Rotational Equilibrium

An object is in rotational equilibrium when it is not speeding up or slowing down its spin around a pivot point. This does not mean the object is standing still; it means the turning effects acting on it are perfectly balanced, so its angular velocity stays constant (often zero). The turning effect of a force is called torque, and rotational equilibrium is defined by one simple condition: the net torque acting on the object is zero.

\( \sum \tau = 0 \)

This is the rotational counterpart to Newton's first law. Just as an object with zero net force keeps a constant velocity, an object with zero net torque keeps a constant angular velocity. If a seesaw, a signpost, or a bridge beam is not rotating, then every torque trying to turn it one way must be cancelled by torques turning it the other way.

Torque depends on three things: the size of the force, how far the force is applied from the pivot, and the angle between the force and the lever arm. The formula is

\( \tau = rF\sin\theta \)

where \( r \) is the distance from the pivot to the point where the force acts, \( F \) is the size of the force, and \( \theta \) is the angle between the force vector and the line \( r \). When the force is applied perpendicular to the lever (\( \theta = 90^\circ \)), \( \sin\theta = 1 \) and the formula simplifies to \( \tau = rF \). The distance \( r \), measured perpendicular to the line of the force, is called the moment arm.

A key idea for rotational equilibrium is direction. Torques that would spin the object counterclockwise are usually treated as positive, and torques that would spin it clockwise are treated as negative. Setting the sum of these signed torques equal to zero is exactly how balance problems are solved.

A uniform 3 metre plank rests on a pivot at its center. A 20 kilogram child sits 1.5 metres from the pivot on one side. How far from the pivot must a 30 kilogram adult sit on the other side to keep the plank in rotational equilibrium? Use \( g = 9.8 \).

The child's torque about the pivot is \( \tau_1 = m_1 g d_1 = (20)(9.8)(1.5) = 294 \). For rotational equilibrium, the adult must produce an equal and opposite torque, so \( m_2 g d_2 = 294 \). Solving for \( d_2 \) gives \( d_2 = \dfrac{294}{(30)(9.8)} = 1.0 \). The adult must sit 1.0 metre from the pivot, on the opposite side from the child, so the two torques cancel and \( \sum \tau = 0 \).

Rotational equilibrium only guarantees that an object is not starting to spin; it says nothing about whether the object is starting to slide or accelerate as a whole. A fully balanced, stationary object must also satisfy translational equilibrium, meaning the net force in every direction is zero. Checking both conditions together is how you confirm an object is truly at rest and staying that way.

Real problems, such as ladders leaning on walls, hanging signs, or loaded beams, usually require both conditions at once: zero net force and zero net torque. Once you are comfortable picking a pivot, finding moment arms, and summing signed torques, you are ready to apply the same reasoning to more involved static equilibrium problems that combine several forces acting at different points and angles.

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