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Topic 11 of 19

Rotational Dynamics & Torque

Rotational dynamics extends Newton's Second Law to rotating bodies. Force → torque, mass → moment of inertia, acceleration → angular acceleration. The analog $\tau = I\alpha$ governs how quickly a torque changes the rotation of an object, and the distribution of mass around the rotation axis — captured by $I$ — plays a crucial role.

Key Concepts

Torque

The rotational effect of a force:

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

, with magnitude

\tau = rF\sin\phi = r_\perp F

, where

r_\perp

is the moment arm (perpendicular distance from the rotation axis to the line of action of the force).

Moment of Inertia

The rotational analog of mass:

I = \sum m_i r_i^2 = \int r^2\,dm

. It depends on both the total mass and how it is distributed relative to the axis. Units: kg·m².

Newton's Second Law for Rotation

\sum\tau = I\alpha

. The net torque about an axis equals the moment of inertia about that axis times the angular acceleration.

Common Moments of Inertia

Solid sphere:

\frac{2}{5}mR^2

; hollow sphere:

\frac{2}{3}mR^2

; solid cylinder/disk:

\frac{1}{2}mR^2

; thin rod about center:

\frac{1}{12}mL^2

; thin rod about end:

\frac{1}{3}mL^2

Parallel Axis Theorem

Moment of inertia about any axis parallel to one through the CM:

I = I_\text{cm} + Md^2

, where

d

is the perpendicular distance between the axes.

Rotational Kinetic Energy

K_\text{rot} = \frac{1}{2}I\omega^2

. For a rolling object (no slip), total KE is

K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2

with

v = R\omega

Key Equations

Torque

\tau = r F\sin\phi = r_{\perp} F = r F_{\perp}

φ is the angle between r and F. The moment arm r⊥ is the perpendicular distance from the pivot to the line of action.

Newton's 2nd law for rotation

\sum \tau = I\alpha

Net torque about an axis = moment of inertia about that axis × angular acceleration.

Rotational kinetic energy

K_\text{rot} = \tfrac{1}{2}I\omega^2

Analogous to translational KE = ½mv².

Parallel axis theorem

I = I_\text{cm} + Md^2

d is the distance between the new axis and the CM axis.

Rolling without slipping

v_\text{cm} = R\omega, \quad a_\text{cm} = R\alpha, \quad K = \tfrac{1}{2}mv^2 + \tfrac{1}{2}I\omega^2

Constraint relating linear and angular quantities for rolling without slip.

Worked Example

Disk Rolling Down a Ramp

Problem

A solid disk (mass $m$ , radius $R$ ) rolls without slipping down a frictionless ramp of height $h$ . Find its speed at the bottom.

Solution

Use energy conservation. Initial: $K = 0$ , $U = mgh$ . Final: $K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$ , $U = 0$ .

For a solid disk $I = \frac{1}{2}mR^2$ and rolling constraint $\omega = v/R$ :

mgh = \tfrac{1}{2}mv^2 + \tfrac{1}{2}\left(\tfrac{1}{2}mR^2\right)\left(\frac{v}{R}\right)^2 = \tfrac{1}{2}mv^2 + \tfrac{1}{4}mv^2 = \tfrac{3}{4}mv^2

v = \sqrt{\frac{4gh}{3}}

Answer

v = \sqrt{4gh/3}

— slower than a sliding block (

\sqrt{2gh}

) because energy goes into rotation.

Practice

Exercises

7 problems

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2 free · 5 Pro

Exercise 1 / 7 Free

+10 XP

Drag the yellow handle to position the force on the wrench. The torque bar shows τ = F × r. What is the maximum torque when the force is at the farthest point (r = 4 m)?

τ_max = N·m

Exercise 2 / 7 Free

+10 XP

Drag the yellow mass (2 kg) on the right side to balance the seesaw. When balanced, τ_net = 0. Find the correct distance r₂ from the pivot.

r₂ = m

3 of 7

A net torque of $12 \text{ N·m}$ acts on an object with $I = 3 \text{ kg·m}^2$ . What is the angular acceleration?

Unlock Exercise 3