, the fact that the derivative of any vector of constant length with respect to time is perpendicular to this vector itself is a fairly simple fact; in this case, we apply this statement to d e τ d t (\displaystyle (\frac (d\mathbf (e) _(\tau ))(dt)))

Remarks

It is easy to see that the absolute value of tangential acceleration depends only on ground acceleration, coinciding with its absolute value, in contrast to the absolute value of normal acceleration, which does not depend on ground acceleration, but depends on ground speed.
The methods presented here, or variations thereof, can be used to introduce concepts such as the curvature of a curve and the radius of curvature of a curve (because in the case when the curve is a circle, R coincides with the radius of such a circle; it is also not too difficult to show that the circle is in the plane e τ , e n (\displaystyle \mathbf (e) _(\tau ),e_(n)\ ) centered in the direction e n (\displaystyle e_(n)\ ) away from this point R from it - will coincide with the given curve - the trajectory - up to the second order of smallness in the distance to the given point).

Story

First correct formulas for centripetal acceleration (or centrifugal force) was apparently obtained by Huygens. Practically since that time, the consideration of centripetal acceleration has been a common technique for solving mechanical problems, etc.
Somewhat later, these formulas played a significant role in the discovery of the law of universal gravitation (the centripetal acceleration formula was used to obtain the law of the dependence of the gravitational force on the distance to the source of gravity, based on the third Kepler law derived from observations).
To XIX century consideration of centripetal acceleration is already becoming quite routine for both pure science and engineering applications.

Allows us to exist on this planet. How can you understand what constitutes centripetal acceleration? Definition of this physical quantity presented below.
Observations
The simplest example of the acceleration of a body moving in a circle can be observed by rotating a stone on a rope. You pull the rope, and the rope pulls the rock towards the center. At each moment in time, the rope gives the stone a certain amount of movement, and each time in a new direction. You can imagine the movement of the rope as a series of weak jerks. A jerk - and the rope changes its direction, another jerk - another change, and so on in a circle. If you suddenly let go of the rope, the jerks will stop, and with them the change in direction of speed will stop. The stone will move in the direction tangent to the circle. The question arises: "With what acceleration will the body move at this instant?"
formula for centripetal acceleration
First of all, it is worth noting that the movement of the body in a circle is complex. The stone participates in two types of movement at the same time: under the action of a force, it moves towards the center of rotation, and at the same time, tangentially to the circle, it moves away from this center. According to Newton's Second Law, the force holding a stone on a string is directed toward the center of rotation along that string. The acceleration vector will also be directed there.
Let for some time t, our stone, moving uniformly at a speed V, gets from point A to point B. Suppose that at the moment when the body crossed point B, the centripetal force ceased to act on it. Then for a period of time it would hit the point K. It lies on the tangent. If at the same moment of time only centripetal forces acted on the body, then in time t, moving with the same acceleration, it would end up at point O, which is located on a straight line representing the diameter of a circle. Both segments are vectors and obey the vector addition rule. As a result of the summation of these two movements for a period of time t, we obtain the resulting movement along the arc AB.
If the time interval t is taken negligibly small, then the arc AB will differ little from the chord AB. Thus, it is possible to replace movement along an arc with movement along a chord. In this case, the movement of the stone along the chord will obey the laws of rectilinear motion, that is, the distance AB traveled will be equal to the product of the speed of the stone and the time of its movement. AB = V x t.
Let us denote the desired centripetal acceleration by the letter a. Then the path traveled only under the action of centripetal acceleration can be calculated by the formula uniformly accelerated motion:
Distance AB is equal to the product of speed and time, i.e. AB = V x t,
AO - calculated earlier using the uniformly accelerated motion formula for moving in a straight line: AO = at 2 / 2.
Substituting these data into the formula and transforming them, we get a simple and elegant formula for centripetal acceleration:
In words, this can be expressed as follows: the centripetal acceleration of a body moving in a circle is equal to the quotient of dividing the linear velocity squared by the radius of the circle along which the body rotates. The centripetal force in this case will look like the picture below.
Angular velocity
The angular velocity is equal to the linear velocity divided by the radius of the circle. The converse is also true: V = ωR, where ω is the angular velocity
If we substitute this value into the formula, we can get the expression for the centrifugal acceleration for the angular velocity. It will look like this:
Acceleration without speed change
And yet, why doesn't a body with acceleration directed towards the center move faster and move closer to the center of rotation? The answer lies in the wording of acceleration itself. The facts show that circular motion is real, but that it requires acceleration towards the center to maintain it. Under the action of the force caused by this acceleration, there is a change in the momentum, as a result of which the trajectory of motion is constantly curved, all the time changing the direction of the velocity vector, but not changing its absolute value. Moving in a circle, our long-suffering stone rushes inward, otherwise it would continue to move tangentially. Every moment of time, leaving on a tangent, the stone is attracted to the center, but does not fall into it. Another example of centripetal acceleration would be a water skier making small circles on the water. The figure of the athlete is tilted; he seems to be falling, continuing to move and leaning forward.
Thus, we can conclude that acceleration does not increase the speed of the body, since the velocity and acceleration vectors are perpendicular to each other. Added to the velocity vector, acceleration only changes the direction of motion and keeps the body in orbit.
Safety margin exceeded
In the previous experience, we were dealing with an ideal rope that did not break. But, let's say our rope is the most common, and you can even calculate the effort after which it will simply break. In order to calculate this force, it is enough to compare the safety margin of the rope with the load that it experiences during the rotation of the stone. By rotating the stone at a higher speed, you give it more movement, and therefore more acceleration.
With a jute rope diameter of about 20 mm, its tensile strength is about 26 kN. It is noteworthy that the length of the rope does not appear anywhere. Rotating a 1 kg load on a rope with a radius of 1 m, we can calculate that the linear speed required to break it is 26 x 10 3 = 1kg x V 2 / 1 m. Thus, the speed that is dangerous to exceed will be equal to √ 26 x 10 3 \u003d 161 m / s.
Gravity
When considering the experiment, we neglected the action of gravity, since at such high speeds its influence is negligibly small. But you can see that when unwinding a long rope, the body describes a more complex trajectory and gradually approaches the ground.
celestial bodies
If we transfer the laws of circular motion into space and apply them to the motion of celestial bodies, we can rediscover several long-familiar formulas. For example, the force with which a body is attracted to the Earth is known by the formula:
In our case, the factor g is the very centripetal acceleration that was derived from the previous formula. Only in this case, the role of a stone will be played by a celestial body attracted to the Earth, and the role of a rope will be the force of earth's attraction. The factor g will be expressed in terms of the radius of our planet and the speed of its rotation.
Results
The essence of centripetal acceleration is the hard and thankless work of keeping a moving body in orbit. A paradoxical case is observed when, with constant acceleration, the body does not change its velocity. To the untrained mind, such a statement is rather paradoxical. Nevertheless, both when calculating the motion of an electron around the nucleus, and when calculating the speed of rotation of a star around a black hole, centripetal acceleration plays an important role.

Since the linear speed uniformly changes direction, then the movement along the circle cannot be called uniform, it is uniformly accelerated.

Angular velocity

Pick a point on the circle 1 . Let's build a radius. For a unit of time, the point will move to the point 2 . In this case, the radius describes the angle. The angular velocity is numerically equal to the angle of rotation of the radius per unit time.

Period and frequency

Rotation period T is the time it takes the body to make one revolution.

RPM is the number of revolutions per second.

The frequency and period are related by the relationship

Relationship with angular velocity

Line speed

Each point on the circle moves at some speed. This speed is called linear. The direction of the linear velocity vector always coincides with the tangent to the circle. For example, sparks from under a grinder move, repeating the direction of instantaneous speed.

Consider a point on a circle that makes one revolution, the time that is spent - this is the period T. The path traveled by a point is the circumference of a circle.

centripetal acceleration

When moving along a circle, the acceleration vector is always perpendicular to the velocity vector, directed to the center of the circle.

Using the previous formulas, we can derive the following relations

Points lying on the same straight line emanating from the center of the circle (for example, these can be points that lie on the wheel spoke) will have the same angular velocities, period and frequency. That is, they will rotate in the same way, but with different linear speeds. The farther the point is from the center, the faster it will move.

The law of addition of velocities is also valid for rotational motion. If the motion of a body or frame of reference is not uniform, then the law applies to instant speeds. For example, the speed of a person walking along the edge of a rotating carousel is equal to the vector sum of the linear speed of rotation of the edge of the carousel and the speed of the person.

The earth is involved in two main rotational movements: diurnal (around its own axis) and orbital (around the Sun). The period of rotation of the Earth around the Sun is 1 year or 365 days. The Earth rotates around its axis from west to east, the period of this rotation is 1 day or 24 hours. Latitude is the angle between the plane of the equator and the direction from the center of the Earth to a point on its surface.

According to Newton's second law, the cause of any acceleration is a force. If a moving body experiences centripetal acceleration, then the nature of the forces that cause this acceleration may be different. For example, if a body moves in a circle on a rope tied to it, then active force is the elastic force.

If a body lying on a disk rotates along with the disk around its axis, then such a force is the force of friction. If the force ceases to act, then the body will continue to move in a straight line

Consider the movement of a point on a circle from A to B. The linear velocity is equal to v A and v B respectively. Acceleration is the change in speed per unit of time. Let's find the difference of vectors.

Allows us to exist on this planet. How can you understand what constitutes centripetal acceleration? The definition of this physical quantity is presented below.
Observations
The simplest example of the acceleration of a body moving in a circle can be observed by rotating a stone on a rope. You pull the rope, and the rope pulls the rock towards the center. At each moment in time, the rope gives the stone a certain amount of movement, and each time in a new direction. You can imagine the movement of the rope as a series of weak jerks. A jerk - and the rope changes its direction, another jerk - another change, and so on in a circle. If you suddenly let go of the rope, the jerks will stop, and with them the change in direction of speed will stop. The stone will move in the direction tangent to the circle. The question arises: "With what acceleration will the body move at this instant?"
formula for centripetal acceleration
First of all, it is worth noting that the movement of the body in a circle is complex. The stone participates in two types of movement at the same time: under the action of a force, it moves towards the center of rotation, and at the same time, tangentially to the circle, it moves away from this center. According to Newton's Second Law, the force holding a stone on a string is directed toward the center of rotation along that string. The acceleration vector will also be directed there.
Let for some time t, our stone, moving uniformly at a speed V, gets from point A to point B. Suppose that at the moment when the body crossed point B, the centripetal force ceased to act on it. Then for a period of time it would hit the point K. It lies on the tangent. If at the same moment of time only centripetal forces acted on the body, then in time t, moving with the same acceleration, it would end up at point O, which is located on a straight line representing the diameter of a circle. Both segments are vectors and obey the vector addition rule. As a result of the summation of these two movements for a period of time t, we obtain the resulting movement along the arc AB.
If the time interval t is taken negligibly small, then the arc AB will differ little from the chord AB. Thus, it is possible to replace movement along an arc with movement along a chord. In this case, the movement of the stone along the chord will obey the laws of rectilinear motion, that is, the distance AB traveled will be equal to the product of the speed of the stone and the time of its movement. AB = V x t.
Let us denote the desired centripetal acceleration by the letter a. Then the path traveled only under the action of centripetal acceleration can be calculated using the formula of uniformly accelerated motion:
Distance AB is equal to the product of speed and time, i.e. AB = V x t,
AO - calculated earlier using the uniformly accelerated motion formula for moving in a straight line: AO = at 2 / 2.
Substituting these data into the formula and transforming them, we get a simple and elegant formula for centripetal acceleration:
In words, this can be expressed as follows: the centripetal acceleration of a body moving in a circle is equal to the quotient of dividing the linear velocity squared by the radius of the circle along which the body rotates. The centripetal force in this case will look like the picture below.
Angular velocity
The angular velocity is equal to the linear velocity divided by the radius of the circle. The converse is also true: V = ωR, where ω is the angular velocity
If we substitute this value into the formula, we can get the expression for the centrifugal acceleration for the angular velocity. It will look like this:
Acceleration without speed change
And yet, why doesn't a body with acceleration directed towards the center move faster and move closer to the center of rotation? The answer lies in the wording of acceleration itself. The facts show that circular motion is real, but that it requires acceleration towards the center to maintain it. Under the action of the force caused by this acceleration, there is a change in the momentum, as a result of which the trajectory of motion is constantly curved, all the time changing the direction of the velocity vector, but not changing its absolute value. Moving in a circle, our long-suffering stone rushes inward, otherwise it would continue to move tangentially. Every moment of time, leaving on a tangent, the stone is attracted to the center, but does not fall into it. Another example of centripetal acceleration would be a water skier making small circles on the water. The figure of the athlete is tilted; he seems to be falling, continuing to move and leaning forward.
Thus, we can conclude that acceleration does not increase the speed of the body, since the velocity and acceleration vectors are perpendicular to each other. Added to the velocity vector, acceleration only changes the direction of motion and keeps the body in orbit.
Safety margin exceeded
In the previous experience, we were dealing with an ideal rope that did not break. But, let's say our rope is the most common, and you can even calculate the effort after which it will simply break. In order to calculate this force, it is enough to compare the safety margin of the rope with the load that it experiences during the rotation of the stone. By rotating the stone at a higher speed, you give it more movement, and therefore more acceleration.
With a jute rope diameter of about 20 mm, its tensile strength is about 26 kN. It is noteworthy that the length of the rope does not appear anywhere. Rotating a 1 kg load on a rope with a radius of 1 m, we can calculate that the linear speed required to break it is 26 x 10 3 = 1kg x V 2 / 1 m. Thus, the speed that is dangerous to exceed will be equal to √ 26 x 10 3 \u003d 161 m / s.
Gravity
When considering the experiment, we neglected the action of gravity, since at such high speeds its influence is negligibly small. But you can see that when unwinding a long rope, the body describes a more complex trajectory and gradually approaches the ground.
celestial bodies
If we transfer the laws of circular motion into space and apply them to the motion of celestial bodies, we can rediscover several long-familiar formulas. For example, the force with which a body is attracted to the Earth is known by the formula:
In our case, the factor g is the very centripetal acceleration that was derived from the previous formula. Only in this case, the role of a stone will be played by a celestial body attracted to the Earth, and the role of a rope will be the force of earth's attraction. The factor g will be expressed in terms of the radius of our planet and the speed of its rotation.
Results
The essence of centripetal acceleration is the hard and thankless work of keeping a moving body in orbit. A paradoxical case is observed when, with constant acceleration, the body does not change its velocity. To the untrained mind, such a statement is rather paradoxical. Nevertheless, both when calculating the motion of an electron around the nucleus, and when calculating the speed of rotation of a star around a black hole, centripetal acceleration plays an important role.

An object that moves in a circular orbit of radius r with uniform tangential speed u is the velocity vector v, whose magnitude is constant, but whose direction is constantly changing. It follows that the object must have acceleration, since (vector) is the rate of change of (vector) speed, and (vector) speed are really different in time.

Suppose an object is moving from a point P to the point Q between time t and, t + δ t as shown in the picture above. Suppose further that the object is rotated by δθ radians during this period of time. The vector , as shown in the diagram, is identical to the vector . Also, the angle between the vectors and this δθ . The vector represents the change in the velocity vector, δ v, between time t and t + δ t. From this it is clear that this vector is directed towards the center of the circle. From standard trigonometry, vector length:

However, at small angles sin θ ≃ θ , provided that θ measured in radians. Consequently,

δv ≃ v δθ.

where is the angular velocity of the object in radians per second. Thus, an object moving in a circular orbit with a radius r, at a uniform tangential velocity v, and uniform angular velocity , has an acceleration directed towards the center of the circle - that is, centripetal acceleration- value:

Let's assume that the body, mass m, attached to the end of the cable, length r, and rotates in such a way that the body describes a horizontal circle of radius r, with uniform tangential velocity v. As we have just learned, the body has a centripetal acceleration of magnitude . Therefore, the body experiences a centripetal force

What gives this power? Well, in this example, the force is provided by the tension in the cable. Consequently, .

Let us assume that the cable is such that it breaks when the voltage in it exceeds some critical value. It follows from this that there is maximum speed, with which the body can move, namely:

If a v exceeds vmax, the cable will break. As soon as the cable breaks, the body will no longer experience centripetal force, so it will move with a speed vmax in a straight line that is tangent to the pre-existing circular orbit.