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mechanics, physics, mechanical engineering

Mechanics, physics, mechanical engineering

What makes a body moving?


"From our perception, we can say, a body moves if we apply either a push or a pull on the body. There are several instances, when after applying push or pull, the body still doesn't move. What is the exact reason behind this?"


"Almost to apply a push or pull on a body, a physical contact is needed, with out any physical contact it is not possible to exert push/pull on a body, although there are exceptions too.

(i) We know on every material body existing on earth experience a downward pull towards the center of Earth and to exert this pull, the object and the earth don't need any physical contact. This phenomena is aptly named as the 'Force of Gravity.'

(ii) The second option is Magnet. A magnet can pull as well as push another magnet from a distance with out any physical contact.

(iii) When we place a charged particle near another charged particle, we watch the particles can exert push as well as pull without any physical contact. Like a proton repels another proton, ie they exert push on each other. But a proton attracts an electron by pulling each other to come close.

So there are two types of motion. Type one is the example of a Cricket ball going to boundary after being hit by the bat. So, here impact is the driver of the motion.

But an apple falling from a tree is also in motion, but for this no impact is there. In fact the attraction between the apple and the earth is responsible for the motion. By attraction, it means the tendency of going closer in two objects. Here motion of apple occurs without being hit.


We know that Force is a kind of physical quantity having both magnitude as well as direction. So, force is a vector quantity. We also know that applying triangle's law or parallelogram law of forces addition, we can add two forces to get an equivalent force which is known Resultant force.

So, application of force on an object brings a change in position of the object. This change is position is called displacement.

An object in a coordinate system has a position vector to define the position vector. Any change of position will bring change in its position vector.

Suppose in a coordinate system we have an object at a position vector (r). Let after a time interval of (dt), the object changes its position by an amount (dr). Hence, the rate of change of position will be (dr)/(dt). Rate of change of position means the change of position in unit time. Rate of change of position is called velocity of the object. It is denoted by (v).

Hence, (v) = (dr)/(dt).
The unit of velocity is m/s in SI units and cm/s in cgs system. The most popular unit is km/hr.

A velocity may change, it may change in direction or it may change in magnitude. Suppose, we have an object moving with a velocity (v) at any instant. Suppose after an interval of time (dt), its new velocity becomes (v + dv), where (dv) is the change in
velocity. Hence (dv)/(dt) will be the rate of change of velocity and it indicates the change of velocity per unit time. This physical quantity is called acceleration. Negative acceleration which is rate of decrease in velocity is called retardation or deacceleration too.

Acceleration may be changed; the rate of change of acceleration is called impulse. Like when a bat touches a moving ball, it has an impact and this changes its acceleration due to this magnitude and direction of the ball changes. If a large magnitude of force act on a body for a very short period of time, it is called impulse.

Free falling under the forces of gravity is a case of constant acceleration and that is denoted by (g) and it is equal to g = 9.81 m/s^2.


These three equations are valid only in the case of constant acceleration. Every particle falling under the gravity will satisfy these equations.

We know (dv)/(dt) = a,
hence, (dv) = a(dt) or by integrating both side from initial state t = 0; v = u to a t = t; v = v,
we get,
v - u = a(t - 0) or v = u + at

Again, a = (dv)/(dt)
a = {(dv)/(dx)}.{(dx)/(dt)}
a = v.(dv)/(dx)
hence, v.dv = a.dx
Integrating both sides from initial v= u, x = 0 to final v = v, x = s
we get,
v² - u² = 2 a (s - 0)
v² - u² = 2as

From the defination of velocity, we get
v = (dx)/(dt)
but, v = u + at
u + at = (dx)/(dt)
dx = (u +at). dt
Integrating both sides of the equation from initial condition t = 0, x = 0 and final condition t = t, x = s, we get

(s - 0) = u(t - 0) + (a.t²)/2
s = ut + (at²)/2

Average Velocity = (u + v)/2
= (u + u + at)/2 = u + at/2

Total distance, s = Average Velocity x time

s = (u + at/2) x t
s = ut + (at²)/2

Uses of these equations:

1) Suppose we have a particle travelling with 5 m/s and an acceleration of 1 m/s^2 is applied on the body. What will be the velocity after a time of 10 s?

Ans: Here, u = 5 m/s, a = 1 m/s^2 and t = 10 s, then

velocity after 10 s,
v = u + at
v = 5 + 1 x 10 = 15 m/s

2) A car moving with a velocity 60 km/hr suddenly applies the brake. As a result, the car comes to a halt after running 50 m after applying brake. What will the value of retardation? Find the time it needs to come to rest after the application of brake.

Ans: Here, initial velocity u = 60 km/hr = (60 x 1000)/(60 x 60) m/s = 16.67 m/s
final veloity, v = 0 m/s
Total distance travelled s = 50 m
a = (v² - u²/2s
a = (0 - 16.67²)/(2 x 50)
a = - 2.78 m/s²

again time, t = (v - u)/a
t = (0 - 16.67)/(-2.78)
t = 5.99 s

The Concept of Relative Velocity:

Suppose in a road two car is moving. The faster car at 40 km/hr and the slower car at 30 km/hr in the same direction. Now, if anyone watches the faster car from the slower car, he won't be see it running at 40 km/hr, in stead he will see the velocity at 15 km/hr. When we watch from a moving body or better we say moving reference frame, the velocity of other bodies seem be reduced. Again if we take the same two cars running in opposite directions to each other, the velocity of the each car will be at 55 km/hr as seen from the other. This is due to relative velocity. The relative velocity of a body is the velocity of the body relative to an observer.

Suppose, a car is moving with a velocity Vc and a train is moving with a velocity Vt. Then velocity of car with respect to a person sitting in the train will be Vct = Vc - Vt and velocity of the train to a person in the car will be
Vtc = Vt - Vc.

3) A train is moving towards east with a velocity 120 km/hr and wind is blowing towards west with a velocity 20 km/hr. What will the velocity of the wind to an observer in the train?

Ans: We shall take towards east direction as positive and towards west direction as negative.
Let the velocity of train is (Vt) and velocity of wind is (Vw)

So, Vt = 120 km/hr and
Vw = - 20 km/hr

Vwt = Vw - Vt
Vwt = - 20 - 120 = - 140 m/s and it means wind is flowing from west.

4) A car is moving on a horizontal road at 40 km/hr. Suddenly rain started to pour down at a velocity 30 km/hr. Find the velocity of the rain drops with rest to an observer in the car. Also find the angle with which rain drops would appear to strike the car.

Ans: Let the velocity of the car be Vc = 40 km/hr and rain drop velocity is Vr = - 30 km/hr. The angle between them is θ = 90.
(Vrc) = (Vr) - (Vc)
(Vrc)² = (Vr)² + (Vc)² - 2(Vr)(Vc) cos θ
= 30² + 40² + 0
= 2500
(Vrc) = 50 m/s


Suppose a line AB displaces side wise such that A remains at same point, but the other end B comes to new position C. This is an angular displacement. Angles are measured in radian. 1 radian is the angle formed by an arc equals to magnitude of radius (r). A full circle produces 2π.

Suppose, a line of length (r) makes an angular displacement of (dθ) in time (dt). Then the rate of change of angular displacement is given by (dθ)/(dt) and it is called angular velocity, (w). Hence, (w) = (dθ)/(dt).

If we apply torque or moment in the line, the angular velocity will be changed. Let during the time interval, (dt), the change in angular velocity be (dω). The rate of change of angular velocity will (dω)/(dt) and it is called as angular acceleration and denoted as α.

α = (dω)/(dt)
(dω) = α.(dt)
Taking initial value ω = ωₒ, t = 0, to final value ω = ω, t = t and integrating both side, we get,
ω - ωₒ = α.(t - 0)
ω = ωₒ + αt

angular velocity ω = (dθ)/(dt)
ωₒ + αt = (dθ)/(dt)
(ωₒ + αt)(dt) = dθ
Taking initial value θ = 0, t = 0, to final value, θ = θ, t = t, and integrating both side we get,

ωₒ.(t - 0) + (α/2)(t² - 0) = θ - 0

θ = ωₒt + (αt²/2)

α = (dω)/(dt)
= {(dω)/(dθ)}.{(dθ)/(dt)}
α = ω.(dω)/(dθ)
ω.(dω) = α.(dθ)
taking the initial value ω = ωₒ, θ = 0 and final value ω =ω, θ = θ and integrating both sides,
(ω²)/2 - (ωₒ²)/2 = α.(θ - 0)
(ω²) = (ωₒ²) + 2.α.θ

(5) A wheel is spinning at a RPM of 250. After 10 s the wheel RPM becomes 100. If the angular acceleration remains constant, then find the time needed to come to rest and the angular acceleration.

w0 = initial angular velocity
w0 = (2*pi*N)/60 = (2*pi*250)/60 = 26.18 rad/s

w = final angular velocity
w = (2*pi*100)/60 = 10.47 rad/s

Hence, angular acceleration (alpha) = (w - w0)/t
= (26.18 - 10.47)/10 rad/s2
= 1.57 rad/s2

(6) A

Plane Motion:

A rigid body is said to be in plane motion when all parts of the body move in parallel planes. The plane motion of a rigid body may be classified into several categories like :

1) Translation
2) Rotation
3) General plane motion.


Translation is the types of motion where

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