Uniform Circular Motion

If a particle travels along a circle or circular arc of radius \(r\) at constant speed \(v\), it is said to be in uniform circular motion and has an acceleration of constant magnitude \(a\). The force acting on the particle is, $$ \text {Force} = \text {mass} \times \text {acceleration}$$ $$ \vec{F} = m\vec{a} $$ For uniform circular motion, $$ \vec{F} = {m \vec{v}^2 \over r}$$ where $$\vec{a} = {\vec{v}^2 \over r} \text {(centripetal acceleration)}$$

Force acting on the object moving in a circular loop

At the bottom of the loop,

$$\begin{aligned} \vec{F}_{net}&=\vec{F}_{normal} - \vec{F}_{gravity} \\ \vec{F}_{normal}&=\vec{F}_{net} + \vec{F}_{gravity} \end{aligned}$$

At the top of the loop,

$$\begin{aligned} \vec{F}_{net}&=\vec{F}_{normal} + \vec{F}_{gravity} \\ \vec{F}_{normal}&=\vec{F}_{net} - \vec{F}_{gravity} \end{aligned}$$

Force on the body due to centripetal acceleration, \(\vec{F}_{net}\) depends on the velocity of the moving body, $$ \vec{F}_{net} = m\vec{a}$$ $$ \vec{F}_{net} = {m \vec{v}^2 \over r}$$

Understanding Circular Motion Mathematically

    ➤ Constant velocity which changes its direction, results in centripetal acceleration

In the simplest case of circular motion at radius \(r\) with position given by the angular displacement \(\phi(t)\) from the x-axis, the orbital angular velocity is the rate of change of angle with respect to time: \[\omega ={\frac {d\phi }{dt}}\] where \(d\phi\) is the angular displacement in time \(dt\).

    ➤ SI units of angular velocity is radians per second or \((rad/sec)\)

If \(\phi\) is measured in radians, the arc-length from the positive x-axis around the circle to the particle is \(l=r\phi\) and the linear velocity is equal to the rate of change in arc length, $$ v = {dl \over dt} = {rd\phi \over dt} = r\omega $$ Hence angular velocity, $$ \omega = {v \over r} $$

    ➤ Angular velocity is equal to linear velocity divided by the circular radius.

The angular acceleration \(\alpha\), on the particle is: $$ \alpha ={d\omega \over dt} = {d \over dt}\left({v \over r}\right) $$ \[ \space = {1 \over r}{dv \over dt} - { v \over r^2}{ dr \over dt} \]

For uniform circular motion with constant radius and speed, \(\alpha = 0 \) and the acceleration due to change in direction is $$\vec{a} = {d \vec{v} \over dt} = {d \vec{v} \over d\vec{r}} {d \vec{r} \over dt}= { v \over r}{ d\vec{r}\over dt} ={ \vec{v} \over r}v ={ \vec{v}^2 \over r} $$

    ➤ Circular motion with constant velocity results in Radial Acceleration (only) directed towards the center

A radial acceleration, also known as centripetal acceleration, acts on a particle in uniform circular motion and, \[ a_{c} = {v^2 \over r} \] The direction of \(\vec{a}\) is towards the center of the circle or circular arc, and the direction of velocity vector \(\vec{v}\) is always tangent to the path.

    ➤ SI units of centripetal acceleration is \(m/s^2\) or \(ms^{-2}\)

The time for the particle to complete a circle, \(T\), is called the period of revolution, or simply the period, of the motion. \[ T = {\text{distance travelled} \over \text{velocity}} \] \[ \space = {\text{Circumference} \over \text{velocity}} \] \[ T={2\pi r \over v} \]

Equations of Centripetal Force

    ➤ In terms of angular velocity, \(\omega\)

$$ \vec{F} = {m \omega ^2 r} $$ $$ \vec{a} = {\omega ^2 r} $$ $$ T = {2 \pi \over \omega} $$

    ➤ In terms of linear velocity, \(v\)

$$ \vec{F} = {m v^2 \over r} $$ $$ \vec{a} = { v^2 \over r} $$ $$ T = {2 \pi r \over v}$$

What is a Centrifugal Force?

In Newtonian mechanics, the centrifugal force is an inertial force (also called a "fictitious" or "pseudo" force) that acts on a particle moving in circular motion. Centrifugal force has magnitude and dimensions same as that of centripetal force which keeps the particle in circular path but points in the opposite direction.