Q.Obtain an expression relating the torque with angular acceleration for a rigid body.

Each particle's tangential force $f_i=m_ir_i\alpha$ contributes torque $\tau_i=m_ir_i^2\alpha$; summing over the rigid body with common $\alpha$ gives $\tau=\le …

Consider the same rigid body of N particles at perpendicular distances $r_1,\ldots,r_N$ from a fixed axis, but now rotating with a common ANGULAR ACCELERATION $\alpha$. Every particle then has its own tangential (linear) acceleration $a_i=r_i\alpha$, requiring a tangential force $f_i=m_ia_i=m_ir_i\alpha$ to produce it. Since this force is tangential (perpendicular to $\vec{r_i}$), its own perpendicular distance from the axis is exactly $r_i$, so the torque this force contributes is $\tau_i=f_ir_i=(m_ir_i\alpha)(r_i)=m_ir_i^2\alpha$ If the rotation is confined to a single plane (a fixed axis), every individual particle's torque contribution again points along the same direction -- the axis -- so, exactly as with angular momentum, the magnitudes add directly: $$\tau=\sum_{i=1}^{N}\tau_i=\ …

Sum the tangential-force-induced torque contributions $\tau_i=m_ir_i^2\alpha$ of every particle of the rigid body (all sharing a common $\alpha$), factoring out $\alpha$ to re …

Writing $\tau=I\alpha$ without deriving it, or confusing the tangential force $f_i=m_ir_i\alpha$ (used to build torque) with the (different) centripetal force -- the tangential force is what changes SPEED, the cent …