Centripetal acceleration (in m/s 2) is calculated by the formula α = ω 2 R, Where ω - angular velocity (in s –1), R- radius of the circle. Using this formula, find the radius R(in meters), if the angular velocity is 10 s –1 and the centripetal acceleration is 54 m/s 2.
Solution.
Let's express the radius from the formula for centripetal acceleration:
Substituting, we get:
Answer: 0.54.
Answer: 0.54
a = ω 2 R, Where ω R R(in meters) if the angular velocity is 9 s −1 and the centripetal acceleration is 243 m/s 2.
Answer: 3
Centripetal acceleration when moving in a circle (in m/s 2) can be calculated using the formula a = ω 2 R, Where ω is the angular velocity (in s −1), and R- radius of the circle. Using this formula, find the distance R(in meters) if the angular velocity is 4 s −1 and the centripetal acceleration is 96 m/s 2.
Answer: 6
Centripetal acceleration when moving in a circle (in m/s 2) can be calculated using the formula a = ω 2 R, Where ω is the angular velocity (in s −1), and R- radius of the circle. Using this formula, find the distance R(in meters) if the angular velocity is 8.5 s −1 and the centripetal acceleration is 650.25 m/s 2 .
Answer: 000
Centripetal acceleration when moving in a circle (in m/s 2) can be calculated using the formula a = ω 2 R, Where ω is the angular velocity (in s −1), and R- radius of the circle. Using this formula, find the distance R(in meters) if the angular velocity is 5.5 s −1 and the centripetal acceleration is 60.5 m/s 2 .
Centripetal acceleration when moving in a circle (in m/s 2) can be calculated using the formula a = ω 2 R, Where ω is the angular velocity (in s −1), and R- radius of the circle. Using this formula, find the distance R(in meters) if the angular velocity is 0.5 s −1 and the centripetal acceleration is 1.75 m/s 2 .
Answer: 7
Centripetal acceleration when moving in a circle (in m/s 2) can be calculated using the formula a = ω 2 R, Where ω is the angular velocity (in s −1), and R- radius of the circle. Using this formula, find the distance R(in meters) if the angular velocity is 3 s −1 and the centripetal acceleration is 81 m/s 2 .
Answer: 9
Centripetal acceleration when moving in a circle (in m/s 2) can be calculated using the formula a=ω 2 R, Where ω is the angular velocity (in s −1), and R- radius of the circle. Using this formula, find the distance R(in meters) if the angular velocity is 4 s −1 and the centripetal acceleration is 64 m/s 2 .
Centripetal acceleration when moving in a circle (in m/s 2) can be calculated using the formula a = ω 2 R, Where ω is the angular velocity (in s −1), and R- radius of the circle. Using this formula, find the distance R(in meters) if the angular velocity is 0.5 s −1 and the centripetal acceleration is 1.5 m/s 2 .
Answer: 6
Centripetal acceleration when moving in a circle (in m/s 2) can be calculated using the formula a = ω 2 R, Where ω is the angular velocity (in s −1), and R- radius of the circle. Using this formula, find the distance R(in meters) if the angular velocity is 0.5 s −1 and the centripetal acceleration is 2.25 m/s 2 .
Centripetal acceleration when moving in a circle (in m/s 2) can be calculated using the formula a = ω 2 R, Where ω is the angular velocity (in s −1), and R- radius of the circle. Using this formula, find the distance R(in meters) if the angular velocity is 4 s −1 and the centripetal acceleration is 48 m/s 2.
Centripetal acceleration when moving in a circle (in m/s 2) can be calculated using the formula a = ω 2 R, Where ω is the angular velocity (in s −1), and R- radius of the circle. Using this formula, find the distance R(in meters) if the angular velocity is 7.5 s −1 and the centripetal acceleration is 337.5 m/s 2 .
Centripetal acceleration when moving in a circle (in m/s 2) can be calculated using the formula a = ω 2 R, Where ω is the angular velocity (in s −1), and R- radius of the circle. Using this formula, find the distance R(in meters) if the angular velocity is 6 s −1 and the centripetal acceleration is 216 m/s 2.
Centripetal acceleration when moving in a circle (in m/s 2) can be calculated using the formula a = ω 2 R, Where ω is the angular velocity (in s −1), and R- radius of the circle. Using this formula, find the distance R(in meters) if the angular velocity is 6 s −1 and the centripetal acceleration is 72 m/s 2 .
Centripetal acceleration when moving in a circle (in m/s 2) can be calculated using the formula a = ω 2 R, Where ω is the angular velocity (in s −1), and R- radius of the circle. Using this formula, find the distance R(in meters) if the angular velocity is 9 s−1 and the centripetal acceleration is 648 m/s 2 .
Answer: 3
Centripetal acceleration when moving in a circle (in m/s 2) can be calculated using the formula a = ω2R, Where ω is the angular velocity (in s −1), and R- radius of the circle. Using this formula, find the distance R(in meters) if the angular velocity is 9.5 s −1 and the centripetal acceleration is 180.5 m/s 2 .
Centripetal acceleration when moving in a circle (in m/s 2) can be calculated using the formula a = ω 2 R, Where ω is the angular velocity (in s −1), and R- radius of the circle. Using this formula, find the distance R(in meters) if the angular velocity is 7.5 s−1 and the centripetal acceleration is 393.75 m/s 2 .
Centripetal acceleration when moving in a circle (in m/s 2) can be calculated using the formula a = ω 2 R, Where ω is the angular velocity (in s −1), and R- radius of the circle. Using this formula, find the distance R(in meters) if the angular velocity is 8.5 s −1 and the centripetal acceleration is 505.75 m/s 2 .
Answer: 7
Centripetal acceleration when moving in a circle (in m/s 2) can be calculated using the formula a = ω 2 R, Where ω is the angular velocity (in s −1), and R- radius of the circle. Using this formula, find the distance R(in meters) if the angular velocity is 8 s−1 and the centripetal acceleration is 128 m/s 2 .
Centripetal acceleration when moving in a circle (in m/s 2) can be calculated using the formula a = ω 2 R, Where ω is the angular velocity (in s −1), and R- radius of the circle. Using this formula, find the distance R(in meters) if the angular velocity is 9 s −1 and the centripetal acceleration is 405 m/s 2.
Centripetal acceleration when moving in a circle (in m/s 2) can be calculated using the formula a = ω 2 R, Where ω is the angular velocity (in s −1), and R- radius of the circle. Using this formula, find the distance R(in meters) if the angular velocity is 8.5 s −1 and the centripetal acceleration is 289 m/s 2 .
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 construct the 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 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 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 stops its action, 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 v A And vB respectively. Acceleration is the change in speed per unit time. Let's find the difference between the vectors.
In nature, body movement often occurs along curved lines. Almost any curvilinear movement can be represented as a sequence of movements along circular arcs. In general, when moving in a circle, the speed of a body changes as in size, so and in direction.
Uniform movement around a circle
Circular motion is called uniform if the speed remains constant.
According to Newton's third law, every action causes an equal and opposite reaction. The centripetal force with which the connection acts on the body is counteracted by an equal in magnitude and oppositely directed force with which the body acts on the connection. This power F 6 called centrifugal, since it is directed radially from the center of the circle. Centrifugal force is equal in magnitude to centripetal force:
Examples
Consider the case where an athlete rotates an object tied to the end of a string around his head. The athlete feels a force applied to the arm and pulling it outward. To hold the object on the circle, the athlete (using a thread) pulls it inward. Therefore, according to Newton’s third law, an object (again through a thread) acts on the hand with an equal and opposite force, and this is the force that the athlete’s hand feels (Fig. 3.23). The force acting on an object is the inward tension of the thread.
Another example: a “hammer” sports equipment is acted upon by a cable held by the athlete (Fig. 3.24).
Let us recall that centrifugal force acts not on a rotating body, but on a thread. If centrifugal force acted on the body then if the thread breaks, it would fly radially away from the center, as shown in Fig. 3.25, a. However, in fact, when the thread breaks, the body begins to move tangentially (Figure 3.25, b) in the direction of the speed that it had at the moment the thread broke.
A centrifuge is a device designed for training and testing pilots, athletes, and astronauts. Large radius(up to 15 m) and high engine power (several MW) make it possible to create centripetal acceleration of up to 400 m/s 2 . The centrifugal force presses the bodies with a force exceeding normal strength gravity on Earth is more than 40 times. A person can withstand a temporary overload of 20-30 times if he lies perpendicular to the direction of the centrifugal force, and 6 times if he lies along the direction of this force.
3.8. Elements of describing human movement
Human movements are complex character and are difficult to describe. However, in a number of cases, it is possible to identify significant points that distinguish one type of movement from another. Consider, for example, the difference between running and walking.
Elements of stepping movements when walking are shown in Fig. 3.26. In walking movements, each leg alternates between supporting and carrying. The support period includes depreciation (braking the movement of the body towards the support) and repulsion, while the transfer period includes acceleration and braking.
The sequential movements of the human body and his legs when walking are shown in Fig. 3.27.
Lines A and B provide a high-quality image of the movement of the feet during walking. The top line A refers to one leg, the bottom line B to the other. Straight sections correspond to the moments of foot support on the ground, arcuate sections correspond to the moments of movement of the feet. During a period of time (a) both feet rest on the ground; then (b)- leg A is in the air, leg B continues to lean; and after (With)- again both legs rest on the ground. The faster you walk, the shorter the intervals become. (A And With).
In Fig. Figure 3.28 shows the sequential movements of the human body when running and a graphical representation of the movements of the feet. As you can see in the figure, when running there are time intervals { b, d, /), when both legs are in the air, and there are no intervals between the legs simultaneously touching the ground. This is the difference between running and walking.
Another common type of movement is pushing off the support during various jumps. The push-off is accomplished by straightening the pushing leg and swinging movements of the arms and torso. The task of repulsion is to ensure the maximum magnitude of the vector initial speed the overall center of mass of the athlete and its optimal direction. In Fig. 3.29 phases are shown
DRIVING DYNAMICSMATERIAL POINT
Dynamics is a branch of mechanics that studies the movement of a body taking into account its interaction with other bodies.
In the “Kinematics” section the concepts were introduced speed And acceleration material point. For real bodies these concepts need clarification, since for different real body points these movement characteristics may vary. For example, a curved soccer ball not only moves forward, but also rotates. The points of a rotating body move at different speeds. For this reason, we first consider the dynamics material point, and then the results obtained are extended to real bodies.
Allows us to exist on this planet. How can we understand what centripetal acceleration is? 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 of 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 jerk - 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 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 a body in a circle is complex. The stone participates in two types of motion simultaneously: under the influence of force it moves towards the center of rotation, and at the same time along a tangent to the circle, moving away from this center. According to Newton's Second Law, the force holding a stone on a rope is directed toward the center of rotation along the rope. The acceleration vector will also be directed there.
Let us assume that after some time t our stone, moving uniformly with speed V, gets from point A to point B. Let us assume that at the moment of time when the body crossed point B, the centripetal force ceased to act on it. Then, in a period of time, it would get to 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 rule vector addition. As a result of summing these two movements over a period of time t, we obtain the resulting movement along the arc AB.
If the time interval t is taken to be 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 rectilinear movement, 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 influence of centripetal acceleration can be calculated using the formula uniformly accelerated motion:
Distance AB is equal to the product of speed and time, that is, AB = V x t,
AO - calculated earlier using the formula of uniformly accelerated motion for moving in a straight line: AO = at 2 / 2.
Substituting this data into the formula and transforming it, 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 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
Angular velocity is equal to the linear velocity divided by the radius of the circle. The converse statement is also true: V = ωR, where ω is the angular velocity
If we substitute this value into the formula, we can obtain an expression for the centrifugal acceleration for 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 formulation of acceleration. The facts show that circular motion is real, but to maintain it requires acceleration directed towards the center. Under the influence of the force caused by this acceleration, a change in the amount of motion occurs, as a result of which the trajectory of movement is constantly curved, all the time changing the direction of the velocity vector, but without changing it absolute value. Moving in a circle, our long-suffering stone rushes inward, otherwise it would continue to move tangentially. Every moment of time, going 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, 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 movement and keeps the body in orbit.
Exceeding the safety factor
In the previous experiment we were dealing with a perfect rope that did not break. But let’s say our rope is the most ordinary, and you can even calculate the force after which it will simply break. In order to calculate this force, it is enough to compare the strength of the rope with the load it experiences during the rotation of the stone. By rotating the stone at a faster speed, you tell it more movement, and therefore greater 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. By 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 = 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 effect of gravity, since at such high speeds its influence is negligible. But you can notice 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 movement 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 same centripetal acceleration that was derived from the previous formula. Only in this case will the role of the stone be played by a celestial body attracted to the Earth, and the role of the rope will be played by force gravity. The g factor will be expressed in terms of the radius of our planet and its rotation speed.
Results
The essence of centripetal acceleration is the hard and thankless work of keeping a moving body in orbit. There is a paradoxical case when constant acceleration the body does not change its speed. To the untrained mind, such a statement is quite 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 not the least role.
Any object that rotates in a circular path experiences a force. It is directed towards the central point of the circle described by the trajectory. This force is called centripetal.
Centrifugal force is often referred to as or fictitious force. It is primarily used to refer to forces that are associated with motion in a non-inertial frame.
According to Newton's third law, every action has an opposite direction and an equal reaction. And in this concept, centrifugal force acts upon the action of centripetal force.
Both forces are inertial, so only when an object moves. They also always appear in pairs and balance each other out. Therefore, in practice they can often be neglected.
Examples of centrifugal and centripetal force
If you take a stone and tie a rope to it, and then begin to rotate the rope above your head, a centripetal force will arise. It will act through the rope on the stone and not allow it to move away to a distance greater than the length of the rope itself, as would happen with a normal throw. Centrifugal force will act in the opposite way. It will be quantitatively equal and opposite in direction to the centripetal force. This power is greater the more more massive body, moving along a closed trajectory.
It is well known that the Moon revolves around the Earth in a circular orbit. The force of attraction that exists between the Earth and the Moon is the result of the centripetal force. Centrifugal force, in this case, is virtual and does not actually exist. This follows from Newton's third law. However, despite the abstraction, centrifugal force performs very important role in the interaction of two celestial bodies. Thanks to it, the Earth and its satellite do not move away or get closer to each other, but move along stationary orbits. Without centrifugal force they would have collided long ago.
Conclusion
1. While the centripetal force is directed towards the center of the circle, the centrifugal force is opposite to it.
2. Centrifugal force is often called inertial or fictitious force.
3. Centrifugal force is always equal to quantitative value and opposite in direction to the centripetal force.
5. The word "centripetal" was derived from Latin words. "Centrum" means center and "petere" means "to seek." The concept "centrifugal" is derived from the Latin words "centrum" and "fugere",
