Rotational Dynamics: Class 12 NEB Physics Note

Equation of Angular Velocity

	Linear	Angular
Displacement	s	$$ \theta $$
Velocity	$$ v=\frac{d s}{d t} $$	$$ \omega=\frac{d \theta}{d t} $$
Acceleration	$$ a=\frac{d v}{d t}=\frac{d^2 s}{d t^2} $$	$$ \alpha=\frac{d \omega}{d t}=\frac{d^2 \theta}{d t^2} $$

 

Relationship between v and ω

$$v=\frac{ds}{dt}=\frac{d(\theta r)}{dt}=r\frac{d \theta}{dt}$$

$$\therefore v=\omega r $$

Relationship between a and α

It is the branch of mechanics in which we study about the causes of rotational motion.

$$a=\frac{dv}{dt}=\frac{d(wr)}{dt}=r\frac{dw}{dt}=rα$$

$$\therefore a=αr$$

Rigid body

The body which cannot be deformed even by applying large force is called rigid body. A rigid body can undergo rotational and translational motion without changing shape and size.

Rotational Kinetic Energy of a rigid rotating body

Consider a body rotating about an axis YY' with a constant angular velocity '\(\omega \)' as shown in Fig. Each and every particle of the body rotates along the circular path with the same angular velocity '\(\omega \)'. The linear velocity of the particle along its circular path is different since they are situated at different perpendicular distances from the axis of rotation.

Suppose the body consists of 'n' number of particles having masses \(m_1,m_2,m_3....m_n\) which are situated at perpendicular distances \(r_1,r_2,r_3....r_n\) from the axis of rotation respectively.

    Let, \(v_1\) be the linear velocity of the particle of mass \(m1\) along its circular path. Now, rotational KE is given by, $$E_1=\frac{1}{2} m_1v_1^2=\frac{1}{2}m_1r_1^2w^2 (\because v_1 =r_1w)$$ Similarly, KE of particle of mass \(m_2\) is given by, $$E_2= \frac{1}{2}m_2r_2^2w^2$$ Likewise, \(E_3= \frac{1}{2}m_3r_3^2w^2\) and \(E_n= \frac{1}{2}m_nr_n^2w^2\) \begin{align} E &=E_1+E_2+E_3+...+E_n\\ &=\frac{1}{2}m_1r_1^2w^2+\frac{1}{2}m_2r_2^2w^2+\frac{1}{2}m_3r_3^2w^2+...+\frac{1}{2}m_nr_n^2w^2\\ &=\frac{1}{2}w^2\left(m_1r_1^2+m_2r_2^2+m_3r_3^2+...+m_nr_n^2\\\right)\\ &=\frac{1}{2}\left(\sum_{i=1}^nm_ir_i^2\right)w^2....(i) \end{align} But, \(\sum_{i=1}^nm_ir_i^2\)=I....(ii) known as moment of inertia of the body about its axis of rotation. It is defined as the sum of the products of masses of various particles of the body and square of their perpendicular distance from their axis of rotation. It plays same role in rotational motion as the mass does in linear motion. From (i) and (ii) $$E=\frac{1}{2}Iw^2$$ This is the required rotational KE of rigid rotating body. This relation is analogous to \(KE=\frac{1}{2}mv^2\)