PHYSICS IS EASY !: August 2011

In physics, circular motion is rotation along a circle: a circular path or a circular orbit. It can be uniform, that is, with constant angular rate of rotation, or non-uniform, that is, with a changing rate of rotation. The rotation around a fixed axis of a three-dimensional body involves circular motion of its parts. The equations describing circular motion of an object do not take size or geometry into account, rather, the motion of a point mass in a plane is assumed. In practice, the center of mass of a body can be considered to undergo circular motion.
Examples of circular motion include: an artificial satellite orbiting the Earth in geosynchronous orbit, a stone which is tied to a rope and is being swung in circles (cf. hammer throw), a racecar turning through a curve in a race track, an electron moving perpendicular to a uniform magnetic field, and a gear turning inside a mechanism.
Circular motion is accelerated even if the angular rate of rotation is constant, because the object's velocity vector is constantly changing direction. Such change in direction of velocity involves acceleration of the moving object by a centripetal force, which pulls the moving object toward the center of the circular orbit. Without this acceleration, the object would move in a straight line, according to Newton's laws of motion.

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[edit] Formulas for uniform circular motion

Figure 1: Vector relationships for uniform circular motion; vector Ω representing the rotation is normal to the plane of the orbit.

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:

$\omega = \frac {2 \pi}{T} \$

The speed of the object traveling the circle is:

$v\, = \frac {2 \pi r } {T} = \omega r$

The angle θ swept out in a time t is:

$\theta = 2 \pi \frac{t}{T} = \omega t\,$

The acceleration due to change in the direction is:

$a\, = \frac {v^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 right-hand rule. With this convention for depicting rotation, the velocity is given by a vector cross product as

$\mathbf{v} = \boldsymbol \Omega \times \mathbf r \ ,$

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

$\mathbf{a} = \boldsymbol \Omega \times \mathbf v = \boldsymbol \Omega \times \left( \boldsymbol \Omega \times \mathbf r \right) \ ,$

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]

[edit] Constant speed

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.
It is subject to a centripetal force of one kilogram metre per square second, which is one newton.
The momentum of the body is one kg·m·s⁻¹.
The moment of inertia is one kg·m².
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.
From the point of view of quantum mechanics, the system is in an excited state having quantum number ~ 9.48×10³⁵.

Then consider a body of mass m, moving in a circle of radius r, with an angular velocity of ω.

The speed is v = r·ω.
The centripetal (inward) acceleration is a = r·ω ² = r ⁻¹·v ².
The centripetal force is F = m·a = r·m·ω ² = r⁻¹·m·v ².
The momentum of the body is p = m·v = r·m·ω.
The moment of inertia is I = r ²·m.
The angular momentum is L = r·m·v = r ²·m·ω = I·ω.
The kinetic energy is E = 2⁻¹·m·v ² = 2⁻¹·r ²·m·ω ² = (2·m)⁻¹·p ² = 2⁻¹·I·ω ² = (2·I)⁻¹·L ² .
The circumference of the orbit is 2·π·r.
The period of the motion is T = 2·π·ω ⁻¹.
The frequency is f = T ⁻¹ . (Frequency is also often denoted by the Greek letter $ν$ , which however is almost indistinguishable from the letter v used here for velocity).
The quantum number is J = 2·π·L h⁻¹

[edit] Description of circular motion using polar coordinates

Figure 2: Polar coordinates for circular trajectory. On the left is a unit circle showing the changes $\mathbf{d\hat u_R}$ and $\mathbf{d\hat u_\theta}$ in the unit vectors $\mathbf{\hat u_R}$ and $\mathbf{\hat u_\theta}$ for a small increment

dθ

in angle

θ

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 $\stackrel{\vec r}{}$ is the radial vector from the origin to the particle location:

$\vec r=R \hat u_R (t)\ ,$

where $\hat u_R (t)$ is the unit vector parallel to the radius vector at time t and pointing away from the origin. It is handy to introduce the unit vector orthogonal to $\hat u_R$ as well, namely $\hat u_\theta$ . It is customary to orient $\hat u_\theta$ to point in the direction of travel along the orbit.
The velocity is the time derivative of the displacement:

$\vec v = \frac {d}{dt} \vec r(t) = \frac {d R}{dt} \hat u_R + R\frac {d \hat u_R } {dt} \ .$

Because the radius of the circle is constant, the radial component of the velocity is zero. The unit vector $\hat u_R$ has a time-invariant 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 $\vec r (t)$ . If the particle displacement rotates through an angle dθ in time dt, so does $\hat u_R$ , describing an arc on the unit circle of magnitude dθ. See the unit circle at the left of Figure 2. Hence:

$\frac {d \hat u_R } {dt} = \frac {d \theta } {dt} \hat u_\theta \ ,$

where the direction of the change must be perpendicular to $\hat u_R$ (or, in other words, along $\hat u_\theta$ ) because any change d $\hat u_R$ in the direction of $\hat u_R$ would change the size of $\hat u_R$ . The sign is positive, because an increase in dθ implies the object and $\hat u_R$ have moved in the direction of $\hat u_\theta$ . Hence the velocity becomes:

$\vec v = \frac {d}{dt} \vec r(t) = R\frac {d \hat u_R } {dt} = R \frac {d \theta } {dt} \hat u_\theta \ = R \omega \hat u_\theta \ .$

The acceleration of the body can also be broken into radial and tangential components. The acceleration is the time derivative of the velocity:

$\vec a = \frac {d}{dt} \vec v = \frac {d}{dt} \left(R\ \omega \ \hat u_\theta \ \right) \ .$

$=R \left( \frac {d \omega}{dt}\ \hat u_\theta + \omega \ \frac {d \hat u_\theta}{dt} \right) \ .$

The time derivative of $\hat u_\theta$ is found the same way as for $\hat u_R$ . Again, $\hat u_\theta$ 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 $\vec r (t)$ implies $\hat u_\theta$ traces an arc of magnitude dθ, and as $\hat u_\theta$ is orthogonal to $\hat u_R$ , we have:

$\frac {d \hat u_\theta } {dt} = -\frac {d \theta } {dt} \hat u_R = -\omega \hat u_R\ ,$

where a negative sign is necessary to keep $\hat u_\theta$ orthogonal to $\hat u_R$ . (Otherwise, the angle between $\hat u_\theta$ and $\hat u_R$ would decrease with increase in dθ.) See the unit circle at the left of Figure 2. Consequently the acceleration is:

$\vec a = R \left( \frac {d \omega}{dt}\ \hat u_\theta + \omega \ \frac {d \hat u_\theta}{dt} \right)$

$=R \frac {d \omega}{dt}\ \hat u_\theta - \omega^2 R \ \hat u_R \ .$

The centripetal acceleration is the radial component, which is directed radially inward:

$\vec a_R= -\omega ^2R \hat u_R \ ,$

while the tangential component changes the magnitude of the velocity:

$\vec a_{\theta}= R \frac {d \omega}{dt}\ \hat u_\theta = \frac {d R \omega}{dt}\ \hat u_\theta =\frac {d |\vec v|}{dt}\ \hat u_\theta \ .$

[edit] 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":

$z=x+iy=R(\cos \theta +i \sin \theta)=Re^{i\theta}\ ,$

where

i

is the imaginary unit, and

$\theta =\theta (t)\ ,$

is the angle of the complex vector with the real axis and is a function of time t. Since the radius is constant:

$\dot R =\ddot R =0 \ ,$

where a dot indicates time differentiation. With this notation the velocity becomes:

$v=\dot z = \frac {d (R e^{i \theta})}{d t} = R \frac {d \theta}{d t} \frac {d (e^{i \theta})}{d \theta} = iR\dot \theta e^{i\theta} = i\omega \cdot Re^{i\theta}= i\omega z$

and the acceleration becomes:

$a=\dot v =i\dot \omega z +i \omega \dot z =(i\dot \omega -\omega^2)z$

$= \left(i\dot \omega-\omega^2 \right) R e^{i\theta}$

$=-\omega^2 R e^{i\theta} + \dot \omega e^{i\frac{\pi}{2}}R e^{i\theta} \ .$

The first term is opposite to the direction of the displacement vector and the second is perpendicular to it, just like the earlier results shown before.