Circular motion “Radial “Radial motion” redirect redirectss here. here. It is not to be confused confused with radial with radial velocity or velocity or rotational rotational speed. speed. In physics In physics,, circular motion is a movement of an object along the circumference of a circle or rotation or rotation along along a circul circular ar path. It can be uniform, uniform, with con constant stant angular angular rate rate of rotati rotation on and cons constan tantt speed speed,, or non non-un -unif iform orm with with a changin changing g rate of rotation. rotation. The rot rotati ation on aro around und a fix fixed ed axi axiss of a three-dimensional body involves circular motion of its parts. parts. The The equat equatio ions ns of motio motion n desc describ ribee the mo move veme ment nt of the center the center of mass of mass of a body. Examples Examples of circular 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 a race track, track, an electron moving perpendicular pendicular to a uniform uniform mag magnetic netic fiel field d, and a gear a gear turning turning inside a mechanism. Since the object’s velocity vector is constantly changing direction, the moving object is undergoing acceleration by a centripetal a centripetal force in 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. motion.
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Figure 1: Velocity v and acceleration a in uniform circular motion at angular rate ω; the speed is constant, but the velocity is always tangent to the orbit; the acceleration has constant magnitude, but always points toward the center of rotation
Unif Uniform circu circular lar motio motion n
In physics In physics,, uniform circular motion describes the motion of a body traversing traversing a circular a circular path path at constant speed constant speed.. Since Since the body describes describes circular circular motion, motion, its its distance from the axis of rotation remains constant at all times. Though the body’s speed is constant, its velocity velocity is is not constant: velocity, velocity, a vector a vector quantity, quantity, depends on both the body’s body’s speed and its direction direction of travel. travel. This changing changing veloc velocity ity indicate indicatess the presenc presencee of an accele accelerati ration; on; this centripetal acceleration is acceleration is of constant magnitude and directed rected at all times times towa toward rdss the axis axis of rotati rotation. on. This This acce accell- Figure 2: The velocity vectors at time t and time t + dt are moved eration is, in turn, produced by a centripetal a centripetal force which force which from the orbit on the left left to new positions where their tails coinBecause the velocity is fixed in magnitude magnitude at is also constant in magnitude and directed towards the cide, on the right. Because v = r ω, the velocity vectors also sweep out a circular path at axis of rotation. dt → 0, the acceleration vector a becomes angular angular rate rate ω. As dt
In the case of rotation of rotation around a fixed axis of axis of a rigid body perpendicular to v , which means it points toward the center of that is not negligibly small compared to the radius of the the orbit in the circle circle on the left. left. Angle ω dt is the very small path, each particle of the body describes a uniform uniform circu- angle between the two velocities and tends to zero as dt→ 0 lar motio motion n with with the same same angula angularr veloc velocity ity,, but with with veloc veloc-ity and acce accele lerat ratio ion n varyi varying ng with with the positi position on with with resp respec ectt to the axis. 1
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UNIFORM CIRCULAR MOTION
In the case of uniform circular motion α will be zero. The acceleration due to change in the direction is:
a =
v2 = ω 2 r r
The centripetal and centrifugal force can also be found out using acceleration:
F c = ma =
mv r
2
The vector relationships are shown in Figure 1. The axis of rotation is shown as a vector ω perpendicular to the Figure 3: (Left) Ball in circular motion – rope provides centripetal force to keep ball in circle (Right) Rope is cut and ball plane of the orbit and with a magnitude ω = d θ / dt . The continues in straight line with velocity at the time of cutting the direction of ω is chosen using the right-hand rule. With rope, in accord with Newton’s law of inertia, because centripetal this convention for depicting rotation, the velocity is given force is no longer there by a vector cross product as
v =
ω
× r ,
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
a =
Figure 1: Vector relationships for uniform circular motion; vector ω representing the rotation is normal to the plane of the orbit.
1.1
Formulas
For motion in a circleof 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:
ω
× v =
ω
× (ω × r) ,
which is a vector perpendicular to both ω and v ( t ) of magnitude ω |v| = ω2 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.
• The speed is one metre per second. • The inward acceleration is one metre per square second[v^2/r]
ω =
2π T
=
• It is subject to a centripetal force of one kilogram
dθ dt
metre per square second, which is one newton.
The speed of the object travelling the circle is:
• The momentum of the body is one kg·m·s −1 . • The moment of inertia is one kg·m2 .
v =
2πr T
= ωr
The angle θ swept out in a time t is:
θ = 2π
t T
= ωt
The angular acceleration,α of the particle is:
α =
dω dt
• The angular momentum is one kg·m2 ·s−1 . • The kinetic energy is 1/2 joule. • The circumference of the orbit is 2π (~ 6.283) metres.
• The period of the motion is 2π seconds per turn. • The frequency is (2π)−1 hertz.
1.1
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Formulas
uR have moved in the direction of ˆ uθ . Hence object and ˆ the velocity becomes:
v⃗ =
d du dθ ˆR = R ˆ ˆθ . r⃗(t) = R uθ = Rω u dt dt dt
The acceleration of the body can also be broken into radial and tangential components. The acceleration is the time derivative of the velocity: d a⃗ = dt v⃗ =
Figure 4: Polar coordinates for circular trajectory. On the left is uR and d^ uθ in the unit vectors a unit circle showing the changes d^ ^ uR and ^ uθ for a small increment d θ in angle θ .
1.1.1
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 4. The displacement vector ⃗r is the radial vector from the origin to the particle location:
ˆR (t) , r⃗ = R u uR (t) is the unit vector parallel to the radius vecwhere ˆ tor at time t and pointing away from the origin. It is convenient to introduce the unit vector orthogonal to u ˆR as uθ . It is customary to orient ˆ uθ to point in well, namely ˆ the direction of travel along the orbit. The velocity is the time derivative of the displacement:
v⃗ =
ˆR d dR du ˆR + R r⃗(t) = u . dt dt dt
= R
�
d dt
( R ω ˆuθ ) .
ˆθ dω du ˆθ + ω u dt dt
�
.
uθ is found the same way as for ˆ uR The time derivative of ˆ uθ is a unit vector and its tip traces a unit circle . Again, ˆ with an angle that is π/2 + θ. Hence, an increase in angle uθ traces an arc of magnitude d θ, and d θ byr⃗(t) implies ˆ uθ is orthogonal to ˆ uR , we have: as ˆ
ˆθ du dθ = − ˆ ˆR , uR = −ω u dt dt uθ orthogonal where a negative sign is necessary to keep ˆ uR . (Otherwise, the angle between ˆ uθ and ˆ uR would to ˆ decrease with increase in d θ.) See the unit circle at the left of Figure 4. Consequently the acceleration is: a⃗ = R
= R
�
dω dt
u ˆθ + ω
ˆθ du dt
�
dω ˆθ − ω 2 R ˆ u uR . dt
The centripetal acceleration is the radial component, which is directed radially inward:
ˆR , a⃗ R = − ω 2 Ru
Because the radius of the circle is constant, the radial while the tangential component changes the magnitude of ˆR the velocity: component of the velocity is zero. The unit vector u has a time-invariant magnitude of unity, so as time varies its tip always lies on a circle of unit radius, with an angle dω dRω dv|⃗ | θ the same as the angle of⃗r (t) . If the particle displaceˆθ = ˆθ = ˆθ . a⃗ θ = R u u u uR , ment rotates through an angle d θ in time dt , so does ˆ dt dt dt describing an arc on the unit circle of magnitude d θ. See the unit circle at the left of Figure 4. Hence: 1.1.2 Using complex numbers
ˆR du dθ uθ , = ˆ dt dt where the direction of the change must be perpendicular uR (or, in other words, along ˆ uθ ) because any change to ˆ ˆ R in the direction of ˆ uR would change the size of ˆ uR . d u The sign is positive, because an increase in d θ implies the
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 (cos θ + i sin θ) = Re iθ ,
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2
NON-UNIFORM
where i is the imaginary unit, and
1.2.2
θ = θ (t) ,
In this case the three-acceleration vector is perpendicular to the three-velocity vector,
Relativistic circular motion
is the angle of the complex vector with the real axis and is a function of time t . Since the radius is constant: u⃗ ·a⃗ = 0. and the square of proper acceleration, expressed as a scalar invariant, the same in all reference frames,
¨ = 0 , R˙ = R where a dot indicates time differentiation. With this notation the velocity becomes:
α2 = γ 4 a2 + γ 6u(⃗ ·a⃗ )2 , becomes the expression for circular motion,
d(Reiθ ) d(eiθ ) d(iθ) ˙ iθ = iω ·Reiθ = iωz = R = R (eiθ ) = iR θe dt dt dt α2 = γ 4 a2 . and the acceleration becomes: or, taking the positive square root and using the three2 acceleration, we arrive at the proper acceleration for cir˙ + iω ˙z = (iω˙ − ω )z a = v˙ = i ωz cular motion: 2 iθ = iω˙ − ω Re v = z˙ =
�
�
π
= −ω 2 Reiθ + ωe ˙ i Reiθ . 2
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.
α = γ 2
v2 . r
1.2.3
Acceleration
Main article: Acceleration
1.2
Discussion
The left-hand circle in Figure 2 is the orbit showing the velocity vectors at two adjacent times. On the right, these 1.2.1 Velocity two velocities are moved so their tails coincide. Because speed is constant, the velocity vectors on the right sweep Figure 1 illustrates velocity and acceleration vectors for out a circle as time advances. For a swept angle d θ = ω uniform motion at four different points in the orbit. Bedt the change in v is a vector at right angles to v and of cause the velocity v is tangent to the circular path, no two magnitude v d θ, which in turn means that the magnitude velocities point in the same direction. Although the obof the acceleration is given by ject 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 condθ v2 stant, but whose direction also is always changing. The a = v = vω = dt r acceleration points radially inwards (centripetally) and is perpendicular to the velocity. This acceleration is known as centripetal acceleration. 2 Non-uniform For a path of radius r , when an angle θ is swept out, the distance travelled on the periphery of the orbit is s = r θ. Therefore, the speed of travel around the orbit is
v = r
dθ = rω dt
In Non-uniform circular motion an object is moving in a circular path with a varying speed. Since the speed is changing, there is tangential acceleration in addition to normal acceleration.
In non-uniform circular motion the net acceleration (a) is along direction of Δv which is directed inside circle but where the angular rate of rotation is ω. (By rearrange- does not pass through its center (see figure). The net acment, ω = v /r .) Thus, v is a constant, and the velocity celeration may be resolved into two components: tangenvector v also rotates with constant magnitude v, at the tial acceleration and normal acceleration also known as same angular rate ω. the centripetal or radial acceleration. Unlike tangential
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v a
a
ω
aθ
aR
n
mg
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 tangential force and centripetal force. The tangential force is zero at the top (as no work is performed when the motion is perpendicular to the direction of force applied. Here weight force is perpendicular to the direction of motion of the object at the top of the circle) and centripetal force points down,thus normal force will point down as well. From a logical standpoint, a person who is travelling 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.
v
n
v
mg
n
n
mg
mg
acceleration, centripetal acceleration is present in both uniform and non-uniform circular motion. In non-uniform 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.
nx n ny
mg
Fc 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. The component of weight force is responsible for the tangential force here (We have neglected frictional force). The radial force (centripetal force) is due the change in direction of velocity as discussed earlier. In non-uniform 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
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.
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Applications
Solving applications dealing with non-uniform 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 non-uniform circular motion, there are additional forces acting on the object due to a non-zero 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 F net = ma r F net = mv 2 /r F net = F c 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 responsi-
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ble 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 , we can draw free body diagrams to list all the forces acting on an object then set it equal to F c . 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 ) . 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.
√ + a2r
a2t = a
Radial acceleration is still equal to v 2 /r . Tangential acceleration is simply the derivative of the velocity at any given point: at = dv /dt . 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, θ) , the Coriolis term ac = 2(dr/dt)(dθ/dt) should be added to a t , whereas radial acceleration then becomes ar = − v2 /r + d2 r /dt2 .
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See also • Angular momentum • Equations of motion for circular motion • Example: circular motion • Fictitious force • Geostationary orbit • Geosynchronous orbit • Pendulum (mathematics) • Reactive centrifugal force • Reciprocating motion • Simple harmonic motion#Uniform circular motion • Sling (weapon)
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References
[1] Knudsen, Jens M.; Hjorth, Poul G. (2000). Elements of Newtonian mechanics: including nonlinear dynamics (3 ed.). Springer. p. 96. ISBN 3-540-67652-X., Chapter 5 page 96
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EXTERNAL LINKS
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 CM
