5.6 - Newton’s Second Law in Rotational Form

From Equilibrium to Acceleration

Following the progression of Newton’s Laws for rotation: the First Law (previous topic) covers no angular acceleration,

\sum \tau = 0 \implies \alpha = 0 \implies \omega = \text{constant}

. The Second Law (this topic) covers what happens when angular acceleration does occur,

\sum \tau \neq 0 \implies \alpha \neq 0 \implies \omega

changes. The key idea: angular velocity changes whenever a net torque acts on a system.

Newton’s Second Law for Rotation

The rate at which angular velocity changes is directly proportional to the net torque and inversely proportional to the rotational inertia, with the angular acceleration pointing in the same direction as the net torque.

\boxed{\sum \tau = I\alpha}

where

\sum \tau

is net torque in N·m,

I

is rotational inertia in kg·m², and

\alpha

is angular acceleration in rad/s². This is the rotational version of Newton’s Second Law:

Linear Motion	Rotational Motion
$\sum F = ma$	$\sum \tau = I\alpha$
Force causes linear acceleration	Torque causes angular acceleration
Mass resists linear acceleration	Rotational inertia resists angular acceleration

On the AP equation sheet, the same law appears solved for the angular acceleration of the system:

\alpha_{\text{sys}} = \frac{\sum \tau}{I_{\text{sys}}} = \frac{\tau_{\text{net}}}{I_{\text{sys}}}

Proportionality Relationships

At constant

I

\alpha \propto \tau

, directly proportional, double the torque and you double the angular acceleration. For example, pushing harder on a door swings it open with greater angular acceleration.

At constant

\tau

\alpha \propto \frac{1}{I}

, inversely proportional, double the rotational inertia and you halve the angular acceleration. For example, the same push on a heavy door versus a light door: the heavy door (larger

I

) accelerates more slowly.

Finding Rotational Inertia from a Graph

Since

\alpha = \frac{\tau_{\text{net}}}{I}

, a graph of angular acceleration versus net torque for one rigid system is a straight line through the origin, and its slope is

\frac{1}{I}

. That gives a lab method for measuring

I

. Wrap a string around a spool of known radius

r

on the rotating object and pull it with a force sensor, so the applied torque is

\tau = rF

. Measure the angular acceleration for several different pulls (with a rotary motion sensor, or by timing how long the object takes to turn a known angle from rest and using

\Delta\theta = \frac{1}{2}\alpha t^2

). Then plot

\alpha

vs.

\tau

, draw a best-fit line, and use its slope.

Example: a student collects this sample data for a rotating platform.

Net torque $\tau$ (N·m)	Angular acceleration $\alpha$ (rad/s²)
0.020	4.8
0.040	10.3
0.060	14.9
0.080	20.2
0.100	24.8

5.6 - Newton’s Second Law in Rotational Form

Use two points on the best-fit line, not data points, spread far apart: (0.010 N·m, 2.5 rad/s²) and (0.110 N·m, 27.5 rad/s²).

\text{slope} = \frac{\Delta\alpha}{\Delta\tau} = \frac{27.5 - 2.5}{0.110 - 0.010} = \frac{25.0\text{ rad/s}^2}{0.100\text{ N}\cdot\text{m}} = 2.50 \times 10^2\ \frac{1}{\text{kg}\cdot\text{m}^2}

I = \frac{1}{\text{slope}} = \frac{1}{2.50 \times 10^2} = 4.00 \times 10^{-3}\text{ kg}\cdot\text{m}^2

If the best-fit line crossed the torque axis to the right of the origin instead of going through it, that would point to a small friction torque at the axle: the platform doesn’t speed up until the applied torque beats friction. The slope would still give

\frac{1}{I}

Plotting

\alpha

vs.

\tau

works well because the relationship is linear. If you instead kept the torque fixed and changed

I

, a graph of

\alpha

vs.

I

would curve (inverse relationship). To get a straight line from that kind of data, plot

\alpha

vs.

\frac{1}{I}

, which has slope

\tau_{\text{net}}

Analyzing Rotating Systems

For systems combining both translation and rotation, you typically need separate analyses. Draw a rigid body diagram showing all forces at their points of application, and identify the axis of rotation. Write separate equations:

\sum F_x = ma_x

and

\sum F_y = ma_y

for linear motion,

\sum \tau = I\alpha

for rotational motion. Then connect them with

a_t = r\alpha

, linking the translational acceleration at the edge to the angular acceleration.

Connecting Linear and Angular Acceleration

a = r\alpha \qquad \text{or} \qquad \alpha = \frac{a}{r}

where

a

is tangential (linear) acceleration in m/s²,

r

is radius in meters, and

\alpha

is angular acceleration in rad/s². Use this whenever rotational and linear motion are coupled, pulleys with hanging masses, rolling wheels, and similar setups.

Example: Pulley System Analysis

For a pulley with masses hanging on either side, the forces at the axis (weight, support) contribute zero torque since

r = 0

there, while tension forces at the rim create the torques that cause rotation. Tensions on opposite sides create opposite torques, one rotating counterclockwise (positive), the other clockwise (negative), and the net torque is the difference between them.

For the hanging blocks, build separate free body diagrams, apply

\sum F = ma

, and link the tensions between the pulley and block equations.

Worked Example: Modified Atwood Machine with Pulley Mass

A disk-shaped pulley (mass 1.2 kg, radius 0.15 m) has a person pulling downward on the left rope with a steady 28 N force, while a 0.7 kg block hangs on the right side. The pulley rotates frictionlessly. Use

g = 10

m/s². Find the tension

T_2

on the right side.

Left side (the person’s pull): the rope is treated as massless, so the net force on its end is zero even while the system accelerates:

\sum F = 0 \implies F_{\text{pull}} - T_1 = 0 \implies T_1 = F_{\text{pull}} = 28

Right side (the block): forces are tension

T_2

(up) and weight

mg

(down). With up as positive:

\sum F = ma \implies T_2 - mg = ma \implies T_2 = m(g + a)

We need

a

before we can find

T_2

The pulley (rotation): for a disk,

I = \frac{1}{2}Mr^2

. Torques about the center:

T_1

creates counterclockwise torque (

+T_1 r

T_2

creates clockwise torque (

-T_2 r

\sum \tau = I\alpha \implies T_1 r - T_2 r = I\alpha

Linking to linear acceleration with

\alpha = \frac{a}{r}

T_1 r - T_2 r = I \cdot \frac{a}{r} = \frac{Ia}{r}

Substituting

I = \frac{1}{2}Mr^2

T_1 r - T_2 r = \frac{\frac{1}{2}Mr^2 \cdot a}{r} = \frac{1}{2}Mra

Dividing by

r

T_1 - T_2 = \frac{1}{2}Ma

Solving the system: we have

T_1 - T_2 = \frac{1}{2}Ma

and

T_2 = m(g+a)

. Substituting

T_1 = 28

28 - T_2 = \frac{1}{2}(1.2)a = 0.6a \implies a = \frac{28 - T_2}{0.6}

Substituting into

T_2 = m(g+a)

T_2 = 0.7\left(10 + \frac{28 - T_2}{0.6}\right) = 7 + \frac{0.7(28 - T_2)}{0.6} = 7 + \frac{19.6 - 0.7T_2}{0.6}

Multiplying through by 0.6:

0.6T_2 = 4.2 + 19.6 - 0.7T_2 \implies 1.3T_2 = 23.8 \implies T_2 \approx 18.3 \text{ N}

The tension in the right rope is approximately 18 N. As a sanity check, the block’s weight is

mg = (0.7)(10) = 7

N, and

T_2 = 18

N exceeds that, which makes sense since the block is accelerating upward, being pulled up by the rotating pulley.

Problem-Solving Strategy Summary

For combined rotation-translation problems: draw separate diagrams (a rigid body diagram for the rotating object, free body diagrams for translating objects), write

\sum F = ma

for translating objects and

\sum \tau = I\alpha

for rotating ones, connect them with

a = r\alpha

, and solve the resulting system, often through substitution and strategic elimination of variables.

Worked Example: Linear and Rotational Analyses Done Separately

Sometimes the linear and rotational parts of a problem aren’t tied together by

a = r\alpha

, so each analysis has to be done on its own. A uniform 2.0 kg disk of radius 0.25 m lies flat on frictionless ice, with

I = \frac{1}{2}MR^2

about its center. A string wrapped around its rim is pulled horizontally with a constant 6.0 N force. Find the acceleration of the disk’s center of mass and its angular acceleration.

Linear analysis (the disk as an object): the only horizontal force is the 6.0 N pull. Gravity and the normal force cancel vertically, and the ice exerts no friction. Newton’s second law for the center of mass doesn’t care where on the disk the force acts:

a_{\text{cm}} = \frac{F_{\text{net}}}{M} = \frac{6.0\text{ N}}{2.0\text{ kg}} = 3.0\text{ m/s}^2

Rotational analysis (the disk as a rigid system): the string pulls along a line tangent to the rim, so the lever arm about the center is the full radius

R

. Keep the extra digits in

I

for the next step:

I = \tfrac{1}{2}MR^2 = \tfrac{1}{2}(2.0\text{ kg})(0.25\text{ m})^2 = 0.0625\text{ kg}\cdot\text{m}^2

\alpha = \frac{\tau_{\text{net}}}{I} = \frac{(6.0\text{ N})(0.25\text{ m})}{0.0625\text{ kg}\cdot\text{m}^2} = 24\text{ rad/s}^2

Compare: pull with the same 6.0 N through the center (say, with the string tied to a pin at the center) and the center of mass still accelerates at 3.0 m/s², but the lever arm is zero, so the disk doesn’t spin. Moving the force to the rim adds rotation without taking anything away from the linear motion. Also notice that

a_{\text{cm}} = 3.0

m/s² is not equal to

R\alpha = (0.25)(24) = 6.0

m/s². Nothing ties the spinning to the sliding here (the disk isn’t rolling on anything), which is why the two analyses have to be done separately.

Real-World Applications

Yo-yos: the string unwinds from a small-radius axle, creating a torque that drives both rotation and translation. A larger axle radius is easier to accelerate, and the mass distribution (

I

) affects how long the yo-yo “sleeps” at the bottom.

Wheels and axles: applying force at a large radius (the wheel) produces a large torque on a small radius (the axle), the principle behind steering wheels, winches, and gears.

Flywheels: a large

I

resists angular acceleration, which smooths out engine rotation and stores rotational kinetic energy.

Rolling motion: an object rotating and translating at once, with the no-slip condition

v = r\omega

and

a = r\alpha

tying the two together, and both rotational and translational kinetic energy need to be considered.

Common Mistakes

Forgetting to use the correct $I$ : don’t use

I = mr^2

for a disk, use the shape-specific formula,

I = \frac{1}{2}mr^2

Wrong sign on torques: don’t give both tensions positive torque, opposite sides of a pulley create opposite torques, one positive, one negative.

Confusing $a$ and $\alpha$ : don’t set

a = \alpha

directly, it’s

a = r\alpha

, always including the radius.

Including axis forces in torque: don’t include weight or support forces acting right at the axis in the torque equation, they contribute zero torque since

r = 0

Not connecting the systems: don’t treat the pulley and hanging masses as fully independent, the tensions link them together, and their accelerations relate through

a = r\alpha

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