Circular motion

In kinematics, circular motion is the motion of an object along a circular path. Examples of this include a stone tied to a string, a car moving around a curve, and a point on a rotating wheel. Circular motion can be uniform, meaning the speed is constant, or non-uniform, meaning the speed changes. Even in uniform circular motion, the object is accelerating because its velocity changes direction. The object accelerates toward the center of the circle; this inward acceleration is called centripetal acceleration. A force toward the center, called centripetal force, is required to produce this acceleration. Circular motion is also used to describe the motion of points in a rigid body rotating about a fixed axis, in which case each point in the body moves in a circle fixed around the axis of rotation. More examples of circular motion include special satellite orbits around the Earth (circular orbits), a ceiling fan's blades rotating around a hub, an electron moving perpendicular to a uniform magnetic field, and a gear turning inside a mechanism. Without centripetal acceleration, the object would move in a straight line, according to Newton's laws of motion.

In kinematics, circular motion is the motion of an object along a circular path. Examples of this include a stone tied to a string, a car moving around a curve, and a point on a rotating wheel. Circular motion can be uniform, meaning the speed is constant, or non-uniform, meaning the speed changes. Even in uniform circular motion, the object is accelerating because its velocity changes direction. The object accelerates toward the center of the circle; this inward acceleration is called centripetal acceleration. A force toward the center, called centripetal force, is required to produce this acceleration. Circular motion is also used to describe the motion of points in a rigid body rotating about a fixed axis, in which case each point in the body moves in a circle fixed around the axis of rotation. More examples of circular motion include special satellite orbits around the Earth (circular orbits), a ceiling fan's blades rotating around a hub, an electron moving perpendicular to a uniform magnetic field, and a gear turning inside a mechanism. Without centripetal acceleration, the object would move in a straight line, according to Newton's laws of motion. \omega = \frac {2 \pi}{T} = 2\pi f = \frac{d\theta}{dt} and the units are radians/second. The speed of the object traveling the circle is: v = \frac{2 \pi r}{T} = \omega r The angle swept out in a time is: \theta = 2 \pi \frac{t}{T} = \omega t The angular acceleration, , of the particle is: \alpha = \frac{d\omega}{dt} In the case of uniform circular motion, will be zero. The inward acceleration caused by the changing direction of motion is: a_c = \frac{v^2}{r} = \omega^2 r The centripetal and centrifugal force can also be found using acceleration. For a body of mass , the centripetal force is, by Newton's second law, F_c = ma_c = \frac{mv^2}{r} 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 1 metre per second. The inward acceleration is 1 metre per square second, . It is subject to a centripetal force of 1 kilogram metre per square second, which is 1 newton. The momentum of the body is 1 kg·m·s −1 . The moment of inertia is 1 kg·m 2 . The angular momentum is 1 kg·m 2 ·s −1 . The kinetic energy is 0.5 joule. The circumference of the orbit is 2 (~6.283) metres. The period of the motion is 2  seconds. The frequency is (2 ) −1  hertz.

In vector form

The vector relationships are shown in Figure 1. The axis of rotation is shown as a vector perpendicular to the plane of the orbit and with a magnitude . The direction of 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 and , tangential to the orbit, and of magnitude . 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 and of magnitude and directed exactly opposite to . \mathbf{r}(t) = R \hat\mathbf{u}_R(t)\,, where \hat\mathbf{u}_R(t) is the unit vector parallel to the radius vector at time and pointing away from the origin. It is convenient to introduce the unit vector orthogonal to \hat\mathbf{u}_R(t) as well, namely \hat\mathbf{u}_\theta(t) . It is customary to orient \hat\mathbf{u}_\theta(t) to point in the direction of travel along the orbit. The velocity is the time derivative of the displacement: \mathbf{v}(t) = \frac{d}{dt} \mathbf{r}(t) = \frac{d R}{dt} \hat\mathbf{u}_R(t) + R \frac{d \hat\mathbf{u}_R}{dt} \, . Because the radius of the circle is constant, the radial component of the velocity is zero. The unit vector \hat\mathbf{u}_R(t) has a time-invariant magnitude of unity, so as time varies its tip always lies on a circle of unit radius, with an angle the same as the angle of \mathbf{r}(t) . If the particle displacement rotates through an angle in time , so does \hat\mathbf{u}_R(t) , describing an arc on the unit circle of magnitude . See the unit circle at the left of Figure 4. Hence: \frac{d \hat\mathbf{u}_R}{dt} = \frac{d \theta}{dt} \hat\mathbf{u}_\theta(t) \, , where the direction of the change must be perpendicular to \hat\mathbf{u}_R(t) (or, in other words, along \hat\mathbf{u}_\theta(t) ) because any change d\hat\mathbf{u}_R(t) in the direction of \hat\mathbf{u}_R(t) would change the size of \hat\mathbf{u}_R(t) . The sign is positive because an increase in implies the object and \hat\mathbf{u}_R(t) have moved in the direction of \hat\mathbf{u}_\theta(t) . Hence the velocity becomes: \mathbf{v}(t) = \frac{d}{dt} \mathbf{r}(t) = R\frac{d \hat\mathbf{u}_R}{dt} = R \frac{d \theta}{dt} \hat\mathbf{u}_\theta(t) = R \omega \hat\mathbf{u}_\theta(t) \, . The acceleration of the body can also be broken into radial and tangential components. The acceleration is the time derivative of the velocity: \begin{align} \mathbf{a}(t) &= \frac{d}{dt} \mathbf{v}(t) = \frac{d}{dt} \left(R \omega \hat\mathbf{u}_\theta(t) \right) \\ &= R \left( \frac{d \omega}{dt} \hat\mathbf{u}_\theta(t) + \omega \frac{d \hat\mathbf{u}_\theta}{dt} \right) \, . \end{align} The time derivative of \hat\mathbf{u}_\theta(t) is found the same way as for \hat\mathbf{u}_R(t) . Again, \hat\mathbf{u}_\theta(t) is a unit vector and its tip traces a unit circle with an angle that is . Hence, an increase in angle by \mathbf{r}(t) implies \hat\mathbf{u}_\theta(t) traces an arc of magnitude , and as \hat\mathbf{u}_\theta(t) is orthogonal to \hat\mathbf{u}_R(t) , we have: \frac{d \hat\mathbf{u}_\theta}{dt} = -\frac{d \theta}{dt} \hat\mathbf{u}_R(t) = -\omega \hat\mathbf{u}_R(t) \, , where a negative sign is necessary to keep \hat\mathbf{u}_\theta(t) orthogonal to \hat\mathbf{u}_R(t) . (Otherwise, the angle between \hat\mathbf{u}_\theta(t) and \hat\mathbf{u}_R(t) would decrease with an increase in .) See the unit circle at the left of Figure 4. Consequently, the acceleration is: \begin{align} \mathbf{a}(t) &= R \left( \frac{d \omega}{dt} \hat\mathbf{u}_\theta(t) + \omega \frac{d \hat\mathbf{u}_\theta}{dt} \right) \\ &= R \frac{d \omega}{dt} \hat\mathbf{u}_\theta(t) - \omega^2 R \hat\mathbf{u}_R(t) \,. \end{align} The centripetal acceleration is the radial component, which is directed radially inward: \mathbf{a}_R(t) = -\omega^2 R \hat\mathbf{u}_R(t) \, , while the tangential component changes the magnitude of the velocity: \mathbf{a}_\theta(t) = R \frac{d \omega}{dt} \hat\mathbf{u}_\theta(t) = \frac{d R \omega}{dt} \hat\mathbf{u}_\theta(t) = \frac{d \left|\mathbf{v}(t)\right|}{dt} \hat\mathbf{u}_\theta(t) \, . z = x + iy = R\left(\cos[\theta(t)] + i \sin[\theta(t)]\right) = Re^{i\theta(t)}\,, where is the imaginary unit, and \theta(t) is the argument of the complex number as a function of time, . Since the radius is constant: \dot{R} = \ddot R = 0 \, , where a dot indicates differentiation in respect of time. With this notation, the velocity becomes: v = \dot{z} \begin{align} a &= \dot{v} = i\dot{\omega} z + i\omega\dot{z} = \left(i\dot{\omega} - \omega^2\right)z \\ &= \left(i\dot{\omega} - \omega^2 \right) R e^{i\theta(t)} \\ &= -\omega^2 R e^{i\theta(t)} + \dot{\omega} e^{i\frac{\pi}{2}} R e^{i\theta(t)} \, . \end{align} 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. v = r \frac{d\theta}{dt} = r\omega , where the angular rate of rotation is . (By rearrangement, .) Thus, is a constant, and the velocity vector also rotates with constant magnitude , at the same angular rate . \mathbf{u} \cdot \mathbf{a} = 0. and the square of proper acceleration, expressed as a scalar invariant, the same in all reference frames, \alpha^2 = \gamma^4 a^2 + \gamma^6 \left(\mathbf{u} \cdot \mathbf{a}\right)^2/c^2, becomes the expression for circular motion, \alpha^2 = \gamma^4 a^2. or, taking the positive square root and using the three-acceleration, we arrive at the proper acceleration for circular motion: \alpha = \gamma^2 \frac{v^2}{r}. a_c = v \frac{d\theta}{dt} = v\omega = \frac{v^2}{r} a_r = \frac{v^2}{r} where is the instantaneous speed and is the radius of the circular path. The tangential acceleration has magnitude a_t = \frac{dv}{dt} . The total acceleration is the vector sum of the radial and tangential components. In uniform circular motion, the tangential component is zero because the speed is constant. In non-uniform circular motion, both components may be present. F_r = m a_r = \frac{mv^2}{r} Tangential acceleration is not responsible for keeping the object on a circular path; it changes the object's speed. If the speed is changing, the tangential component of the net force is: F_t = m a_t = m\frac{dv}{dt} When drawing a free-body diagram, centripetal force is usually not drawn as a separate force. Instead, the real forces acting on the object are resolved into radial and tangential components. For example, at the top of a vertical circular path, the radial direction points downward toward the center of the circle. If an object is acted on by gravity and a downward normal force, the radial force equation is: F_r = n + mg = \frac{mv^2}{r} In non-uniform circular motion, the total acceleration is the vector sum of the radial and tangential components: a = \sqrt{a_r^2 + a_t^2} Moving in Circles episode of The Mechanical Universe

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