Rotational Equilibrium

A rigid body is in rotational equilibrium when it is not rotating, or is rotating at a constant angular velocity. In most introductory problems this simply means the object is not spinning up or slowing down. The key requirement for this to happen is that the net torque acting on the body about any chosen pivot point must equal zero.

Torque is the rotational equivalent of force. Just as an unbalanced force causes an object to accelerate in a straight line, an unbalanced torque causes an object to angularly accelerate, or start spinning. Rotational equilibrium is the rotational version of Newton's first law: no net twisting effect means no change in rotational motion.

The condition for rotational equilibrium is written as:

\( \sum \tau = 0 \)

This means that when you add up every torque acting on the object, taking direction into account, the total must be zero. Each individual torque is calculated with:

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

where \(r\) is the distance from the pivot to the point where the force is applied (the lever arm), \(F\) is the magnitude of the force, and \(\theta\) is the angle between the force vector and the lever arm. When the force is applied perpendicular to the lever arm, \(\sin\theta = 1\) and the formula simplifies to \(\tau = rF\).

Because torques can rotate an object in opposite directions, you need a sign convention before you can add them. The usual convention is:

• Counterclockwise torques are taken as positive.
• Clockwise torques are taken as negative.

Using this convention consistently is essential. If you assign a torque the wrong sign, the equation \( \sum \tau = 0 \) will give you the wrong answer, even if every magnitude was calculated correctly.

In the diagram above, a child of weight 300 N sits at distance \(d_1 = 1.5\ \)m\(\) on the left side of the pivot. Where should a second child of weight 450 N sit on the right side, at distance \(d_2\), so the seesaw is in rotational equilibrium?

Set the pivot as the point of rotation. The 300 N weight produces a counterclockwise torque, and the 450 N weight produces a clockwise torque. For equilibrium, these must have equal magnitude:

\( \tau_1 = \tau_2 \)

\( (300)(1.5) = (450)(d_2) \)

\( 450 = 450 d_2 \)

\( d_2 = 1.0\ \)m\( \)

The heavier child must sit closer to the pivot to balance the torque from the lighter child sitting farther away.

A uniform beam of weight \(W\) is supported horizontally by two vertical cables. To find the tension in each cable, you choose a pivot at one end of the beam. Only torques matter here, so the weight of the cable at the chosen pivot contributes zero torque (its lever arm is zero), while the beam's weight (acting at its center) and the tension in the far cable both contribute torques. Setting \( \sum \tau = 0 \) about that pivot gives you one equation you can solve directly for the unknown tension, without needing to know the force at the pivot itself. This is the main reason a well-chosen pivot makes rotational equilibrium problems much easier to solve.

Rotational equilibrium only guarantees that an object is not spinning. It says nothing about whether the object is sliding or accelerating in a straight line. A separate condition, translational equilibrium, requires that the net force on the object also equals zero. A rigid body at rest, such as a sign hanging from a bracket or a ladder leaning against a wall, must satisfy both conditions at the same time.

Problems that combine both conditions are usually called static equilibrium problems, and they typically involve writing one equation for the net force in the x-direction, one for the net force in the y-direction, and one for the net torque, then solving the three equations together.

• Forgetting the \(\sin\theta\) term when the force is not applied perpendicular to the lever arm.
• Mixing up clockwise and counterclockwise signs partway through a problem.
• Using the wrong lever arm distance, for example the length along the object instead of the perpendicular distance from the pivot to the line of the force.
• Forgetting that the pivot point can be chosen anywhere. Choosing it at the location of an unknown force can eliminate that force from the torque equation entirely.

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