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.
Examples of central forces
1. uniform circular motion
2. force due to gravitation
3. simple harmonic motion
4. projectile motion
5. uniformly accelerated motion
6. others, like electrostatic , magnetostati c forces, etc.

— By studying central for ces you may
master
— 1. uniform circular motion
2. force due to gravitation
3. simple harmonic motion
4. projectile motion
5. uniformly accelerated motion
at the same time !
—All

Since forces involve mass and acceleration ,
acceleration involves differentiation of velocity,
velocity is differentiation of displacement, we need
to know differentiation prior to it.
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
it.
vectors,
vectors , differentiation and vector
differentiation of course .
course.

differentiation of vector functions of scalar
variable-
variable - time in Cartesian coordinates
vector r of a moving mass point may be resolved
(Position
—
into x and y components in Cartesian coordinates as r cos q
and r sin q respectively. We write
r = x + y = r cos q i + r sin q 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)
dv y
dv x
where ax and ay are respectively
and
dt dt
or 2 a n d respectively as
d 2x d 2y
dt dt 2
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.

DIFFERENTIATION OF VECTORS
POLAR COORDINATES
Y
Q
ds
dr
P
r+dr
dq q
Y
q
p/2+q
s
r y
T
r
q q
X
O x X
R
Fig 1:Resolution of radius vector into
components

DIFFERENTIATION OF VECTORS POLAR
COORDINATES
Instead of di fferentiating displacement and vel ocity
¢
vectors, let us differentiate unit vectors r and Ù
Ù
(taken to each other) . Expressi ng them in
Cartesian coordinates, or resol ving into components
¢ r =cos q i + sin q j and = - sin q i + cos q j .(5)
Ù
Ù
¢ Since magnitudes of both of them uni ty but directions
are both vari ables . (see the figure in the above sl ide,
no 7.
¢ For differentiation of the unit vectors refer to the
figure in the next slide. Later on the formul a for
differentiation of unit vectors shal l be fruitfully
utilised for differentiating displacement and vel ocity
vectors.

The unit vectors , , thei r increments Ù Ù Ù Ù
dr dr
r q
,are shown in the figure.
Ù
q
Q S
Ù
q
Ù Ù
q+ d q Ù
Q
dq Ù
q
Ù Ù
P
r+ dr Ù Ù
r+ dr q
Ù
r T
q q P
Ù Ù
j q+ d q
Ù
r
A A O P
r=1 O S
x
Fig 2 : differentiation of unit vectors

DIFFERENTIATION OF UNIT VECTORS.
as the unit vector makes an angle q with the x axis and the unit
¢
vector makes an angle p/2+q with the x axis and both the unit Ù
Ù
vectors have obviously magnitudes unity. Mind it that r 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
Ù Ù
(5) above) d r = - sin q dq i + cos q dq j and d = - cos q dq i - sin q dq j respectively
dt dt
dt dt dt dt
Ù Ù
or, d r = (- sin qi + cos qj) dq = Ù dq and respectively,
dq dq
Ù
= (- cos qi - sin qj ) = - r
¢ d
dt dt
dt dt dt dt
Ù
dr Ù Ù
or and = - r wrespectively .. .(6)
Ù
=w d
¢
dt dt
dq
w=
where , the magnitude of angular velocity of the moving
¢ dt
particle around the point O, or the time rate of turning of q .
Ù
dr
It is important to see here that dt is parallel to , i.e.,
Ù
¢
Ù
perpendicular to r , i.e., in a direction tangent to the unit circle.
¢ Ù
Also d q is parallel to - r , i.e., along the radius and towards the
Ù
Ù
dt d2 r
Ù
center, and thus it is perpendicular to . Thus dt2 is parallel
¢
Ù
to d q , i.e., parallel to - r .
Ù
¢
dt
Ù
Ù
Thus the derivative of is in the direction of - r or centripetal.
¢

DIFFERENTIATION OF VELOCITY AND ACCELERATION VECTORS

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

Different cases of central force
æ .. öÙ æ. öÙ
.
m ç r - r w2 ÷ r + m ç 2 r w + r w ÷
Ù
F = ma, then Frr + F = è ø è ø
..
1. For uniform circular motion, r =a, is a constant and r = 0
since r is a constant. So F r = - a w 2 F =0 ..
2. For simple harmonic motion, F =0, =0, r = - kr
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.