Examples of central forcesExamples of central forces1.1. uniform circular motionuniform circular motion2.2. force due to gravitationforce due to gravitation3.3. simple harmonic motionsimple harmonic motion4.4. projectile motionprojectile motion5.5. uniformly accelerated motionuniformly accelerated motion6.6. others, like electrostatic , magnetostati c forces, etc.others, like electrostatic , magnetostati c forces, etc.

If the force on a body is alwaystowards a fixed point, it is called acentral force. Take the fixed point asthe origin.

� By studying central for ces you may master � 1. uniform circular motion

2. force due to gravitation3. simple harmonic motion4. projectile motion5. uniformly accelerated motion

�All at the same time !

SinceSince forcesforces involveinvolve massmass andand acceleration ,acceleration ,accelerationacceleration involvesinvolves differentiationdifferentiation ofof velocity,velocity,velocityvelocity isis differentiationdifferentiation ofof displacement,displacement, wewe needneedtoto knowknow differentiationdifferentiation priorprior toto itit..SinceSince displacement,displacement, velocity,velocity, accelerationacceleration andand forceforceareare vectorvector quantities,quantities, wewe needneed toto knowknow vectorsvectorspriorprior toto itit.. ThenThen whatwhat wewe areare requiredrequired toto knowknow isisvectorsvectors,, differentiationdifferentiation andand vectorvectordifferentiationdifferentiation ofof coursecourse..

differentiation of vector functions of scalar differentiation of vector functions of scalar variablevariable-- time in Cartesian coordinatestime 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 θand r sin θ respectively. We write

� r = x + y = r cos θ i + r sin θ 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 CARTESIANCOORDINATES (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 CARTESIANCOORDINATES (CONTINUED FROM PREVIOUS SLIDE)

where ax and ay are respectively or respectively asacceleration 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 dd t d t

x ya n d

DIFFERENTIATION OF VECTORSPOLAR COORDINATES

XO

Y

P

Q

R

r

r+δr

T

x

y

X

Y

δr

δθ

δs

θ

s

r

θ

θ

Fig 1:Resolution of radius vector into components

θπ/2+θ

DIFFERENTIATION OF VECTORS POLARCOORDINATES¢ Instead of differentiating displacement and vel ocity

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

¢ =cos θ i + sin θ j and = - sin θ i + cos θ j ….(5)¢ Since magnitudes of both of them uni ty but directions

are both variables . (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.

∧

r∧

θ

∧

r∧

θ

The unit vectors , , thei r increments ,are shown in the figure.

P

OAA’ O

Q

ϕθ θ

θ

r=1

P

PS

x

Fig 2 : differentiation of unit vectors

QS

T∧

r∧

r

∧ ∧

+ δr r ∧ ∧

+ δr r

∧ ∧

θ + δ θ

∧ ∧

θ + δ θ

∧

θ

∧

θ

∧

δ θ

∧

r∧

δ r∧

δ r

∧

θ

∧

θ

DIFFERENTIATION OF UNIT VECTORS.¢ as the unit vector makes an angle θ with the x – axis and the unit

vector makes an angle π/2+θ 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 θ .

¢ 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

jirdtdcos

dtdsind θ

θ+θ

θ−=

∧

dtjiθ

dtdsin

dtdcosd θ

θ−θ

θ−=

∧

dt

( )dtd

dtdcossind θ

=θ

θ+θ−=∧

∧

θjirdt

( )dtd

dtdsincosd θ

−=θ

θ−θ−=∧

∧

rjiθdt

ω=∧

∧

θrdtd

ω−=∧

∧

rθdtd

dtdθ

=ω

d∧

rdt

∧

θ∧

rd

∧

dtθ ∧

− r∧

θ2d

∧

r2dt

d∧

dtθ ∧

− 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 force

r̂ ∧

θ.. . .

2 2m r r m r r∧ ∧ − ω + ω + ω

r θF = ma, then Fr + Fθ =

1. For uniform circular motion, r =a, ω is a constant andsince r is a constant. So F = - a Fθ=0

2. 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 2ω ..r kr= −