Why does rotational inertia represent the resistance of a body to angular acceleration

I understand how rotational inertia, $$I=mr^2$$, is defined based on $$\tau_z=mr^2 \alpha_z$$ as an analogy to the linear inertia $$m$$. But I'm still not sure why it represents the resistance of a body to angular acceleration. Also, why doesn't it represent the resistance of a body to tangential and radial acceleration?

This is valid when the mass is concentrated in a point that rotates around a pivot. Suppose for example a pendulum of lenght $$L$$ and mass $$m$$.

The forces on the mass are the weight $$P$$ and the tension $$T$$ in the (massless) rod.

The $$2^{nd}$$ Newton's law: $$\mathbf F = m \mathbf a$$ relates 2 vectors, and we can make a dot product of both sides with the same vector, without change the equation. We choose the infinitesimal displacement in the only possible direction for movement, that is perpendicular to the rod.

$$\mathbf{F.}\delta \mathbf r = m \mathbf {a.} \delta \mathbf r$$.

But $$\mathbf F = \mathbf T + \mathbf P$$, and $$\mathbf {T.}\delta \mathbf r = 0$$ because that vectors are perpendicular. So: $$\mathbf{F.}\delta \mathbf r = {\mathbf P.}\delta \mathbf r = Psin{\theta}L\delta \theta = Psin{\theta}\omega L\delta t$$ (1)

, where $$\omega = \frac{d\theta}{dt}$$.

At the right side: $$m \mathbf {a.} \delta \mathbf r = m\frac{\mathbf {dv}}{dt}\mathbf .\frac{\mathbf {dr}}{dt}\delta t = m\frac{\mathbf {dv}}{dt}\mathbf .\mathbf v\delta t = \frac{1}{2}m\frac{d(v^2)}{dt} \delta t$$ (2)

By definition, $$v = \omega L \implies \frac{d(v^2)}{dt} = L^2\frac{d(\omega^2)}{dt} = 2L^2 \omega \frac{d\omega}{dt}$$

Equating (1) and (2), the $$\omega$$'s cancel: $$Psin{\theta}L = mL^2 \frac{d\omega}{dt}$$

$$Psin{\theta}$$ is the force in the tangential direction, so $$Psin{\theta}L$$ is the torque $$\tau$$

We get the expression: $$\tau = mL^2 \frac{d\omega}{dt}$$