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Centripetal acceleration through angular velocity. What is centripetal acceleration

Because linear speed uniformly changes direction, then the circular motion cannot be called uniform, it is uniformly accelerated.

Angular velocity

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

Period and frequency

Rotation period T- this is the time during which the body makes one revolution.

Rotation frequency is the number of revolutions per second.

Frequency and period are interrelated by the relationship

Relationship with angular velocity

Linear speed

Each point on the circle moves at a certain 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 grinding machine move, repeating the direction of instantaneous speed.


Consider a point on a circle that makes one revolution, the time spent is the period T. The path that a point travels is the circumference.

Centripetal acceleration

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

Using the previous formulas, we can derive the following relationships


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

The law of addition of speeds is also valid for rotational motion. If the motion of a body or frame of reference is not uniform, then the law applies to instantaneous velocities. 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 participates in two main rotational movements: diurnal (around its 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 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 acting force is the elastic force.

If a body lying on a disk rotates with the disk around its axis, then such a force is the friction force. 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 speed is equal to vA And vB respectively. Acceleration is the change in speed per unit time. Let's find the difference between the vectors.

Definition

Centripetal acceleration called the component of total acceleration material point, moving along a curved path, which determines the speed of change in the direction of the velocity vector.

Another component of total acceleration is tangential acceleration, which is responsible for the change in velocity. Denotes centripetal acceleration, usually $(\overline(a))_n$. Centripetal acceleration is also called normal acceleration.

Centripetal acceleration is equal to:

\[(\overline(a))_n=\frac(v^2)(r^2)\overline(r\ )=\frac(v^2)(r)(\overline(e))_r\left (1\right),\]

where $(\overline(e))_r=\frac(\overline(r\ ))(r)$ is the unit vector, which is directed from the center of curvature of the trajectory to the point in question; $r$ is the radius of curvature of the trajectory at the location of the material point at the considered moment in time.

H. Huygens was the first to obtain the correct formulas for calculating centripetal acceleration.

The International System of Units unit of centripetal acceleration is the meter divided by the squared second:

\[\left=\frac(m)(s^2).\]

Formula for centripetal acceleration for uniform motion of a point in a circle

Let us consider the uniform motion of a material point along a circle. With such movement, the velocity of the material point remains unchanged ($v=const$). But that doesn't mean that full acceleration of a material point with this type of motion is equal to zero. The instantaneous velocity vector is directed tangentially to the circle along which the point is moving. Consequently, in this movement the speed constantly changes its direction. It follows that the point has acceleration.

Let's consider points A and B that lie on the trajectory of the particle. We find the velocity change vector for points A and B as:

\[\Delta \overline(v)=(\overline(v))"-\overline(v)\left(2\right).\]

If the time spent moving from point A to point B tends to zero, then the arc AB does not differ much from the chord AB. Triangles AOB and BMN are similar, we get:

\[\frac(\Delta v)(v)=\frac(\Delta l)(R)=\alpha \left(3\right).\]

The magnitude of the average acceleration module is determined as:

\[\left\langle a\right\rangle =\frac(\Delta v)(\Delta t)=\frac(v\Delta l)(R\Delta t)\left(4\right).\]

Let's move to the limit at $\Delta t\to 0\ $ from $\left\langle a\right\rangle \ \ $in formula (4):

The average acceleration vector makes an angle equal to the velocity vector:

\[\beta =\frac(\pi +\alpha )(2)\left(6\right).\]

At $\Delta t\to 0\ $ angle $\alpha \to 0.$ It turns out that the instantaneous acceleration vector makes an angle $\frac(\pi )(2)$ with the velocity vector.

And so that a material point moving uniformly around a circle has an acceleration that is directed towards the center of the circle ($(\overline(a))_n\bot \overline(v)$), its value is equal to the speed squared divided by the radius circles:

where $\omega $ is the angular velocity of motion of a material point ($v=\omega \cdot R$). IN vector form the formula for centripetal acceleration can be written based on (7) as:

\[(\overline(a))_n=-(\omega )^2\overline(R)\ \left(8\right),\]

where $\overline(R)$ is the radius vector, equal in length to the radius of the circular arc, directed from the center of curvature to the location of the material point under consideration.

Examples of problems with solutions

Example 1

Exercise. Vector equation $\overline(r)\left(t\right)=\overline(i)(\cos \left(\omega t\right)+\overline(j)(\sin \left(\omega t\right )\ )\ )$, where $\omega =2\ \frac(rad)(s),$ describes the motion of a material point. What trajectory is this point moving along? What is the magnitude of its centripetal acceleration? Consider all quantities in the SI system.

Solution. Consider the equation of motion of a point:

\[\overline(r)\left(t\right)=\overline(i)(\cos \left(\omega t\right)+\overline(j)(\sin (\omega t)\ )\ ) \ \left(1.1\right).\]

IN Cartesian system coordinates, this equation is equivalent to the system of equations:

\[\left\( \begin(array)(c) x=(\cos \left(\omega t\right);;\ ) \\ y=(\sin \left(\omega t\right)\ ) \end(array) \left(1.2\right).\right.\]

In order to understand what trajectory the point is moving along, we should exclude time from the equations of system (1.2). To do this, we square both equations and add them:

From equation (1.3) we see that the trajectory of the point is a circle (Fig. 2) with radius $R=1$ m.

In order to find the centripetal acceleration we use the formula:

Let's determine the velocity module using the system of equations (1.2). Let's find the velocity components that are equal to:

\[\left\( \begin(array)(c) v_x=\frac(dx)(dt)=-\omega (\sin \left(\omega t\right)\ ), \\ v_y=\frac( dy)(dt)=\omega ((\cos \left(\omega t\right)\ ) ,\ ) \end(array) \right.\left(1.5\right).\]

The square of the velocity module will be equal to:

From the resulting modulus of velocity (1.6), we see that our point moves uniformly around the circle, therefore, the centripetal acceleration will coincide with the total acceleration.

Substituting $v^2$ from (1.6) into formula (1.4), we have:

Let's calculate $a_n$:

$a_n=\frac(4)(1)=4\ \left(\frac(m)(s^2)\right).$

Answer. 1) Circle; 2) $a_n=4\ \frac(m)(s^2)$

Example 2

Exercise. What is the centripetal acceleration of points on the rim of the disk at a time equal to $t=2$c, if the disk rotates in accordance with the equation: $\varphi (t)=3+2t^3$? The radius of the disk is $R=0,(\rm 1)$ m.

Solution. We will look for the centripetal acceleration of points on the disk using the formula:

We find the angular velocity using the equation $\varphi (t)=3+2t^3$ as:

\[\omega =\frac(d\varphi )(dt)=6t^2.\ \]

At $t=2\ $c the angular velocity is equal to:

\[\omega \left(t=2\right)=24\ \left(\frac(rad)(s)\right).\]

You can calculate the centripetal acceleration using formula (2.1):

Answer.$a_n=57.6\frac(m)(s^2)$

Centripetal acceleration- component of the acceleration of a point, characterizing the speed 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 module). Directed towards the center of curvature of the trajectory, which is where the term comes from. The value 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 centripetal force.

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

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

Encyclopedic YouTube

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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, one must keep in mind that centripetal acceleration is not the full acceleration vector, but only its component perpendicular to the trajectory (or, what is the same, perpendicular to the instantaneous velocity vector); the full acceleration vector then also includes a tangential component ( tangential acceleration) a τ = d v / d t (\displaystyle a_(\tau )=dv/dt\ ), the direction coinciding with the tangent to the trajectory (or, what is the same, with the instantaneous speed).

    Motivation and conclusion

    The fact that the decomposition of the acceleration vector into components - one along the tangent to the vector trajectory (tangential acceleration) and the other orthogonal to it (normal acceleration) - can be convenient and useful is quite obvious in itself. When moving with a constant modulus 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 clearly defined properties and structure, and normal acceleration contains quite important and non-trivial geometric content in the structure of its formula. Not to mention the important special case of circular motion.

    Formal conclusion

    The decomposition of acceleration into tangential and normal components (the second of which is centripetal or normal acceleration) can be found by differentiating with respect to time the velocity vector, presented in the form 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 vector normal to the trajectory and l (\displaystyle l\ )- for the current trajectory length ( 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. Moreover, its meaning, the meaning of the objects included in it, as well as proof of the fact that it is indeed orthogonal to the tangent vector (that is, that e n (\displaystyle \mathbf (e)_(n)\ )- really a normal vector) - will follow from geometric considerations (however, the fact that the time derivative of any vector of constant length is perpendicular to this vector itself is a fairly simple fact; in in this case we apply this statement to d e τ d t (\displaystyle (\frac (d\mathbf (e) _(\tau ))(dt)))

    Notes

    It is easy to notice that the absolute value of the tangential acceleration depends only on the ground acceleration, coinciding with its absolute value, in contrast to the absolute value of the normal acceleration, which does not depend on the ground acceleration, but depends on the 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 (since in the case where 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)\ ) with center in direction e n (\displaystyle e_(n)\ ) from a given point at a distance R from it - will coincide with the given curve - 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. Almost from this time on, consideration of centripetal acceleration has become part of the usual technique for solving mechanical problems, etc.

    Somewhat later, these formulas played a significant role in the discovery of the law of universal gravitation (the formula of centripetal acceleration was used to obtain the law of the dependence of gravitational force on the distance to the source of gravity, based on Kepler’s third law derived from observations).

    TO 19th century consideration of centripetal acceleration is already becoming completely routine both for pure science and for engineering applications.

    An object that moves in a circular orbit of radius r with uniform tangential speed u is the velocity vector v, the magnitude of which is constant, but whose direction is constantly changing. It follows that an object must have acceleration, since (vector) is the rate of change of (vector) speed, and (vector) speed are indeed 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. Let us further assume that the object is rotated by δθ radians during this period of time. Vector, as shown in the diagram, is identical to vector. In addition, 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, the length of a vector is:

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

    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 uniform tangential speed v, and uniform angular velocity, has an acceleration directed towards the center of the circle - that is, centripetal acceleration- size:

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

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

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

    If v exceeds v max, the cable will break. Once the cable breaks, the body will no longer experience centripetal force, so it will move at speed v max along a straight line that is tangent to the pre-existing circular orbit.

    The characteristics were previously considered rectilinear movement: movement, speed, acceleration. Their analogues at rotational movement are: angular displacement, angular velocity, angular acceleration.

    • The role of displacement in rotational motion is played by corner;
    • The magnitude of the rotation angle per unit time is angular velocity;
    • The change in angular velocity per unit time is angular acceleration.

    During uniform rotational motion, a body moves in a circle with the same speed, but with a changing direction. For example, this movement is made by the hands of a clock on a dial.

    Let's say the ball rotates uniformly on a thread 1 meter long. At the same time, it will describe a circle with a radius of 1 meter. The length of this circle is: C = 2πR = 6.28 m

    The time it takes for the ball to become completely alone full turn around the circumference is called rotation period - T.

    To calculate the linear speed of the ball, it is necessary to divide the displacement by time, i.e. circumference per period of rotation:

    V = C/T = 2πR/T

    Rotation period:

    T = 2πR/V

    If our ball makes one revolution in 1 second (rotation period = 1s), then its linear speed is:
    V = 6.28/1 = 6.28 m/s

    2. Centrifugal acceleration

    At any point in the rotational motion of the ball, its linear velocity vector is directed perpendicular to the radius. It is not difficult to guess that with such a circular rotation, the linear velocity vector of the ball constantly changes its direction. The acceleration characterizing such a change in speed is called centrifugal (centripetal) acceleration.

    During uniform rotational motion, only the direction of the velocity vector changes, but not the magnitude! Therefore linear acceleration = 0 . The change in linear speed is supported by centrifugal acceleration, which is directed towards the center of the circle of rotation perpendicular to the speed vector - a c.

    Centrifugal acceleration can be calculated using the formula: a c = V 2 /R

    The greater the linear speed of the body and the smaller the radius of rotation, the greater the centrifugal acceleration.

    3. Centrifugal force

    From rectilinear motion we know that force is equal to the product of the mass of a body and its acceleration.

    With uniform rotational motion, a centrifugal force acts on a rotating body:

    F c = ma c = mV 2 /R

    If our ball weighs 1 kg, then to keep it on the circle you will need centrifugal force:

    F c = 1 6.28 2 /1 = 39.4 N

    We encounter centrifugal force in everyday life at any turn.

    The friction force must balance the centrifugal force:

    F c = mV 2 / R; F tr = μmg

    F c = F tr; mV 2 /R = μmg

    V = √μmgR/m = √μgR = √0.9 9.8 30 = 16.3 m/s = 58.5 km/h

    Answer: 58.5 km/h

    Please note that turning speed does not depend on body weight!

    Surely you have noticed that some turns on the highway have a slight inclination towards the inside of the turn. Such turns are “easier” to take, or rather, you can take turns at greater speed. Let's consider what forces act on the car in such a tilted turn. In this case, we will not take into account the friction force, and the centrifugal acceleration will be compensated only by the horizontal component of gravity:


    F c = mV 2 /R or F c = F n sinα

    In the vertical direction, the force of gravity acts on the body F g = mg, which is balanced by the vertical component of the normal force F n cosα:

    Fn cosα = mg, hence: Fn = mg/cosα

    We substitute the value of the normal force into the original formula:

    F c = F n sinα = (mg/cosα)sinα = mg sinα/cosα = mg tgα

    Thus, the angle of inclination of the roadway:

    α = arctg(F c /mg) = arctg(mV 2 /mgR) = arctg(V 2 /gR)

    Again, note that body weight is not included in the calculations!

    Task #2: On a certain section of the highway there is a turn with a radius of 100 meters. Average speed driving this section of the road at 108 km/h (30 m/s). What should be the safe angle of inclination of the road surface in this section so that the car does not “fly off” (neglect friction)?

    α = arctan(V 2 /gR) = arctan(30 2 /9.8 100) = 0.91 = 42° Answer: 42°. Pretty decent angle. But, do not forget that in our calculations we do not take into account the friction force of the road surface.

    4. Degrees and radians

    Many people are confused in understanding angular values.

    In rotational motion, the basic unit of measurement for angular movement is radian.

    • 2π radians = 360° - complete circle
    • π radian = 180° - half a circle
    • π/2 radians = 90° - quarter circle

    To convert degrees to radians, divide the angle by 360° and multiply by 2π. For example:

    • 45° = (45°/360°) 2π = π/4 radians
    • 30° = (30°/360°) 2π = π/6 radians

    The table below presents the basic formulas for linear and rotational motion.