Rigid body : A rigid body is a body with a definite and unchanged shape and size i.e. a body is said to be rigid if the distance between any two particles of the body remains invariant. E.g. Fan , Pen , Table , Stone etc.

Our body is not a rigid body , Two blocks with a spring attached between them is also not a rigid body .

For every pair of particles in a rigid body there is no velocity of separation or approach between the particles . Suppose velocities of A and B with respect to ground are V_{A} and V_{B} respectively .

If body is rigid then,

**V _{A} cosθ_{1} = V_{B} cosθ_{2}**

Motion of a rigid body

Translation: A rigid body is said to undergo translation if it moves such that it always remains parallel to itself: this means that a line connecting any two particles of the rigid body always remains parallel to itself throughout its motion.

Rotation: A rigid body is said to undergo rotation if there exists a straight line from which the distance of any particle of the rigid body remains constant throughout its motion. This straight line, whether fixed or moving is known as the axis of rotation. The rigid body is said to undergo rotation about this axis.

__Angular Displacement:__

Consider a rigid body undergoing rotation about an axis, perpendicular to the plane of the paper and passing through O.

Suppose that A and B are any two particles of the rigid body at the position 1 while A’ and B’ are their subsequent locations when the body is at the position 2.

Since the body undergoes rotation, OA = OA’ & OB = OB’

Further AB = A’B’, since the body is rigid.

ΔOAB ≅ ΔOA’B'(congruent)

i.e. ∠AOB = ∠A’OB’

Adding ∠AOB’ to both sides of the above equation, we get,

∠BOB’ = ∠AOA’ = θ (say)

This implies that in a given interval of time the angular displacements of all particles of the rigid body undergoing rotation are identical.

Therefore, a single variable, viz. angular displacement (θ) can be used to describe the rotational motion of the rigid body.

Angular displacement is not a vector quantity. However, for infinitesimal time intervals, the corresponding angular displacement is infinitesimal and behaves like a vector.

Average angular velocity is defined for a fixed axis of rotation by:

$ \displaystyle \omega_{avg} = \frac{\theta_2 – \theta_1}{t_2 – t_1}$

Instantaneous angular velocity is defined by , $ \displaystyle \omega = \frac{d \theta }{dt}$

The direction of ω (instantaneous) is along the axis n^{^}

Average angular accelerations are defined by:

$ \displaystyle \vec{\alpha_{avg}} = \frac{\vec{\omega_2}-\vec{\omega_1}}{t_2 – t_1}$

Instantaneous angular accelerations are defined by :

$ \displaystyle \vec{\alpha} = \frac{\vec{d \omega}}{dt}$

Angular velocity is same for all the particles of a rigid body and the same is true about angular acceleration also as and the reason being equal angular displacement.

__Equations Of Rotational Kinematics__

If the angular acceleration is constant, the following relations hold:

$ \displaystyle (i)\quad \omega = \omega_0 + \alpha t $

$ \displaystyle (ii) \quad \theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2 $

$ \displaystyle (iii) \quad \omega^2 = {\omega_0}^2 + 2\alpha \theta $

Here ω_{o} = magnitude of the initial angular velocity

ω = magnitude of the angular velocity after time t.

θ_{o} = Initial angular position.

θ = Angular position after time t.

Illustration : A fan starts rotating with constant angular acceleration of π radians/s^{2} about a fixed axis perpendicular to its plane and through its centre. Find

(a) the angular velocity of the fan after 4 sec.

(b) the angular displacement of the fan after 4 sec and

(c) number of turns accomplished by the fan in 4 sec.

Solution : Here α = π rad/sec^{2}

ω_{0 }= 0

t = 4 sec

(a) ω(t) = ω_{o} + αt

ω_{(4 sec)} = 0 + ( π ) x 4

= 4π rad/sec.

(b) θ(t) = ω_{o}t + (1/2) αt^{2}

θ_{(4 sec)} = 0 + (1/2) ( π rad/sec^{2} ) x (16 sec^{2} )

= 8 π radian.

(c) Let the number of turns be n

⇒ n (2 π rad ) = 8 π rad

⇒ n = 4

Exercise : A wheel rotates with an angular acceleration given by α = 4at^{3} – 3bt^{2} , where t is the time and a and b are constants. If the wheel has initial angular speed ω_{0} ,

write the equations for the:

(i) angular speed

(ii) angle displacement.

__Relation b/w linear & angular variables__

Consider a particle A of a rigid body undergoing rotation about a fixed axis- , the particle A describing an arc ABA’ of a circle with its centre O on the axis of rotation. Taking the origin at O, the position vector of A ,

$ \displaystyle \vec{r} = \vec{OA} $

OA = OA’ = constant (radius of the circle)

∠A’OA = θ(t) (say)

The arc length, ABA’ , S = r θ

The tangential velocity,

$ \displaystyle v_A = \frac{ds}{dt} $

$ \displaystyle v_A = r \frac{d\theta}{dt} $

= r ω

The direction of the angular velocity vector ω^{→} be taken along the axis of rotation:

$ \displaystyle \vec{\omega} = \omega \hat{n} $

$ \displaystyle \hat{n} $ being the unit vector along the axis of rotation.

Then, v^{→} , instantaneous velocity of A with respect to the axis of rotation, can be written as

$ \displaystyle \vec{v} = \vec{\omega} \times \vec{r}$

The acceleration of the point A with respect to the axis of rotation is

$ \displaystyle \vec{a} = \frac{\vec{dv}}{dt}$

$ \displaystyle \vec{a} = \frac{\vec{d\omega}}{dt}\times \vec{r} + \vec{\omega} \times \frac{\vec{dr}}{dt}$

$ \displaystyle \vec{a} = \vec{\alpha}\times \vec{r} + \vec{\omega} \times \vec{v}$

$\large \vec{a} = \vec{a_t} + \vec{a_c} $

If ω^{→} is constant, then α ^{→} = 0 and,

$ \displaystyle \vec{a} = \vec{\omega} \times \vec{v}$

$ \displaystyle \vec{a} = \vec{\omega} \times (\vec{\omega} \times \vec{r})$

$ \displaystyle \vec{a} = -\omega^2 \vec{r} $

Illustration : A disc rotates with constant angular acceleration of 2 rad/s^{2} about a fixed vertical axis through its centre and perpendicular to its plane. A coin is placed on it at a distance of 1m from the axis of rotation. The coin is always at rest relative to the disc. If at t = 0 the disc was at rest, then find the total acceleration of the coin after one second.

Solution: After 1 second angular velocity of the disc and hence that of the coin about the axis of rotation is

ω = 0 + 2(rad/s^{2}) × 1 sec = 2 rad/s

a_{t} = α r = (2 rad/s^{2}) × 1 m = 2m/s^{2}

a_{r} = ω^{2} r = (2rad/s)^{2} × 1m = 4 m/s^{2}

$ \displaystyle a = \sqrt{a_t^2 + a_r^2} $

$ \displaystyle a = \sqrt{2^2 + 4^2} = 2\sqrt5 m/s^2$