Centripetal acceleration formula through angular. Rotational movement

The date: 11.10.2019

Allows us to exist on this planet. How can you understand what constitutes centripetal acceleration? The 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 stone towards the center. At each moment in time, the rope imparts a certain amount of movement to the stone, and each time in a new direction. You can imagine the movement of the rope as a series of weak jerks. A dash - and the rope changes its direction, another jerk - another change, and so on in a circle. If you suddenly release the rope, the jerking will stop, and with it the change in the direction of speed will stop. The stone will move in the direction of the tangent to the circle. The question arises: "With what acceleration will the body move at this moment?"

Centripetal acceleration formula

First of all, it should be noted 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 force, it moves to the center of rotation, and at the same time it moves away from this center tangentially to the circle. According to Newton's Second Law, the force holding a stone on a rope is directed towards the center of rotation along that rope. The acceleration vector will also be directed there.

Let for some time t our stone, moving uniformly with 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, within a period of time, it would have got to point K. It lies on a tangent line. If at the same moment of time only centripetal forces acted on the body, then during time t, moving with the same acceleration, it would be 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 time interval t, we obtain the resultant 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 the arc movement with the chord movement. In this case, the movement of the stone along the chord will obey the laws of rectilinear movement, that is, the distance traveled AB will be equal to the product of the stone's speed and the time of its movement. AB = V x t.

Let us denote the desired centripetal acceleration by the letter a. Then the path traversed only under the action of centripetal acceleration can be calculated using the formula for uniformly accelerated motion:

Distance AB is equal to the product of speed and time, that is, AB = V x t,

AO - calculated earlier by the formula of uniformly accelerated motion for moving in a straight line: AO = at 2/2.

Substituting this data into the formula and transforming them, we get a simple and graceful formula 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 in a square 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 quotient of the linear velocity divided by the radius of the circle. The converse is also true: V = ωR, where ω is the angular velocity

If you substitute this value into the formula, you can get an expression for the centrifugal acceleration for the angular velocity. It will look like this:

Acceleration without changing speed

And yet, why does a body with acceleration directed towards the center not move faster and move closer to the center of rotation? The answer lies in the very wording of acceleration. Evidence suggests that circular motion is real, but sustaining it requires acceleration toward the center. Under the action of the force caused by this acceleration, the momentum changes, as a result of which the trajectory of movement 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 tangentially, 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 athlete's figure is tilted; he seems to fall, continuing to move and leaning forward.

Thus, it can be concluded that acceleration does not increase the body's velocity, since the vectors of velocity and acceleration are perpendicular to each other. Adding to the velocity vector, the acceleration only changes the direction of motion and keeps the body in orbit.

Exceeding the safety margin

In the previous experiment, we dealt with a perfect rope that did not break. But, let's say, our rope is the most common one, 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 margin of safety of the rope with the load that it experiences during the rotation of the stone. By rotating the stone at a faster speed, you tell it large quantity movement, which means 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 weight of 1 kg 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 = 1 kg x V 2/1 m.Thus, the speed that is dangerous to exceed will be equal to √ 26 x 10 3 = 161 m / s.

Gravity

When considering the experiment, we neglected the action of the force of gravity, since at such high speeds its influence is negligible. But you can see that when untwisting a long rope, the body follows a more complex trajectory and gradually approaches the ground.

Celestial bodies

If you transfer the laws of motion in a circle to space and apply them to the motion of celestial bodies, you 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 same centripetal acceleration that was derived from the previous formula. Only in this case, the role of the stone will be performed heavenly body, gravitating towards the Earth, and the role of the rope is the force of gravity. The factor g will be expressed in terms of the radius of our planet and the speed of its rotation.

Outcomes

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 the magnitude of its velocity. For an untrained mind, such a statement is rather paradoxical. Nevertheless, 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, the movement around the circle cannot be called uniform, it is uniformly accelerated.

Angular velocity

Choose a point on the circle 1 ... Let's build a radius. In 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 of time.

Period and frequency

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

Rotation speed is the number of revolutions per second.

Frequency and period are interrelated by the ratio

Angular Velocity Relationship

Linear Velocity

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 the grinder move, repeating the direction of the instantaneous speed.

Consider a point on a circle that makes one revolution, the time spent is a period T... The path that a point overcomes is the length of a circle.

Centripetal acceleration

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

Using the previous formulas, we can derive the following relations

Points lying on one straight line outgoing from the center of the circle (for example, these can be points that lie on the spoke of the wheel) will have the same angular velocity, period and frequency. That is, they will rotate in the same way, but with different linear speeds. The further 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 movement of the body or the frame of reference is not uniform, then the law is applied for 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 person's movement speed.

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 equatorial plane and the direction from the center of the Earth to a point on its surface.

According to Newton's second law, force is the cause of any acceleration. If a moving body experiences centripetal acceleration, then the nature of the forces that cause this acceleration can be different. For example, if the body moves in a circle on a rope attached 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 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 - the change in speed per unit of time. Let's find the difference in 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

Centripetal acceleration formula

Distance AB is equal to the product of speed and time, that is, AB = V x t,

AO - calculated earlier by the formula of uniformly accelerated motion 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:

Angular velocity

The angular velocity is equal to the quotient of the linear velocity divided by the radius of the circle. The converse is also true: V = ωR, where ω is the angular velocity

If you substitute this value into the formula, you can get an expression for the centrifugal acceleration for the angular velocity. It will look like this:

Acceleration without changing speed

Exceeding the safety margin

In the previous experiment, we dealt with a perfect rope that did not break. But, let's say, our rope is the most common one, 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 margin of safety 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.

Gravity

Celestial bodies

In our case, the factor g is the same 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 gravity. The factor g will be expressed in terms of the radius of our planet and the speed of its rotation.

Outcomes

When moving along a circle with a constant in magnitude linear velocity υ, the body has a constant centripetal acceleration directed to the center of the circle

a c = υ 2 / R, (18)

where R is the radius of the circle.

Derivation of the formula for centripetal acceleration

By definition.

Figure 6 Derivation of the formula for centripetal acceleration

In the figure, the triangles formed by the vectors of displacements and velocities are similar. Considering that == R and == υ, from the similarity of triangles we find:

(20)

(21)

Place the origin at the center of the circle and select the plane in which the circle lies beyond the plane (x, y). The position of a point on a circle at any time is uniquely determined by the polar angle φ, measured in radians (rad), and

x = R cos (φ + φ 0), y = R sin (φ + φ 0), (22)

where φ 0 defines initial phase(the initial position of a point on the circle at the zero moment of time).

In the case of uniform rotation, the angle φ, measured in radians, increases linearly with time:

φ = ωt, (23)

where ω is called the cyclic (circular) frequency. Cyclic frequency dimension: [ω] = s –1 = Hz.

Cyclic frequency is equal to the value of the angle of rotation (measured in rad) per unit of time, so it is otherwise called angular velocity.

The dependence of the coordinates of a point on a circle on time in the case of uniform rotation with a given frequency can be written as:

x = R cos (ωt + φ 0), (24)

y = R sin (ωt + φ 0).

The time during which one revolution is completed is called the period T.

Frequency ν = 1 / T. (25)

Frequency dimension: [ν] = s –1 = Hz.

Relationship of the cyclic frequency with the period and frequency: 2π = ωT, whence

ω = 2π / T = 2πν. (26)

The relationship between linear velocity and angular velocity is found from the equality:

2πR = υT, whence

υ = 2πR / T = ωR. (27)

The expression for centripetal acceleration can be written different ways using the links between speed, frequency and period:

a q = υ 2 / R = ω 2 R = 4π 2 ν 2 R = 4π 2 R / T 2. (28)

4.6 Relationship between translational and rotational movements

Basic kinematic characteristics of motion in a straight line with constant acceleration: displacement s, velocity υ and acceleration a... Corresponding characteristics when moving along a circle of radius R: angular displacement φ, angular velocity ω and angular acceleration ε (in case the body rotates at variable speed).

From geometric considerations, the following relationships arise between these characteristics:

displacement s → angular displacement φ = s / R;

speed υ → angular speed ω = υ / R;

acceleration a→ angular acceleration ε = a/ R.

All the formulas for the kinematics of uniformly accelerated motion along a straight line can be transformed into formulas for the kinematics of rotation along a circle, if we make the indicated changes. For example:

s = υt → φ = ωt, (29)

υ = υ 0 + a t → ω = ω 0 + ε t. (29a)

The relationship between the linear and angular velocities of a point when rotating around a circle can be written in vector form. Indeed, let the circle centered at the origin be located in the (x, y) plane. At any moment in time, the vector drawn from the origin to the point on the circle where the body is located is perpendicular to the body's velocity vector tangential to the circle at this point. Let us define a vector , which is equal in magnitude to the angular velocity ω and is directed along the axis of rotation in the direction that is determined by the rule of the right screw: if the screw is screwed in so that the direction of its rotation coincides with the direction of rotation of the point around the circumference, then the direction of movement of the screw shows the direction of the vector ... Then the connection of three mutually perpendicular vectors ,and can be written using the cross product of vectors.

Previously, the characteristics of rectilinear motion were considered: movement, speed, acceleration... Their counterparts in rotary motion are: angular movement, angular velocity, angular acceleration.

The role of displacement in rotary motion is played by injection;
The value of the angle of rotation per unit of time is angular velocity;
The change in the angular velocity per unit of time is angular acceleration.

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

Let's say the ball rotates uniformly on a thread 1 meter long. In this case, he will describe a circle with a radius of 1 meter. The length of such a circle: C = 2πR = 6.28 m

The time during which the ball completely makes one complete revolution around the circumference is called rotation period - T.

To calculate the linear velocity of the ball, it is necessary to divide the displacement by the time, i.e. circumference for the 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:
V = 6.28 / 1 = 6.28 m / s

2. Centrifugal acceleration

At any point of the ball's rotational motion, its linear velocity vector is directed perpendicular to the radius. It is easy to guess that with such a rotation around a circle, the vector of the linear velocity of the ball is constantly changing its direction. The acceleration characterizing such a change in speed is called centrifugal (centripetal) acceleration.

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

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

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

3. Centrifugal force

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

With a 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 centrifugal force is needed to hold it on a circle:

F c = 1 6.28 2/1 = 39.4 N

WITH centrifugal force we face 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 cornering speed is independent of body weight!

You've probably noticed that some turns on the highway have a slight inward slope. Such turns are "easier" to pass, or rather, you can pass at a greater speed. Let's consider what forces act on the car in such a bend with an inclination. 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 the gravity force:

F c = mV 2 / R or F c = F n sinα

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

F n cosα = mg, hence: F n = 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 = arctan (F c / mg) = arctan (mV 2 / mgR) = arctan (V 2 / gR)

Note again that body weight is not included in the calculation!

Task number 2: on some section of the highway there is a turn with a radius of 100 meters. The average speed of passing this section of the road by cars is 108 km / h (30 m / s). What should be the safe angle of inclination of the roadbed on this section so that the car does not "take 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 frictional force of the roadway.

4. Degrees and radians

Many are confused about the understanding of angular values.

In rotary motion, the basic unit of measure for angular displacement is radian.

2π radians = 360 ° full circle
π radians = 180 ° - half 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 shows the basic formulas for straight and rotary motion.