Motion Under Central Forces (part 1)

central forces notes

This article covers an introduction to central forces , equation of motion under central forces and

Introduction

If the force \mathop f\limits^ \to acting on a body has following characteristics then it is a central force
(i) it depends on the distance between two particles
(ii) it is always directed towards or away from a fixed point.
Gravitational force is an example of central forces. Mathematically if we
consider central point as origin

    \[\overrightarrow f (r) = f(r)( \pm \hat r)\]

where + stands for repulsion forces and – for attractive forces , f(r) is
the magnitude of the central forces and \hat r is unit vector in the
direction of central forces.

Theorem:- Under the action of central forces particle
always move in the same plane.
Multiplying equation (1) by \vec r on both the sides we get

    \[\vec r \times \vec f(r) = f(r)\left\{ {\vec r \times \hat r} \right\}\]

This gives

    \[f(r) = 0\]

since {\vec r \times \hat r = 0}
but we know that

    \[\vec f(r) = m\frac{{{d^2}\vec r}}{{d{t^2}}}\]

therefore

    \[\vec r \times m\frac{{{d^2}\vec r}}{{d{t^2}}} = 0\]

or,

    \[\vec r \times \frac{{{d^2}\vec r}}{{d{t^2}}} = 0\]

now,

    \[\frac{d}{{dt}}\left( {\vec r \times \frac{{d\vec r}}{{dt}}} \right) = \vec r \times \frac{{{d^2}\vec r}}{{d{t^2}}} + 0\]

    \[\frac{d}{{dt}}\left( {\vec r \times \frac{{d\vec r}}{{dt}}} \right) = 0\]

    \[\left( {\vec r \times \frac{{d\vec r}}{{dt}}} \right) = cons\tan t(\vec h)\]

where {\vec h} is a vector which does not depend on time and is perpandicular to the plane formed by the position vector {\vec r} and velocity {\frac{{d\vec r}}{{dt}}} . Thus the plane formed by {\vec r} and velocity {\frac{{d\vec r}}{{dt}}} will also remains constant. Therefore the particle will always move in the same plane.
The torque due to central forces can be found out by the above expression. Torque

    \[\tau = \vec r \times \vec f(r) = \vec r \times f(r)\hat r = 0\]

but

    \[\tau = \frac{{d\vec J}}{{dt}}\]

This implies thet {\vec J} is a constant quantity and it is known as angular momentum. Thus angular momentum is also conserved. But

    \[\vec J = \vec r \times \vec p = \vec r \times m\frac{{d\vec r}}{{dt}} = \vec h\]

Equation of motion under central forces

The equation of particle moving in a plane under
central forces can be obtained in terms of polar co-ordinates. Let m be the
mass of the particle, (x,y) be its cartesian co-ordinates and (r,\theta ) be the polar co-ordinates of the particle.
therefore, x = r\cos \theta and y = r\sin \theta
Differentiating it with respect to time we get

    \[\frac{{dx}}{{dt}} = \frac{{dr}}{{dt}}\cos \theta - r\sin \theta \frac{{d\theta }}{{dt}}\]

    \[\frac{{dy}}{{dt}} = \frac{{dr}}{{dt}}\sin \theta + r\cos \theta \frac{{d\theta }}{{dt}}\]

again differentiating above two equations with respect to time we get

    \[\frac{{{d^2}x}}{{d{t^2}}} = \left[ {\frac{{{d^2}r}}{{d{t^2}}} - r{{\left( {\frac{{d\theta }}{{dt}}} \right)}^2}} \right]\cos \theta - \left[ {r\frac{{{d^2}\theta }}{{d{t^2}}} + 2\frac{{dr}}{{dt}}\frac{{d\theta }}{{dt}}} \right]\sin \theta \]

similarly,





    \[\frac{{{d^2}y}}{{d{t^2}}} = \left[ {\frac{{{d^2}r}}{{d{t^2}}} - r{{\left( {\frac{{d\theta }}{{dt}}} \right)}^2}} \right]\sin \theta + \left[ {r\frac{{{d^2}\theta }}{{d{t^2}}} + 2\frac{{dr}}{{dt}}\frac{{d\theta }}{{dt}}} \right]\cos \theta \]

now acceleration

    \[\vec a = \left( {\hat i\frac{{{d^2}x}}{{d{t^2}}} + \hat j\frac{{{d^2}y}}{{d{t^2}}}} \right) = \left[ {\frac{{{d^2}r}}{{d{t^2}}} - r{{\left( {\frac{{d\theta }}{{dt}}} \right)}^2}} \right]\left( {\hat i\cos \theta + \hat j\sin \theta } \right) - \left[ {r\frac{{{d^2}\theta }}{{d{t^2}}} + 2\frac{{dr}}{{dt}}\frac{{d\theta }}{{dt}}} \right]\left( { - \hat i\sin \theta + \hat j\cos \theta } \right)\]

We know that in radial direction unit vector is
\hat r = \left( {\hat i\cos \theta + \hat j\sin \theta } \right)
and in transverse direction unit vector is
\hat \theta = \left( { - \hat i\sin \theta + \hat j\cos \theta } \right)

this vector is perpandicular to the radial vector.
Now acceleration is \vec a = (\hat r{a_r} + \hat \theta {a_\theta }) and
this implies that

    \[\vec a = \left[ {\frac{{{d^2}r}}{{d{t^2}}} - r{{\left( {\frac{{d\theta }}{{dt}}} \right)}^2}} \right]\hat r - \left[ {r\frac{{{d^2}\theta }}{{d{t^2}}} + 2\frac{{dr}}{{dt}}\frac{{d\theta }}{{dt}}} \right]\hat \theta \]

(2)   \[{a_r} = \left[ {\frac{{{d^2}r}}{{d{t^2}}} - r{{\left( {\frac{{d\theta }}{{dt}}} \right)}^2}} \right]\]

(3)   \[{a_\theta } = \left[ {r\frac{{{d^2}\theta }}{{d{t^2}}} + 2\frac{{dr}}{{dt}}\frac{{d\theta }}{{dt}}} \right]\]

According to Newton’s second law of motion

    \[\vec f(r) = f(r)\hat r\]

    \[f(r)\hat r = m(\hat r{a_r} + \hat \theta {a_\theta })\]

On comparison

(4)   \[m\left[ {\frac{{{d^2}r}}{{d{t^2}}} - r{{\left( {\frac{{d\theta }}{{dt}}} \right)}^2}} \right] = f(r)\]

and

(5)   \[m\left[ {r\frac{{{d^2}\theta }}{{d{t^2}}} + 2\frac{{dr}}{{dt}}\frac{{d\theta }}{{dt}}} \right] = 0\]

Equations 4 and 5 are called the equation of motion of the particle moving under central forces.

Law of conservation of angular momentum

Multiplying equation (5) of previous page by r on
both the sides we get

    \[m\left[ {{r^2}\frac{{{d^2}\theta }}{{d{t^2}}} + 2r\frac{{dr}}{{dt}}\frac{{d\theta }}{{dt}}} \right] = 0\]

This implies that

(6)   \[\frac{d}{{dt}}\left( {m{r^2}\frac{{d\theta }}{{dt}}} \right) = 0\]

or,

(7)   \[\frac{{dJ}}{{dt}} = 0\]

where,J = \left( {m{r^2}\frac{{d\theta }}{{dt}}} \right) = constant\tag{8}

Angular momentum of a particle moving with rotational motion is \left( {m{r^2}\frac{{d\theta }}{{dt}}} \right) therefore equation 7 shows law of conservation of angular momentum of moving particle under the influence of central force.
J = \left( {m{r^2}\frac{{d\theta }}{{dt}}} \right) is always conserved. Now if equation (6) is multiplied by the quantity \frac{1}{{2m}} then we get another physical quantity which is AREAL VELOCITY. It also remains constant for the particle moving under central forces
This implies that

    \[\frac{d}{{dt}}\left( {\frac{1}{2}{r^2}\frac{{d\theta }}{{dt}}} \right) = 0\]

or,

(9)   \[\frac{1}{2}{r^2}\frac{{d\theta }}{{dt}} = constant(invarient)\]

It defines areal velocity as the area swept by position vector {\vec r} of particle in unit time. Therefore we come across to know that the law of conservation of angular momentum of central forces expresses the invariance of areal velocity. It is actually Kepler’s Second law of planetary motion which states that “Every planet moves with a constant areal velocity around the sun”