Newton's Second Law for Rotation

From linear to rotational

• In linear motion, the force required to give an object a certain acceleration depends on its mass: $$F = ma$$
• In rotational motion, the torque required to give a rotating object a certain angular acceleration depends on its moment of inertia: $$\tau = I\alpha$$
• This is Newton's Second Law of rotational motion, where: $\tau$ = torque (N m), $I$ = moment of inertia (kg m$^2$), $\alpha$ = angular acceleration (rad s$^{-2}$).

Deriving $\tau = I\alpha$

• This equation comes from the fact that torque is the rotational equivalent of force. Starting from: $$\tau = Fr \quad \text{and} \quad F = ma$$
• Combining gives: $\tau = r(ma)$
• Since $\alpha = \frac{a}{r}$, we can write $a = r\alpha$, and since $I = mr^2$ for a point mass: $$\tau = r(mr\alpha) = (mr^2)\alpha = I\alpha$$

Comparison of linear and rotational variables

• Force $F$ corresponds to Torque $\tau$
• Mass $m$ corresponds to Moment of inertia $I$
• Acceleration $a$ corresponds to Angular acceleration $\alpha$
• $F = ma$ corresponds to $\tau = I\alpha$

Worked example: pulley and falling block

• A block of mass $m$ hangs from a string wrapped around a cylindrical pulley of mass $M$ and radius $R$, with moment of inertia $I = \tfrac{1}{2}MR^2$.
• For the block (linear motion): $mg - T = ma$ ... (1)
• For the pulley (rotational motion): $TR = I\alpha$, and since $\alpha = \frac{a}{R}$: $$TR = \left(\tfrac{1}{2}MR^2\right)\frac{a}{R}$$ $$T = \tfrac{1}{2}Ma \quad \text{... (2)}$$
• Substituting (2) into (1) and rearranging: $$mg - \tfrac{1}{2}Ma = ma$$ $$mg = a\left(m + \frac{M}{2}\right)$$ $$a = \frac{mg}{m + \frac{M}{2}}$$
• The key part is that the pulley's moment of inertia effectively adds $\frac{M}{2}$ to the system's inertia, slowing the block's acceleration compared to a massless pulley.