Question

Prove the relation τ = Iα for a rigid body rotating about a fixed axis?​

Answer 1

Explanation:

If a point rotate around a fixed axis with a distance r and the force applied on the point is F, then

 $\vec{\tau} = \vec{r} \times \vec{F}$

we should keep in mind , a body is combination of a number of points.

Let, the body is a combination of $m_1, m_2, m_3,............................,m_n$  points.

the distances from fixed axis are $r_1, r_2, r_3, ................................., r_n$ respectively.

Now,  $F_1 = m_1 \times linear\ acceleration$

               $=m_1 \times angular\ acceleration \times r_1$

               $=m_1 \times \frac{d\omega}{dt} \times r_1$

Next, $\tau = \sum\limits_{x=1}^n r_x \times F_x = \sum\limits_{x=1}^n r_x \times m_x \times \frac{d\omega}{dt} \times r_x= \sum\limits_{x=1}^n m_x\times(r_x)^2 \times \frac{d\omega}{dt}$

here, $m_x \times r_x^2 =moment \ of \ inertia= I_x$

So, $\tau = \sum\limits_{x=1}^n I_x \times \frac{d\omega}{dt} = \frac{d\omega}{dt}(I_1+I_2+I_3+ ..............................I_n) = \frac{d\omega}{dt}\times I = \alpha \times I = I\alpha$

The relation τ = Iα for a rigid body rotating around a fixed axis is proved.