# Inertial frame and Non – Inertial frame of Reference

##### Frame of Reference :

A frame of reference is a set of coordinates with respect to which the motion of a body is analysed. Different types of coordinates may be used – Rectangular Cartesian coordinates or Polar coordinates.

A frame of reference may be inertial or non-inertial and this has to be borne in mind when Newton’s 2nd law of motion is applied to a body.

### Inertial frame of reference :

Any frame of reference in which Newton’s Ist law is observed to hold is inertial i.e. a body on which no net force is acting moves with constant velocity with respect to this frame.
All frames of reference moving with constant velocity w.r.t. any inertial frame of reference (IFR) are also inertial.

Any two inertial frames of reference (i.e. any two frames in which Newton’s Ist law holds) move with constant velocity relative to each other.

### Non – Inertial frame of reference :

A frame of reference which is moving with non-zero acceleration w.r.t. any IFR is non-inertial.

Newton’s 2nd law requires appropriate modification before it is applied to an object in a non-IFR.
For analysis of different types of terrestrial motion (e.g. motion of a car), the earth is found to be a suitable inertial frame of reference.

Example : A block of mass m = 10 kg is pulled by a force F = 100 N at an angle θ = 30° with the horizontal along a smooth horizontal surface. What is the acceleration of the block? (g = 10 m/s2) Solution: The F.B.D. of the block shows the force components. Along X-axis, let the acceleration of the block be ‘a’

F cos θ = ma ….(1)

a = F cos θ/m …..(2)

$\displaystyle = \frac{100 cos30}{10} = 5\sqrt{3} m/s^2$

The acceleration of the block is 5√3 m/s2 directed towards right. Since F sinθ < mg & the surface is rigid, the block remains in equilibrium along y-axis.

Exercise : Two balls of masses m1 = 5 kg, m2 = 10 kg are connected by a light inextensible string. A horizontal force F = 15 N is acting on the ball m2. What is the (a) acceleration of the balls (b) tension in the string? Example : A block of mass m1 = 2 kg is kept on a smooth horizontal table. Another block of mass m= 1 kg is connected to m1 by an light inextensible string passing over a smooth pulleys as shown in the figure. If the blocks are released what will be the (a) Acceleration of each block (b) tension in the string ? Solution: From the F.B.D. of m1 N = m1g   ………(i)

T = m1a …………..(ii)

From the F.B.D. of m2 m2g – T = m2a    ……….(iii)

(a) From (ii) and (iii),

m2g = (m1+m2)a

$\displaystyle a = \frac{ m_2 g}{m_1 + m_2 }$  ……….(iv)

= 3.26 m/s2 (b) From (ii) and (iv),

T = m1a

$\displaystyle T = \frac{m_1 m_2 g}{m_1 + m_2 }$

= 6.52 N

Example : A man of mass M is standing on a plank kept in a box. The plank and box as a whole has mass m. A light string passing over a fixed smooth pulley connects the man and box. If the box remains stationary, find the tension in the string and the force exerted by the man on the Plank. Solution: The fixed pulley is taken as frame of reference. The forces on man and box with plank are shown in figure. The forces are as follows:

i. Weight of the man = Mg

ii. The tension in the string = T.

iii. The normal contact force between the man and the plank = N.

iv. The weight of the plank and box = mg

Referring to figure. The equation of motion of the man is given as
T + N – Mg = M a

since M > m and the box remains at rest, the man will had to be at rest T + N – Mg = 0 ….(1)

Similarly referring to figure,

T – N – mg = ma = 0 . . . (2)

Solving (1) and (2) we obtain,

T = (M + m)g/2

and N = (M – m)g/2

Example : A pendulum is hanging from the ceiling of a car having an acceleration a0 with respect to the road. Find the angle made by the string with the vertical.

Solution : For the vertical equilibrium of the bob T cos θ – mg = 0

T cos θ = mg   …(i)

Since the bob is accelerating horizontally,

T sinθ = ma0   …(ii)

On dividing (ii) by (i)

tan θ = a0/g

θ = tan-1 (a0/g)

Exercise : With what acceleration ‘ a ‘ should a box descend so that a block of mass M on it exerts a force Mg/4 on the floor of the box ?

Exercise : The time period of a simple pendulum in a lift , moving upwards with constant velocity , is 4 sec. What will be its time period, if the lift starts moving downwards with constant acceleration of 4 m/sec2 ?