KAHIBARO
Table of Contents
From Linear Motion to Rotational Motion
In linear motion, Newton's second law connects net force and acceleration through $F = ma$. In rotational motion, the matching idea is that net torque produces angular acceleration. The rotational form of Newton's second law is
$$
\sum \tau = I\alpha
$$
where $\sum \tau$ is the net torque about a chosen axis, $I$ is the moment of inertia about that axis, and $\alpha$ is the angular acceleration.
This equation tells us that rotation changes when a torque acts, just as translation changes when a force acts. A larger torque produces a larger angular acceleration, but a larger moment of inertia makes the object harder to spin up.
The rotational version of Newton's second law is
$$
\sum \tau = I\alpha
$$
This is valid for rotation about a fixed axis for a rigid body, when $I$ is taken about that same axis.
Meaning of Each Quantity
Torque measures how strongly a force tends to rotate an object. Moment of inertia measures how difficult it is to change the rotational motion of that object. Angular acceleration tells us how quickly the angular velocity changes.
There is a strong analogy with linear motion:
| Linear motion | Rotational motion |
|---|---|
| Force $F$ | Torque $\tau$ |
| Mass $m$ | Moment of inertia $I$ |
| Acceleration $a$ | Angular acceleration $\alpha$ |
| $F = ma$ | $\sum \tau = I\alpha$ |
Mass resists linear acceleration. Moment of inertia resists angular acceleration. But unlike mass, moment of inertia depends not only on how much matter there is, but also on how that matter is distributed relative to the axis of rotation.
How the Law Arises
For a single particle at distance $r$ from an axis, the tangential force is related to tangential acceleration by
$$
F_t = ma_t
$$
Since torque is $\tau = rF_t$ and tangential acceleration is $a_t = r\alpha$, we get
$$
\tau = r(ma_t) = r(mr\alpha) = mr^2\alpha
$$
For many particles making up a rigid body, we add all contributions:
$$
\sum \tau = \left(\sum mr^2\right)\alpha
$$
The quantity in parentheses is the moment of inertia:
$$
I = \sum mr^2
$$
So the rotational equation becomes
$$
\sum \tau = I\alpha
$$
This shows clearly why mass farther from the axis has a larger effect on rotational inertia.
Choosing the Axis
The axis matters. Both torque and moment of inertia must be calculated about the same axis. If you change the axis, both quantities may change, and the equation must be written again for the new axis.
For example, a force may produce a large torque about one point and zero torque about another. Likewise, the same object can have different moments of inertia about different axes.
Always calculate $\sum \tau$ and $I$ about the same axis.
Using different axes in the same equation gives a wrong result.
Sign Convention
Because torque and angular acceleration can act in opposite rotational senses, we need a sign convention. In many problems, counterclockwise is taken as positive and clockwise as negative. Then the torque equation is written algebraically:
$$
\sum \tau = I\alpha
$$
A positive net torque gives positive angular acceleration. A negative net torque gives negative angular acceleration.
Example, A Force on a Door
Imagine pushing a door at its edge. The door rotates about its hinges. If a force $F$ is applied perpendicular to the door at distance $r$ from the hinges, then
$$
\tau = rF
$$
and the angular acceleration is
$$
\alpha = \frac{\tau}{I} = \frac{rF}{I}
$$
If you push farther from the hinges, the torque is larger and the door starts turning more easily. If the same door had more mass concentrated far from the hinges, its moment of inertia would be larger and it would accelerate less.
Net Torque and Multiple Forces
If several forces act, we add all torques about the axis:
$$
\sum \tau = \tau_1 + \tau_2 + \tau_3 + \cdots
$$
Then we use
$$
\alpha = \frac{\sum \tau}{I}
$$
Some torques may help the rotation and others may oppose it. The net result determines the angular acceleration.
As in linear mechanics, it is the net effect that matters. A body can have several forces acting on it, but only the total torque determines how its rotational motion changes.
Translational and Rotational Motion Together
An object can translate and rotate at the same time. In that case, linear and rotational versions of Newton's second law may both be needed:
$$
\sum F = ma
$$
and
$$
\sum \tau = I\alpha
$$
These equations describe different aspects of the same motion. The first governs motion of the center of mass, and the second governs rotation about an axis.
For example, when a wheel is driven by a force at its rim, the force may both accelerate the wheel's center and spin the wheel.
Static and Dynamic Cases
If the net torque is zero, then
$$
\sum \tau = 0
$$
and therefore
$$
\alpha = 0
$$
This means the angular velocity does not change. The object may remain at rest, or it may continue rotating at constant angular velocity.
If the net torque is not zero, then angular acceleration is not zero, and the rotation changes.
Zero net torque does not always mean zero angular velocity.
It means zero angular acceleration.
Units
Torque is measured in newton meters, $\mathrm{N \, m}$. Moment of inertia is measured in $\mathrm{kg \, m^2}$. Angular acceleration is measured in $\mathrm{rad/s^2}$.
Checking units in the equation:
$$
I\alpha = (\mathrm{kg \, m^2})(\mathrm{rad/s^2})
$$
Since the radian is dimensionless, this becomes
$$
\mathrm{kg \, m^2/s^2} = \mathrm{N \, m}
$$
which matches the unit of torque.
Solving Problems
A typical rotational dynamics problem follows a simple pattern. First choose the axis of rotation. Then calculate each torque about that axis with the correct sign. Add them to get the net torque. Find or use the appropriate moment of inertia about that axis. Finally apply
$$
\sum \tau = I\alpha
$$
to solve for the unknown.
| Step | Action |
|---|---|
| 1 | Choose the axis |
| 2 | Identify all torques about that axis |
| 3 | Assign signs consistently |
| 4 | Add torques to get $\sum \tau$ |
| 5 | Use the correct moment of inertia $I$ |
| 6 | Apply $\sum \tau = I\alpha$ |
A Simple Worked Example
Suppose a rigid wheel has moment of inertia $I = 2.0\ \mathrm{kg\,m^2}$. A net torque of $6.0\ \mathrm{N\,m}$ acts on it. Its angular acceleration is
$$
\alpha = \frac{\sum \tau}{I} = \frac{6.0}{2.0} = 3.0\ \mathrm{rad/s^2}
$$
If instead the same torque acts on a wheel with $I = 6.0\ \mathrm{kg\,m^2}$, then
$$
\alpha = \frac{6.0}{6.0} = 1.0\ \mathrm{rad/s^2}
$$
So for the same torque, the larger moment of inertia gives a smaller angular acceleration.
Physical Insight
Newton's second law for rotation captures a central idea of rotational dynamics. To change how fast something spins, you need torque. How much the spin changes depends not only on the applied torque but also on how resistant the object is to rotational change. That resistance is the moment of inertia.
A light bicycle wheel and a heavy flywheel can experience the same torque, but the bicycle wheel usually speeds up its rotation much more quickly because its moment of inertia is smaller.
Important physical idea:
A larger torque makes rotation change more quickly.
A larger moment of inertia makes rotation change more slowly.
Final Connection
Rotational dynamics is the rotational counterpart of translational dynamics. The equation
$$
\sum \tau = I\alpha
$$
plays the same role for spinning objects that
$$
\sum F = ma
$$
plays for moving objects. Once torque and moment of inertia are known, the angular acceleration follows directly.