In physics, circular motion is a movement of an object along the circumference of a circle or rotation along a circular path. It can be uniform, with constant angular rate of rotation (and constant speed), or nonuniform with a changing rate of rotation. The rotation around a fixed axis of a threedimensional body involves circular motion of its parts. The equations of motion describe the movement of the center of mass of a body.
Examples of circular motion include: an artificial satellite orbiting the Earth at constant height, a stone which is tied to a rope and is being swung in circles, a car turning through a curve in a race track, an electron moving perpendicular to a uniform magnetic field, and a gear turning inside a mechanism.
Since the object's velocity vector is constantly changing direction, the moving object is undergoing acceleration by a centripetal force in the direction of the center of rotation. Without this acceleration, the object would move in a straight line, according to Newton's laws of motion.
Uniform
In physics, uniform circular motion describes the motion of a body traversing a circular path at constant speed. The distance of the body from the axis of rotation remains constant at all times. Though the body's speed is constant, its velocity is not constant: velocity, a vector quantity, depends on both the body's speed and its direction of travel. This changing velocity indicates the presence of an acceleration; this centripetal acceleration is of constant magnitude and directed at all times towards the axis of rotation. This acceleration is, in turn, produced by a centripetal force which is also constant in magnitude and directed towards the axis of rotation.
In the case of rotation around a fixed axis of a rigid body that is not negligibly small compared to the radius of the path, each particle of the body describes a uniform circular motion with the same angular velocity, but with velocity and acceleration varying with the position with respect to the axis.
Formulas
For motion in a circle of radius r, the circumference of the circle is C = 2π r. If the period for one rotation is T, the angular rate of rotation, also known as angular velocity, ω is:
 $\backslash omega\; =\; \backslash frac\; \{2\; \backslash pi\}\{T\}\; \backslash $ and the units are radians/sec
The speed of the object traveling the circle is:
 $v\backslash ,\; =\; \backslash frac\; \{2\; \backslash pi\; r\; \}\; \{T\}\; =\; \backslash omega\; r$
The angle θ swept out in a time t is:
 $\backslash theta\; =\; 2\; \backslash pi\; \backslash frac\{t\}\{T\}\; =\; \backslash omega\; t\backslash ,$
The acceleration due to change in the direction is:
 $a\backslash ,\; =\; \backslash frac\; \{v^2\}\; \{r\}\; \backslash ,\; =\; \{\backslash omega^2\}\; \{r\}$
The vector relationships are shown in Figure 1. The axis of rotation is shown as a vector Ω perpendicular to the plane of the orbit and with a magnitude ω = dθ / dt. The direction of Ω is chosen using the righthand rule. With this convention for depicting rotation, the velocity is given by a vector cross product as
 $\backslash mathbf\{v\}\; =\; \backslash boldsymbol\; \backslash Omega\; \backslash times\; \backslash mathbf\; r\; \backslash \; ,$
which is a vector perpendicular to both Ω and r ( t ), tangential to the orbit, and of magnitude ω r. Likewise, the acceleration is given by
 $\backslash mathbf\{a\}\; =\; \backslash boldsymbol\; \backslash Omega\; \backslash times\; \backslash mathbf\; v\; =\; \backslash boldsymbol\; \backslash Omega\; \backslash times\; \backslash left(\; \backslash boldsymbol\; \backslash Omega\; \backslash times\; \backslash mathbf\; r\; \backslash right)\; \backslash \; ,$
which is a vector perpendicular to both Ω and v ( t ) of magnitude ω v = ω^{2} r and directed exactly opposite to r ( t ).^{[1]}
In the simplest case the speed, mass and radius are constant.
Consider a body of one kilogram, moving in a circle of radius one metre, with an angular velocity of one radian per second.
In polar coordinates
During circular motion the body moves on a curve that can be described in polar coordinate system as a fixed distance R from the center of the orbit taken as origin, oriented at an angle θ (t) from some reference direction. See Figure 2. The displacement vector $\backslash stackrel\{\backslash vec\; r\}\{\}$ is the radial vector from the origin to the particle location:
 $\backslash vec\; r=R\; \backslash hat\; u\_R\; (t)\backslash \; ,$
where $\backslash hat\; u\_R\; (t)$ is the unit vector parallel to the radius vector at time t and pointing away from the origin. It is convenient to introduce the unit vector orthogonal to $\backslash hat\; u\_R$ as well, namely $\backslash hat\; u\_\backslash theta$. It is customary to orient $\backslash hat\; u\_\backslash theta$ to point in the direction of travel along the orbit.
The velocity is the time derivative of the displacement:
 $\backslash vec\; v\; =\; \backslash frac\; \{d\}\{dt\}\; \backslash vec\; r(t)\; =\; \backslash frac\; \{d\; R\}\{dt\}\; \backslash hat\; u\_R\; +\; R\backslash frac\; \{d\; \backslash hat\; u\_R\; \}\; \{dt\}\; \backslash \; .$
Because the radius of the circle is constant, the radial component of the velocity is zero. The unit vector $\backslash hat\; u\_R$ has a timeinvariant magnitude of unity, so as time varies its tip always lies on a circle of unit radius, with an angle θ the same as the angle of $\backslash vec\; r\; (t)$. If the particle displacement rotates through an angle dθ in time dt, so does $\backslash hat\; u\_R$, describing an arc on the unit circle of magnitude dθ. See the unit circle at the left of Figure 2. Hence:
 $\backslash frac\; \{d\; \backslash hat\; u\_R\; \}\; \{dt\}\; =\; \backslash frac\; \{d\; \backslash theta\; \}\; \{dt\}\; \backslash hat\; u\_\backslash theta\; \backslash \; ,$
where the direction of the change must be perpendicular to $\backslash hat\; u\_R$ (or, in other words, along $\backslash hat\; u\_\backslash theta$) because any change d$\backslash hat\; u\_R$ in the direction of $\backslash hat\; u\_R$ would change the size of $\backslash hat\; u\_R$. The sign is positive, because an increase in dθ implies the object and $\backslash hat\; u\_R$ have moved in the direction of $\backslash hat\; u\_\backslash theta$.
Hence the velocity becomes:
 $\backslash vec\; v\; =\; \backslash frac\; \{d\}\{dt\}\; \backslash vec\; r(t)\; =\; R\backslash frac\; \{d\; \backslash hat\; u\_R\; \}\; \{dt\}\; =\; R\; \backslash frac\; \{d\; \backslash theta\; \}\; \{dt\}\; \backslash hat\; u\_\backslash theta\; \backslash \; =\; R\; \backslash omega\; \backslash hat\; u\_\backslash theta\; \backslash \; .$
The acceleration of the body can also be broken into radial and tangential components. The acceleration is the time derivative of the velocity:
 $\backslash vec\; a\; =\; \backslash frac\; \{d\}\{dt\}\; \backslash vec\; v\; =\; \backslash frac\; \{d\}\{dt\}\; \backslash left(R\backslash \; \backslash omega\; \backslash \; \backslash hat\; u\_\backslash theta\; \backslash \; \backslash right)\; \backslash \; .$
 $=R\; \backslash left(\; \backslash frac\; \{d\; \backslash omega\}\{dt\}\backslash \; \backslash hat\; u\_\backslash theta\; +\; \backslash omega\; \backslash \; \backslash frac\; \{d\; \backslash hat\; u\_\backslash theta\}\{dt\}\; \backslash right)\; \backslash \; .$
The time derivative of $\backslash hat\; u\_\backslash theta$ is found the same way as for $\backslash hat\; u\_R$. Again, $\backslash hat\; u\_\backslash theta$ is a unit vector and its tip traces a unit circle with an angle that is π/2 + θ. Hence, an increase in angle dθ by $\backslash vec\; r\; (t)$ implies $\backslash hat\; u\_\backslash theta$ traces an arc of magnitude dθ, and as $\backslash hat\; u\_\backslash theta$ is orthogonal to $\backslash hat\; u\_R$, we have:
 $\backslash frac\; \{d\; \backslash hat\; u\_\backslash theta\; \}\; \{dt\}\; =\; \backslash frac\; \{d\; \backslash theta\; \}\; \{dt\}\; \backslash hat\; u\_R\; =\; \backslash omega\; \backslash hat\; u\_R\backslash \; ,$
where a negative sign is necessary to keep $\backslash hat\; u\_\backslash theta$ orthogonal to $\backslash hat\; u\_R$. (Otherwise, the angle between $\backslash hat\; u\_\backslash theta$ and $\backslash hat\; u\_R$ would decrease with increase in dθ.) See the unit circle at the left of Figure 2. Consequently the acceleration is:
 $\backslash vec\; a\; =\; R\; \backslash left(\; \backslash frac\; \{d\; \backslash omega\}\{dt\}\backslash \; \backslash hat\; u\_\backslash theta\; +\; \backslash omega\; \backslash \; \backslash frac\; \{d\; \backslash hat\; u\_\backslash theta\}\{dt\}\; \backslash right)$
 $=R\; \backslash frac\; \{d\; \backslash omega\}\{dt\}\backslash \; \backslash hat\; u\_\backslash theta\; \; \backslash omega^2\; R\; \backslash \; \backslash hat\; u\_R\; \backslash \; .$
The centripetal acceleration is the radial component, which is directed radially inward:
 $\backslash vec\; a\_R=\; \backslash omega\; ^2R\; \backslash hat\; u\_R\; \backslash \; ,$
while the tangential component changes the magnitude of the velocity:
 $\backslash vec\; a\_\{\backslash theta\}=\; R\; \backslash frac\; \{d\; \backslash omega\}\{dt\}\backslash \; \backslash hat\; u\_\backslash theta\; =\; \backslash frac\; \{d\; R\; \backslash omega\}\{dt\}\backslash \; \backslash hat\; u\_\backslash theta\; =\backslash frac\; \{d\; \backslash vec\; v\}\{dt\}\backslash \; \backslash hat\; u\_\backslash theta\; \backslash \; .$
Using complex numbers
Circular motion can be described using complex numbers. Let the $x$ axis be the real axis and the $y$ axis be the imaginary axis. The position of the body can then be given as $z$, a complex "vector":
 $z=x+iy=R(\backslash cos\; \backslash theta\; +i\; \backslash sin\; \backslash theta)=Re^\{i\backslash theta\}\backslash \; ,$
where $i$ is the imaginary unit, and
 $\backslash theta\; =\backslash theta\; (t)\backslash \; ,$
is the angle of the complex vector with the real axis and is a function of time t.
Since the radius is constant:
 $\backslash dot\; R\; =\backslash ddot\; R\; =0\; \backslash \; ,$
where a dot indicates time differentiation.
With this notation the velocity becomes:
 $v=\backslash dot\; z\; =\; \backslash frac\; \{d\; (R\; e^\{i\; \backslash theta\})\}\{d\; t\}\; =\; R\; \backslash frac\; \{d\; \backslash theta\}\{d\; t\}\; \backslash frac\; \{d\; (e^\{i\; \backslash theta\})\}\{d\; \backslash theta\}\; =\; iR\backslash dot\; \backslash theta\; e^\{i\backslash theta\}\; =\; i\backslash omega\; \backslash cdot\; Re^\{i\backslash theta\}=\; i\backslash omega\; z$
and the acceleration becomes:
 $a=\backslash dot\; v\; =i\backslash dot\; \backslash omega\; z\; +i\; \backslash omega\; \backslash dot\; z\; =(i\backslash dot\; \backslash omega\; \backslash omega^2)z$
 $=\; \backslash left(i\backslash dot\; \backslash omega\backslash omega^2\; \backslash right)\; R\; e^\{i\backslash theta\}$
 $=\backslash omega^2\; R\; e^\{i\backslash theta\}\; +\; \backslash dot\; \backslash omega\; e^\{i\backslash frac\{\backslash pi\}\{2\}\}R\; e^\{i\backslash theta\}\; \backslash \; .$
The first term is opposite in direction to the displacement vector and the second is perpendicular to it, just like the earlier results shown before.
Discussion
Velocity
Figure 1 illustrates velocity and acceleration vectors for uniform motion at four different points in the orbit. Because the velocity v is tangent to the circular path, no two velocities point in the same direction. Although the object has a constant speed, its direction is always changing. This change in velocity is caused by an acceleration a, whose magnitude is (like that of the velocity) held constant, but whose direction also is always changing. The acceleration points radially inwards (centripetally) and is perpendicular to the velocity. This acceleration is known as centripetal acceleration.
For a path of radius r, when an angle θ is swept out, the distance travelled on the periphery of the orbit is s = rθ. Therefore, the speed of travel around the orbit is
 $v\; =\; r\; \backslash frac\{d\backslash theta\}\{dt\}\; =\; r\backslash omega$,
where the angular rate of rotation is ω. (By rearrangement, ω = v/r.) Thus, v is a constant, and the velocity vector v also rotates with constant magnitude v, at the same angular rate ω.
Acceleration
Main article:
Acceleration
The lefthand circle in Figure 2 is the orbit showing the velocity vectors at two adjacent times. On the right, these two velocities are moved so their tails coincide. Because speed is constant, the velocity vectors on the right sweep out a circle as time advances. For a swept angle dθ = ω dt the change in v is a vector at right angles to v and of magnitude v dθ, which in turn means that the magnitude of the acceleration is given by
 $a\; =\; v\; \backslash frac\{d\backslash theta\}\{dt\}\; =\; v\backslash omega\; =\; \backslash frac\{v^2\}\{r\}$
Centripetal acceleration for some values of radius and magnitude of velocity
v
r

1 m/s 3.6 km/h 2.2 mph

2 m/s 7.2 km/h 4.5 mph

5 m/s 18 km/h 11 mph

10 m/s 36 km/h 22 mph

20 m/s 72 km/h 45 mph

50 m/s 180 km/h 110 mph

100 m/s 360 km/h 220 mph

Slow walk


Bicycle


City car


Aerobatics

10 cm 3.9 in

Laboratory centrifuge

10 m/s² 1.0 g

40 m/s² 4.1 g

250 m/s² 25 g

1.0 km/s² 100 g

4.0 km/s² 410 g

25 km/s² 2500 g

100 km/s² 10000 g

20 cm 7.9 in


5.0 m/s² 0.51 g

20 m/s² 2.0 g

130 m/s² 13 g

500 m/s² 51 g

2.0 km/s² 200 g

13 km/s² 1300 g

50 km/s² 5100 g

50 cm 1.6 ft


2.0 m/s² 0.20 g

8.0 m/s² 0.82 g

50 m/s² 5.1 g

200 m/s² 20 g

800 m/s² 82 g

5.0 km/s² 510 g

20 km/s² 2000 g

1 m 3.3 ft

Playground carousel

1.0 m/s² 0.10 g

4.0 m/s² 0.41 g

25 m/s² 2.5 g

100 m/s² 10 g

400 m/s² 41 g

2.5 km/s² 250 g

10 km/s² 1000 g

2 m 6.6 ft


500 mm/s² 0.051 g

2.0 m/s² 0.20 g

13 m/s² 1.3 g

50 m/s² 5.1 g

200 m/s² 20 g

1.3 km/s² 130 g

5.0 km/s² 510 g

5 m 16 ft


200 mm/s² 0.020 g

800 mm/s² 0.082 g

5.0 m/s² 0.51 g

20 m/s² 2.0 g

80 m/s² 8.2 g

500 m/s² 51 g

2.0 km/s² 200 g

10 m 33 ft

Rollercoaster vertical loop

100 mm/s² 0.010 g

400 mm/s² 0.041 g

2.5 m/s² 0.25 g

10 m/s² 1.0 g

40 m/s² 4.1 g

250 m/s² 25 g

1.0 km/s² 100 g

20 m 66 ft


50 mm/s² 0.0051 g

200 mm/s² 0.020 g

1.3 m/s² 0.13 g

5.0 m/s² 0.51 g

20 m/s² 2 g

130 m/s² 13 g

500 m/s² 51 g

50 m 160 ft


20 mm/s² 0.0020 g

80 mm/s² 0.0082 g

500 mm/s² 0.051 g

2.0 m/s² 0.20 g

8.0 m/s² 0.82 g

50 m/s² 5.1 g

200 m/s² 20 g

100 m 330 ft

Freeway onramp

10 mm/s² 0.0010 g

40 mm/s² 0.0041 g

250 mm/s² 0.025 g

1.0 m/s² 0.10 g

4.0 m/s² 0.41 g

25 m/s² 2.5 g

100 m/s² 10 g

200 m 660 ft


5.0 mm/s² 0.00051 g

20 mm/s² 0.0020 g

130 m/s² 0.013 g

500 mm/s² 0.051 g

2.0 m/s² 0.20 g

13 m/s² 1.3 g

50 m/s² 5.1 g

500 m 1600 ft


2.0 mm/s² 0.00020 g

8.0 mm/s² 0.00082 g

50 mm/s² 0.0051 g

200 mm/s² 0.020 g

800 mm/s² 0.082 g

5.0 m/s² 0.51 g

20 m/s² 2.0 g

1 km 3300 ft

Highspeed railway

1.0 mm/s² 0.00010 g

4.0 mm/s² 0.00041 g

25 mm/s² 0.0025 g

100 mm/s² 0.010 g

400 mm/s² 0.041 g

2.5 m/s² 0.25 g

10 m/s² 1.0 g

Nonuniform
Nonuniform circular motion is any case in which an object moving in a circular path has a varying speed. The tangential acceleration is nonzero; the speed is changing.
Since there is a nonzero tangential acceleration, there are forces that act on an object in addition to its centripetal force (composed of the mass and radial acceleration). These forces include weight, normal force, and friction.
In nonuniform circular motion, normal force does not always point in the opposite direction of weight. Here is an example with an object traveling in a straight path then loops a loop back into a straight path again.
This diagram shows the normal force pointing in other directions rather than opposite to the weight force. The normal force is actually the sum of the radial and tangential forces that help to counteract the weight force and contribute to the centripetal force. The horizontal component of normal force is what contributes to the centripetal force. The vertical component of the normal force is what counteracts the weight of the object.
In nonuniform circular motion, normal force and weight may point in the same direction. Both forces can point down, yet the object will remain in a circular path without falling straight down. First let’s see why normal force can point down in the first place. In the first diagram, let's say the object is a person sitting inside a plane, the two forces point down only when it reaches the top of the circle. The reason for this is that the normal force is the sum of the weight and centripetal force. Since both weight and centripetal force points down at the top of the circle, normal force will point down as well. From a logical standpoint, a person who is traveling in the plane will be upside down at the top of the circle. At that moment, the person’s seat is actually pushing down on the person, which is the normal force.
The reason why the object does not fall down when subjected to only downward forces is a simple one. Think about what keeps an object up after it is thrown. Once an object is thrown into the air, there is only the downward force of earth’s gravity that acts on the object. That does not mean that once an object is thrown in the air, it will fall instantly. What keeps that object up in the air is its velocity. The first of Newton's laws of motion states that an object’s inertia keeps it in motion, and since the object in the air has a velocity, it will tend to keep moving in that direction.
Applications
Solving applications dealing with nonuniform circular motion involves force analysis. With uniform circular motion, the only force acting upon an object traveling in a circle is the centripetal force. In nonuniform circular motion, there are additional forces acting on the object due to a nonzero tangential acceleration. Although there are additional forces acting upon the object, the sum of all the forces acting on the object will have to equal to the centripetal force.
$F\_\{net\}\; =\; ma\backslash ,$
$F\_\{net\}\; =\; ma\_r\backslash ,$
$F\_\{net\}\; =\; mv^2/r\backslash ,$
$F\_\{net\}\; =\; F\_c\backslash ,$
Radial acceleration is used when calculating the total force. Tangential acceleration is not used in calculating total force because it is not responsible for keeping the object in a circular path. The only acceleration responsible for keeping an object moving in a circle is the radial acceleration. Since the sum of all forces is the centripetal force, drawing centripetal force into a free body diagram is not necessary and usually not recommended.
Using $F\_\{net\}\; =\; F\_c\backslash ,$, we can draw free body diagrams to list all the forces acting on an object then set it equal to $F\_c\backslash ,$. Afterwards, we can solve for what ever is unknown (this can be mass, velocity, radius of curvature, coefficient of friction, normal force, etc.). For example, the visual above showing an object at the top of a semicircle would be expressed as $F\_c\; =\; (n+mg)\backslash ,$.
In uniform circular motion, total acceleration of an object in a circular path is equal to the radial acceleration. Due to the presence of tangential acceleration in non uniform circular motion, that does not hold true any more. To find the total acceleration of an object in non uniform circular, find the vector sum of the tangential acceleration and the radial acceleration.
$\backslash sqrt\{a\_r^2+a\_t^2\}=a$
Radial acceleration is still equal to $v^2/r$.
Tangential acceleration is simply the derivative of the velocity at any given point: $a\_t\; =\; dv/dt\; \backslash ,$.
This root sum of squares of separate radial and tangential accelerations is only correct for circular motion; for general motion within a plane with polar coordinates $(r,\backslash theta)$, the Coriolis term $a\_c\; =\; 2(dr/dt)(d\backslash theta/dt)$ should be added to $a\_t$, whereas radial acceleration then becomes $a\_r=\; v^2/r+d^2\; r/dt^2$.
See also
References
External links
 Physclips: Mechanics with animations and video clips from the University of New South Wales
 Circular Motion – a chapter from an online textbook
 Circular Motion Lecture – a video lecture on CMml:വര്ത്തുളചലനം
vi:Chuyển động quay
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