Physics Demonstrations

Torque

Mass (kg)
Force (N)
Force Angle (degrees)
Beam Length (m)

In this demonstration, we explore torque acting on a uniform beam of mass $m$ and length $L$ , pivoted at one end. Two torques act about the pivot: the torque due to the beam's weight (acting at its centre of mass, $L / 2$ ), and a user-applied force $F$ at angle $φ$ relative to the beam at its free end. Torque is defined as the cross product of the position vector and the force: $τ = r \cdot F \cdot \sin (φ)$ The two individual torques about the pivot are: $τ_{weight} = - m g \frac{L}{2}$ $τ_{F} = F \cdot L \cdot \sin (φ)$ The net torque produces an angular acceleration via Newton's second law for rotation, where $I$ is the moment of inertia of the beam about its pivot end: $τ_{net} = I \cdot α I = \frac{1}{3} m L^{2} α = \frac{τ_{net}}{I}$ The beam rotates counter-clockwise when $τ_{net} > 0$ and clockwise when $τ_{net} < 0$ . The title displays all three torque values. The beam stops when it reaches ±90°. Some questions to consider while viewing the demonstration: At what force angle $φ$ is the applied torque maximised for a given force magnitude? Find a combination of inputs where the beam is in rotational equilibrium ( $τ_{net} = 0$ ). How does doubling the beam length affect the moment of inertia and the angular acceleration? Identify a real-world example where controlling torque about a pivot is important.