What is a central force ?

Examples of central forces1. uniform circular motion2. force due to gravitation3. simple harmonic motion4. projectile motion5. uniformly accelerated motion6. others, like electrostatic , magnetostatic forces, etc.

If the force on a body is always towards a fixed point, it is called a central force. Take the fixed point as the origin.

WHY STUDY CENTRAL FORCES

By studying central forces you may master

1. uniform circular motion2. force due to gravitation3. simple harmonic motion4. projectile motion5. uniformly accelerated motion

All at the same time !

Since forces involve mass and acceleration, acceleration involves differentiation of velocity, velocity is differentiation of displacement, we need to know differentiation prior to it.Since displacement, velocity, acceleration and force are vector quantities, we need to know vectors prior to it. Then what we are required to know is vectors, differentiation and vector differentiation of course.

What is required for study of central forces

differentiation of vector functions of scalar variable- time in Cartesian coordinates

(Position vector r of a moving mass point may be resolved into x and y components in Cartesian coordinates as r cos and r sin respectively. We write

r = x + y = r cos i + r sin j ……………………….(1)

 where i and j are unit vectors in x and y directions respectively.

On differentiation, we get, or , v = vx + vy………………………...

………….(2)where vx and vy as respectively and velocity is

vector differentiation of position vector.

DIFFERENTIATION OF VECTORS CARTESIAN COORDINATES (CONTINUED FROM PREVIOUS SLIDE)

where vx and vy as respectively and velocity is vector differentiation of position vector.

Eqn.(2) makes an important statement that the components of velocity in Cartesian coordinates are time derivatives of the components of position vectors. This result appears too obvious, but as we would see later, it may not hold in other system of coordinates .A second differentiation gives

or , a = ax + ay………………………….….(3)

DIFFERENTIATION OF VECTORS CARTESIAN COORDINATES (CONTINUED FROM PREVIOUS SLIDE)

where ax and ay are respectively or respectively as acceleration is vector differentiation of velocity vector.  Eqn.(3)similarly states that the components of acceleration in Cartesian coordinates are time derivatives of the components of velocity vectors. Again it may not hold in other system of coordinates.

dd

dt dtx

vv yand2 2

2 2

d d

dt dt

x yand

DIFFERENTIATION OF VECTORS POLAR COORDINATES

XO

Y

P

Q

R

r

r+r

T

x

y

X

Y

r

s

s

r

Fig 1:Resolution of radius vector into components

/2+

DIFFERENTIATION OF VECTORS POLAR COORDINATES Instead of differentiating displacement and velocity

vectors, let us differentiate unit vectors and (taken ┴ to each other) . Expressing them in Cartesian coordinates, or resolving into components

=cos i + sin j and = - sin i + cos j ….(5) Since magnitudes of both of them unity but

directions are both variables . (see the figure in the above slide, no 7.

For differentiation of the unit vectors refer to the figure in the next slide. Later on the formula for differentiation of unit vectors shall be fruitfully utilised for differentiating displacement and velocity vectors.

r

θ

r

θ

The unit vectors , , their increments ,are shown in the figure.

P

OAA’ O

Q

r=1

P

PS

x

Fig 2 : differentiation of unit vectors

Q

S

T

r

r

r r

r r

r

r

r

DIFFERENTIATION OF UNIT VECTORS.

as the unit vector makes an angle with the x – axis and the unit vector makes an angle /2+ with the x – axis and both the unit vectors have obviously magnitudes unity. Mind it that and are unit vectors continuously changing in direction and are not constant vectors as such; whereas i and j are constant vectors.

Differentiating the unit vectors with respect to time t, we have,(from (5) above) and respectively

or, and respectively,

or and respectively…………………..……….(6)

where , the magnitude of angular velocity of the moving particle around the point O, or the time rate of turning of .

It is important to see here that is parallel to , i.e., perpendicular to , i.e., in a direction tangent to the unit circle.

Also is parallel to , i.e., along the radius and towards the center, and thus it is perpendicular to . Thus is parallel to , i.e., parallel to .

Thus the derivative of is in the direction of or centripetal.

r

θ

jir

dt

dcos

dt

dsin

d

dtji

θdt

dsin

dt

dcos

d

dt

dt

d

dt

dcossin

d

θjir

dt

dt

d

dt

dsincos

d

rjiθ

dt

θr

dtd

rθ

dtd

dt

d

d

rdt

θ

rd

dt

r

θ2d

r2dt

d

dt

r

θ

r

DIFFERENTIATION OF VELOCITY AND ACCELERATION VECTORS

WHAT IF THE FORCE IS ALWAYS TOWARDS A FIXED POINT, I.E., CENTRAL FORCE

Different cases of central forcer̂

θ

.. . .2 2m r r m r r

r θ

F = ma, then Fr + Fθ =

1. For uniform circular motion, r =a, ω is a constant and since r is a constant. So F = - a Fθ=02. For simple harmonic motion, Fθ=0, ω =0, 3. For projectile motion, simpler will be Cartesian coordinates, ax =0,

and ay =-g, and uniform acceleration is a particular case of projectile motion where the horizontal velocity is 0 always.

..

0r

r̂ 2 ..

r kr