1styear PHYSICS Notes Chapter # 06 GRAVITATION

Gravitation
The property of all objects in the universe which carry mass, by virtue of which they attract one another, is called Gravitation.



Centripetal Acceleration of the Moon
Newton, after determining the centripetal acceleration of the moon, formulated the law of universal gravitation.
Suppose that the moon is orbiting around the earth in a circular orbit.
If V = velocity of the moon in its orbit,
Rm = distance between the centres of earth and moon,
T = time taken by the moon to complete one revolution around the earth.
For determining the centripetal acceleration of the moon,. Newton applied Huygen's formula which is
a(c) = v2 / r
For moon, am = v2 / Rm ..................... (1)
But v = s/t = circumference / time period = 2πRm/T
Therefore,
v2 = 4π2Rm2 / T2
Therefore,
=> a(m) = (4p2Rm2/T2) x (1/Rm)
a(m) = 4p2Rm / T2
Put Rm = 3.84 x 10(8) m
T = 2.36 x 10(6) sec
Therefore,
a(m) = 2.72 x 10(-3) m/s2

Comparison Between 'am' AND 'g'


Newton compared the centripetal acceleration of the moon 'am' with the gravitational acceleration 'g'.
i.e., am / g = 1 / (60)2 ................. (1)
If Re = radius of the earth, he found that
Re2 / Rm2 - 1 / (60)2 ......................... (2)
Comparing (1) and (2),
am / g = Re2 / Rm2 ..................................... (3)
From equation (3), Newton concluded that at any point gravitational acceleration is inversely to the square of the distance of that point from the centre of the earth. It is true of all bodies in the universe. This conclusion provided the basis for the Newton' Law of Universal Gravitation.

Newton's Law of Universal Gravitation
Consider tow bodies A and B having masses mA and mB respectively.

Let,
F(AB) = Force on A by B
F(BA) = Force on B by A
r(AB) = displacement from A to B
r(BA) = displacement from B to A
r(AB) = unit vector in the direction of r(AB).
r(BA) = unit vector in the direction of r(BA).

From a(m) / g = Re2 / Rm2, we have
F(AB) ∞1 / r(BA)2 ......................... (1)
Also,
F(AB) ∞ m(A) ............................... (2)
F(BA) ∞ m(B)
According to the Newton's third law of motion
F(AB) = F(BA) .................... (for magnitudes)
Therefore,
F(AB) ∞ m(B) ................................ (3)
Combining (1), (2) and (3), we get
F(AB) ∞ m(A)m(B) / r(BA)2
F(AB) = G m(A)m(B) / r(BA)2 ........................... (G = 6.67 x 10(-11) N - m2 / kg2)

Vector Form
F(AB) = - (G m(A)m(B) / r(BA)2) r(BA)
F(BA) = - (G m(B)m(A) / r(AB)22) r(AB)
Negative sign indicates that gravitational force is attractive.

Statement of the Law
"Every body in the universe attracts every other body with a force which is directly proportional to the products of their masses and inversely proportional to the square of the distance between their centres."

Mass and Average Density of Earth
Let,
M = Mass of an object placed near the surface of earth
M(e) = Mass of earth
R(e) = Radius of earth
G = Acceleration due to gravity
According to the Newton's Law of Universal Gravitation.
F = G M Me / Re2 ............................. (1)
But the force with which earth attracts a body towards its centre is called weight of that body.
Therefore,
F = W = Mg
(1) => M g = G M Me / Re2
g = G Me / Re2
Me = g Re2 / G .................................. (2)
Put
g = 9.8 m/sec2,
Re = 6.38 x 10(6) m,
G = 6.67 x 10(-11) N-m2/kg2, in equation (2)
(1) => Me = 5.98 x 10(24) kg. ................... (In S.I system)
Me = 5.98 x 10(27) gm .................................... (In C.G.S system)
Me = 6.6 x 10(21) tons
For determining the average density of earth (?),
Let Ve be the volume of the earth.
We know that
Density = mass / volume
Therefore,
ρ= Me / Ve ........................... [Ve = volume of earth]
ρ = Me / (4/3 ? Re3) .............. [since Ve = 4/3 ? Re3]
ρ = 3 Me / 4 ? Re3
Put,
Me = 5.98 x 10(24) kg
& Re = 6.38 x 10(6) m
Therefore,
ρ = 5.52 x 10(3) kg / m3

Mass of Sun
Let earth is orbiting round the sun in a circular orbit with velocity V.
Me = Mass of earth
Ms = Mass of the sun
R = Distance between the centres of the sun and the earth
T = Period of revolution of earth around sun
G = Gravitational constant
According to the Law of Universal Gravitation
F = G Ms Me / R2 .................................... (1)
This force 'F' provides the earth the necessary centripetal force
F = Me V2 / R ............................................ (2)
(1) & (2) => Me V2 / R = G Ms Me / R2
=> Ms = V2 R / G ........................................ (3)
V = s / Π = 2Π R / T
=> V2 = 4Π2 R2 / T2
Therefore,
(3) => Ms = (4Π R2 / T2) x (R / G)
Ms = 4Π2 R3 / GT2 ........................................ (4)
Substituting the value of
R = 1.49 x 10(11),
G = 6.67 x 10(-11) N-m2 / kg2,
T = 365.3 x 24 x 60 x 60 seconds, in equation (4)
We get
Ms = 1.99 x 10(30) kg

Variation of 'g' with Altitude
Suppose earth is perfectly spherical in shape with uniform density ?. We know that at the surface of earth
g = G Me / Re2
where
G = Gravitational constant
Me = Mass of earth
Re = Radius of earth
At a height 'h' above the surface of earth, gravitational acceleration is
g = G Me / (Re + h)2
Dividing (1) by (2)
g / g = [G Me / Re2] x [(Re + h)2 / G Me)
g / g = (Re + h)2 / Re2
g / g = [Re + h) / Re]2
g / g = [1 + h/Re]2
g / g = [1 + h/Re]-2
We expand R.H.S using Binomial Formula,
(1 + x)n = 1 + nx + n(n-1) x2 / 1.2 + n(n + 1)(n-2)x3 / 1.2.3 + ...
If h / Re < 1, then we can neglect higher powers of h / Re.
Therefore
g / g = 1 - 2 h / Re
g = g (1 - 2h / Re) ................................. (3)
Equation (3) gives the value of acceleration due to gravity at a height 'h' above the surface of earth.
From (3), we can conclude that as the value of 'h' increases, the value of 'g' decreases.

Variation of 'g' with Depth
Suppose earth is perfectly spherical in shape with uniform density ?.
Let
Re = Radius of earth
Me = Mass of earth
d = Depth (between P and Q)
Me = Mass of earth at a depth 'd'
At the surface of earth,
g = G Me / Re2 ...................................... (1)
At a depth 'd', acceleration due to gravity is
g = G Me / (Re - d)2 ........................... (2)
Me = ρ x Ve = ρ x (4/3) π Re3 = 4/3 π Re3 ρ
Me = ρ x Ve = ρ x (4/3) π (Re - d)3 = 4/3 π (Re - d)3 ρ
Ve = Volume of earth
Substitute the value of Me in (1),
(1) => g = (G / Re2) x (4/3) π Re3 ρ
g = 4/3 π Re ρ G ................................. (3)
Substitute the value of Me in (2)
g = [G / Re - d)2] x (4/3) π (Re - d)3 ρ
g = 4/3 π (Re - d) ρ G
Dividing (4) by (3)
g / g = [4/3 π (Re - d) ρ G] / [4/3 π Re ρ G]
g / g = (Re - d) / Re
g / g = 1 - d/Re
g / g = g (1 - d / Re) ........................... (5)
Equation (5) gives the value of acceleration due to gravity at a depth 'd' below the surface of earth
From (5), we can conclude that as the value of 'd' increases, value of 'g' decreases.
At the centre of the earth.
d = Re => Re / d = 1
Therefore,
(5) => g = g (1-1)
g = 0
Thus at the centre of the earth, the value of gravitational acceleration is zero.

Weightlessness in Satellites
An apparent loss of weight experienced by a body in a spacecraft in orbit is called weightlessness.
To discuss weightlessness in artificial satellites, let us take the example of an elevator having a block of mass ;m; suspended by a spring balance attached to the coiling of the elevator. The tension in the thread indicates the weight of the block.

Consider following cases.

1. When Elevator is at Rest
T = m g

2. When Elevator is Ascending with an Acceleration 'a'
In this case
T > m g
Therefore, Net force = T - mg
m a = T - m g
T = m g + m a
In this case of the block appears "heavier".

3. When Elevator is Descending with an Acceleration 'a'
In this case
m g > T
Therefore
Net force = m g - T
m a = m g - T
T = m g - m a
In this case, the body appears lighter

4. When the Elevator is Falling Freely Under the Action of Gravity
If the cable supporting the elevator breaks, the elevator will fall down with an acceleration equal to 'g'
From (3)
T = m g - m a
But a = g
Therefore
T = m g - m g
T = 0
In this case, spring balance will read zero. This is the state of "weightlessness".
In this case gravitation force still acts on the block due to the reason that elevator block, spring balance and string all have same acceleration when they fall freely, the weight of the block appears zero.

Artificial Gravity
In artificial satellites, artificial gravity can be created by spinning the space craft about its own axis.
Now we calculate frequency of revolution (v) of a space craft of length 2R to produce artificial gravity in it. Its time period be 'T' and velocity is V.

1styear PHYSICS Notes Chapter # 05 TORQUE,ANGULAR,MOMENTUM AND EQUILIBRIUM

Torque or Moment of Force
Definition
If a body is capable of rotating about an axis, then force applied properly on this body will rotate it about the axis (axis of rotating). This turning effect of the force about the axis of rotation is called torque.
Torque is the physical quantity which produces angular acceleration in the body.

Explanation
Consider a body which can rotate about O (axis of rotation). A force F acts on point P whose position vector w.r.t O is r.

F is resolved into F1 and F2. θ is the angle between F and extended line of r.
The component of F which produces rotation in the body is F1.
The magnitude of torgue (π) is the product of the magnitudes of r and F1.
Equation (1) shows that torque is the cross-product of displacement r and force F.
Torque → positive if directed outward from paper
Torque → negative if directed inward from paper
The direction of torque can be found by using Right Hand Rule and is always perpendicular to the plane containing r & F.
Thus
Clockwise torque → negative
Counter-Clockwise torque → positive

Alternate Definition of Torque
π = r x F
|π| = r F sin θ
|π| = F x r sin θ
But r sin ? = L (momentum arm) (from figure)
Therefore,
|π| = F L
Magnitude of Torque = Magnitude of force x Moment Arm
Note
If line of action of force passes through the axis of rotation then this force cannot produce torque.
The unit of torque is N.m.

Couple
Two forces are said to constitute a couple if they have
1. Same magnitudes
2. Opposite directions
3. Different lines of action
These forces cannot produces transiatory motion, but produce rotatory motion.

Moment (Torque) of a Couple
Consider a couple composed of two forces F and -F acting at points A and B (on a body) respectively, having position vectors r1 & r2.
If p1 is the torque due to force F, then
π1 = r1 x F
Similarly if p2 is the torque due to force - F, then
π2 = r2 x (-F)
The total torque due to the two forces is
π = π1 + π2
π = r1 x F + r2 x (-F)
π = r1 x F - r2 x (-F)
π = (r1 - r2) x F
π = r x F
where r is the displacement vector from B to A.
The magnitude of torque is
π = r F sin (180 - θ)
π = r F sin ? .................... {since sin (180 - θ) = sin θ}
Where ? is the angle between r and -F.
π = F (r sin θ)
But r sin θ is the perpendicular distance between the lines of action of forces F and -F is called moment arm of the couple denoted by d.
π = Fd
Thus
[Mag. of the moment of a couple] = [Mag. of any of the forces forming the couple] x [Moment arm of the couple]
Moment (torque) of a given couple is independent of the location of origin.

Centre of Mass
Definition
The centre of mass of a body, or a system of particles, is a point on the body that moves in the same way that a single particle would move under the influence of the same external forces. The whole mass of the body is supposed to be concentrated at this point.

Explanation
During translational motion each point of a body moves in the same manner i.e., different particles of the body do not change their position w.r.t each other. Each point on the body undergoes the same displacement as any other point as time goes on. So the motion of one particle represents the motion of the whole body. But in rotating or vibrating bodies different particles move in different manners except one point called centre of mass. The centre of mass of a body or a system of particle is a point which represents the movement of the entire system. It moves in the same way that a single particle would move under the influence of same external forces.

Centre of Mass and Centre of Gravity
In a completely uniform gravitational field, the centre of mass and centre of gravity of an extended body coincides. But if gravitational field is not uniform, these points are different.

Determination of the Centre of Mass
Consider a system of particles having masses m1, m2, m3, ................. mn. Suppose x1, and z1, z2, z3 are their distances on z-axis, all measured from origin.

Equilibrium
A body is said to be in equilibrium if it is
1. At rest, or
2. Moving with uniform velocity
A body in equilibrium possess no acceleration.

Static Equilibrium
The equilibrium of bodies at rest is called static equilibrium. For example,
1. A book lying on a table
2. A block hung from a string

Dynamic Equilibrium
The equilibrium of bodies moving with uniform velocity is called dynamic equilibrium. For example,
1. The jumping of a paratrooper by a parachute is an example of uniform motion. In this case, weight is balanced by the reaction of the air on the parachute acts in the vertically upward direction.
2. The motion of a small steel ball through a viscous liquid. Initially the ball has acceleration but after covering a certain distance, its velocity becomes uniform because weight of the ball is balanced by upward thrust and viscous force of the liquid. Therefore, ball is in dynamic equilibrium.

Angular Momentum
Definition
The quantity of rotational motion in a body is called its angular momentum. Thus angular momentum plays same role in rotational motion as played by linear momentum in translational motion.
Mathematically, angular momentum is the cross-product of position vector and the linear momentum, both measured in an inertial frame of reference.
ρ= r x P

Explanation
Consider a mass 'm' rotating anti-clockwise in an inertial frame of reference. At any point, let P be the linear momentum and r be the position vector.
ρ = r x P
ρ = r P sinθ ........... (magnitude)
ρ = r m V sinθ .......... {since P = m v)
where,
V is linear speed
θ is the angle between r and P
θ = 90º in circular motion (special case)
The direction of the angular momentum can be determined by the Righ-Hand Rule.
Also
ρ = r m (r ω) sin θ
ρ = m r2 ω sin θ

Units of Angular Momentum
The units of angular momentum in S.I system are kgm2/s or Js.
1. ρ = r m V sin θ
= m x kg x m/s
= kg.m2/s

2. ρ = r P sin θ
= m x Ns
= (Nm) x s
= J.s

Dimensions of Angular Momentum
[ρ] = [r] [P]
= [r] [m] [V]
= L . M . L/T
= L2 M T-1

Relation Between Torque and Angular Momentum
OR
Prove that the rate of change of angular momentum is equal to the external torque acting on the body.
Proof
We know that rate of change of linear momentum is equal to the applied force.
F = dP / dt
Taking cross product with r on both sides, we get
R x F = r x dP / dt
τ= r x dP / dt ............................. {since r x P = τ}
Now, according to the definition of angular momentum
ρ = r x P
Taking derivative w.r.t time, we get
d? / dt = d / dt (r x P)
=> dρ / dt = r x dP / dt + dr / dt x P
=> dρ / dt = τ + V x P .................. {since dr / dt = V}
=> dρ / dt = τ + V x mV
=> dρ / dt = τ+ m (V x V)
=> dρ / dt = τ + 0 ................. {since V x V = 0}
=> dρ / dt = τ
Or, Rate of change of Angular Momentum = External Torque .

1styear PHYSICS Notes Chapter# 04 MOTION IN TWO DIMENSION

Projectile Motion
A body moving horizontally as well as vertically under the action of gravity simultaneously is called a projectile. The motion of projectile is called projectile motion. The path followed by a projectile is called its trajectory.

Examples of projectile motion are
1. Kicked or thrown balls
2. Jumping animals
3. A bomb released from a bomber plane
4. A shell of a gun.

Analysis of Projectile Motion
Let us consider a body of mass m, projected an angle ? with the horizontal with a velocity V0. We made the following three assumptions.
1. The value of g remains constant throughout the motion.
2. The effect of air resistance is negligible.
3. The rotation of earth does not affect the motion.

Horizontal Motion
Acceleration : ax = 0
Velocity : Vx = Vox
Displacement : X = Vox t

Vertical Motion
Acceleration : ay = - g
Velocity : Vy = Voy - gt
Displacement : Y = Voy t - 1/2 gt2

Initial Horizontal Velocity
Vox = Vo cos θ...................... (1)

Initial Vertical Velocity
Voy = Vo sin θ...................... (2)
Net force W is acting on the body in downward vertical direction, therefore, vertical velocity continuously changes due to the acceleration g produced by the weight W.
There is no net force acting on the projectile in horizontal direction, therefore, its horizontal velocity remains constant throughout the motion.

X - Component of Velocity at Time t (Vx)
Vx = Vox = Vo cos θ .................... (3)

Y - Component of Velocity at Time t (Vy)
Data for vertical motion
Vi = Voy = Vo sin θ
a = ay = - g
t = t
Vf = Vy = ?
Using Vf = Vi + at
Vy = Vo sin θ- gt .................... (4)



Range of the Projectile (R)
The total distance covered by the projectile in horizontal direction (X-axis) is called is range
Let T be the time of flight of the projectile.
Therefore,
R = Vox x T .............. {since S = Vt}
T = 2 (time taken by the projectile to reach the highest point)
T = 2 Vo sin θ/ g
Vox = Vo cos θ
Therefore,
R = Vo cos θx 2 Vo sin θ/ g
R = Vo2 (2 sin θcos θ) / g
R = Vo2 sin 2 θ/ g .................. { since 2 sin θcos θ= sin2 θ}
Thus the range of the projectile depends on
(a) The square of the initial velocity
(b) Sine of twice the projection angle θ

The Maximum Range
For a given value of Vo, range will be maximum when sin2 θ in R = Vo2 sin2 θ / g has maximum value. Since
0 = sin2 θ= 1
Hence maximum value of sin2 θ is 1.
Sin2 θ = 1
2θ = sin(-1) (1)
2θ = 90º
θ= 45º
Therefore,
R(max) = Vo2 / g ; at θ= 45º
Hence the projectile must be launched at an angle of 45º with the horizontal to attain maximum range.

Projectile Trajectory
The path followed by a projectile is referred as its trajectory.
We known that
S = Vit + 1/2 at2
For vertical motion
S = Y
a = - g
Vi = Voy = Vo sin ?
Therefore,
Y = Vo sinθ t - 1/2 g t2 ....................... (1)
Also
X = Vox t
X = Vo cos θ t ............ { since Vox = Vo cos?}
t = X / Vo cos θ
(1) => Y = Vo sinθ (X / Vo cos θ) - 1/2 g (X / Vo cos θ)2
Y = X tan θ - gX2 / 2Vo2 cos2 θ
For a given value of Vo and θ, the quantities tanθ, cosθ, and g are constant, therefore, put
a = tan θ
b = g / Vo2 cos2θ
Therefore
Y = a X - 1/2 b X2
Which shows that trajectory is parabola.

Uniform Circular Motion
If an object moves along a circular path with uniform speed then its motion is said to be uniform circular motion.

Recitilinear Motion
Displacement →R
Velocity → V
Acceleration → a

Circular Motion
Angular Displacement → θ
Angular Velocity → ω
Angular Acceleration → a

Angular Displacement
The angle through which a body moves, while moving along a circular path is called its angular displacement.
The angular displacement is measured in degrees, revolutions and most commonly in radian.

s = arc length
r = radius of the circular path
θ = amgular displacement

It is obvious,
s ∞ θ
s = r θ
θ = s / r = arc length / radius

Radian
It is the angle subtended at the centre of a circle by an arc equal in length to its radius.
Therefore,
When s = r
θ = 1 radian = 57.3º

Angular Velocity
When a body is moving along a circular path, then the angle traversed by it in a unit time is called its angular velocity.
Diagram Coming Soon
Suppose a particle P is moving anticlockwise in a circle of radius r, then its angular displacement at P(t1) is θ1 at time t1 and at P(t2) is θ 2 at time t2.
Average angular velocity = change in angular displacement / time interval
Change in angular displacement = θ2 - θ1 = ?θ
Time interval = t2 - t1 = Δt
Therefore,
ω = Δθ / Δt
Angular velocity is usually measured in rad/sec.
Angular velocity is a vector quantity. Its direction can be determined by using right hand rule according to which if the axis of rotation is grasped in right hand with fingers curled in the direction of rotation then the thumb indicates the direction of angular velocity.

Angular Acceleration
It is defined as the rate of change of angular velocity with respect to time.
Thus, if ω1 and ω2 be the initial and final angular velocity of a rotating body, then average angular acceleration "aav" is defined as
aav = (ω2 - ω1) / (t2 - t1) = Δω/ Δt
The units of angular acceleration are degrees/sec2, and radian/sec2.
Instantaneous angular acceleration at any instant for a rotating body is given by
Angular acceleration is a vector quantity. When ω is increasing, α has same direction as ω. When ω is decreasing, a has direction opposite to ω.

Relation Between Linear Velocity And Angular Velocity
Consider a particle P in an object in X-Y plane rotating along a circular path of radius r about an axis through O, perpendicular to the plane of figure as shown here (z-axis).
If the particle P rotates through an angle Δθ in time Δt,
Then according to the definition of angular displacement.
Δθ = Δs / r
Dividing both sides by θt,
Δθ /Δt = (Δs / Δt) (1/r)
=> Δs / Δt = r Δ? / Δt
For a very small interval of time
Δt → 0

Alternate Method
We know that for linear motion
S = v t .............. (1)
And for angular motion
S = r θ ................. (2)
Comparing (1) & (2), we get
V t = r θ
v = r θ/t
V = r ω ........................... {since θ/t = ω}

Relation Between Linear Acceleration And Angular Acceleration
Suppose an object rotating about a fixed axis, changes its angular velocity by Δω in Δt. Then the change in tangential velocity, ΔVt, at the end of this interval is
ΔVt = r Δω
Dividing both sides by Δt, we get
ΔVt / Δt = r Δω / Δt
If the time interval is very small i.e., Δt → 0 then

Alternate Method
Linear acceleration of a body is given by
a = (Vr - Vi) / t
But Vr = r ω r and Vi = r ω i
Therefore,
a = (r ω r - r ω i) / t
=> a = r (ωr- ωi) / t
a = r α .................................... {since (ωr = ωi) / t = ω}

Time Period
When an object is rotating in a circular path, the time taken by it to complete one revolution or cycle is called its time period, (T).
We know that
ω= Δθ / Δt OR Δt = Δθ / ω
For one complete rotation
Δθ = 2 π
Δt = T
Therefore,
T = 2 π / ω
If ω = 2π f ........................ {since f = frequency of revolution}
Therefore,
T = 2π / 2π f
=> T = 1 / f

Tangential Velocity
When a body is moving along a circle or circular path, the velocity of the body along the tangent of the circle is called its tangential velocity.
Vt = r ω
Tangential velocity is not same for every point on the circular path.

Centripetal Acceleration
A body moving along a circular path changes its direction at every instant. Due to this change, the velocity of the body 'V' is changing at every instant. Thus body has an acceleration which is called its centripetal acceleration. It is denoted by a(c) or a1 and always directed towards the centre of the circle. The magnitude of the centripetal acceleration a(c) is given as follows
a(c) = V2 / r, ........................... r = radius of the circular path

Prove That a(c) = V2 / r
Proof
Consider a body moving along a circular path of radius of r with a constant speed V. Suppose the body moves from a point P to a point Q in a small time Δt. Let the velocity of the body at P is V1 and at Q is V2. Let the angular displacement made in this time be ΔO .
Since V1 and V2 are perpendicular to the radial lines at P and Q, therefore, the angle between V1 and V2 is also Δ0, Triangles OPQ and ABC are similar.
Therefore,
|?V| / |V1| = Δs / r
Since the body is moving with constant speed
Therefore,
|V1| = |V2| = V
Therefore,
ΔV / V = Vs / r
ΔV = (V / r) Δs
Dividing both sides by Δt
Therefore,
ΔV / Δt = (V/r) (V/r) (Δs / Δt)
taking limit Δt → 0.

Proof That a(c) = 4π2r / T2
Proof
We know that
a(c) = V2 / r
But V = r ω
Therefore,
a(c) = r2 ω2 / r
a(c) = r ω2 ...................... (1)
But ω = Δθ / Δt
For one complete rotation Δ? = 2π, Δt = T (Time Period)
Therefore,
ω = 2π / T
(1) => a(c) = r (2π / T)2
a(c) = 4 π2 r / T2 .................. Proved

Tangential Acceleration
The acceleration possessed by a body moving along a circular path due to its changing speed during its motion is called tangential acceleration. Its direction is along the tangent of the circular path. It is denoted by a(t). If the speed is uniform (unchanging) the body do not passes tangential acceleration.

Total Or Resultant Acceleration
The resultant of centripetal acceleration a(c) and tangential acceleration a(t) is called total or resultant acceleration denoted by a.

Centripetal Force
If a body is moving along a circular path with a constant speed, a force must be acting upon it. Direction of the force is along the radius towards the centre. This force is called the centripetal force by F(c).
F(c) = m a(c)
F(c) = m v(2) / r ..................... {since a(c) = v2 / r}
F(c) = mr2 ω2 r ....................... {since v = r ω}
F(c) = mrω2