AcademyRotational Dynamics
Academy
Torque
Level 1 - Physics topic page in Rotational Dynamics.
Principle
Torque measures how a force tends to rotate a body about a chosen axis.
Notation
\(\vec\tau\)
torque vector
\(\mathrm{N\,m}\)
\(\vec r\)
position from axis to force point
\(\mathrm{m}\)
\(\vec F\)
applied force
\(\mathrm{N}\)
\(\phi\)
angle between radius and force
\(\mathrm{rad}\)
\(\ell\)
perpendicular lever arm
\(\mathrm{m}\)
\(\tau_z\)
signed torque about the z-axis
\(\mathrm{N\,m}\)
Method
Derivation 1: Build torque from the perpendicular force
A force can rotate a body only through the part perpendicular to the radius from the axis.
Choose the axis
\[\vec r\ \text{starts at the rotation axis}\]
Perpendicular force
\[F_\perp=F\sin\phi\]
Torque magnitude
\[\tau=rF_\perp=rF\sin\phi\]
Lever-arm form
\[\tau=F\ell\]
The beam sketch shows why only the perpendicular part matters. A force line passing through the axis has zero lever arm.
Derivation 2: Use the cross product for sign and direction
The vector definition packages magnitude and direction together. In a flat xy problem, the z-component gives the signed torque.
Vector definition
\[\vec\tau=\vec r\times\vec F\]
Plane component
\[\tau_z=xF_y-yF_x\]
Add torques
\[\sum\tau_z=\tau_{1z}+\tau_{2z}+\cdots\]
Rules
These are the compact results from the method above.
Vector torque
\[\vec\tau=\vec r\times\vec F\]
Torque magnitude
\[\tau=rF\sin\phi\]
Lever arm
\[\tau=F\ell\]
Plane component
\[\tau_z=xF_y-yF_x\]
Examples
Question
A
\[30\,\mathrm{N}\]
force acts perpendicular to a handle \[0.40\,\mathrm{m}\]
from the axis. Find the torque magnitude.Answer
Use the perpendicular lever arm:
\[\tau=rF=0.40(30)=12\,\mathrm{N\,m}\]
Checks
- Torque depends on the chosen axis.
- A force through the axis gives zero torque.
- Units are newton metres, not joules in this context.
- The sign comes from rotation sense, not force direction alone.