5.5 Rotational Equilibrium and Newton's First Law in Rotational Form

Rotational equilibrium occurs when an object maintains constant angular velocity, even if it's not in translational equilibrium. Torque plays a role analogous to force in linear motion, and understanding this helps us analyze spinning objects like wheels, gears, and pulley systems.

• Objects can be in rotational equilibrium while moving translationally
• Net torque of zero is required for constant angular velocity
• Newton's laws have rotational equivalents that govern angular motion

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Conditions for Constant Angular Velocity

Rotational vs Translational Equilibrium

A system may exhibit rotational equilibrium (constant angular velocity) without being in translational equilibrium, and vice versa. Rotational and translational motions can be truly independent of each other.

• A spinning top maintains constant angular velocity while its center of mass moves 🌀
• An object can have zero net force but nonzero net torque, or zero net torque but nonzero net force, depending on how forces are applied
• An object can be in rotational equilibrium without being in translational equilibrium. For example, a wheel can rotate with constant angular velocity while being pulled so that its center of mass accelerates forward, as long as the net torque about its center is zero.

Free-body and force diagrams help us visualize the forces and torques acting on a system. These diagrams are essential tools for analyzing both types of equilibrium.

• Free-body diagrams show all external forces
• Force diagrams focus on a specific point or axis of rotation
• Arrows should indicate both magnitude and direction of forces

Zero Net Torque

Rotational equilibrium is a configuration of torques such that the net torque exerted on the system is zero. When all torques on an object balance perfectly, the object maintains its rotational state. This is similar to how balanced forces prevent acceleration in linear motion.

$\sum \tau_{i}=0$

• Balanced clockwise and counterclockwise torques result in rotational equilibrium ⚖️
• Net torque of zero means angular velocity remains constant
• Objects in rotational equilibrium don't speed up or slow down their rotation

Calculating net torque requires vector addition of all individual torques about the axis of rotation.

• Torque $\tau = r \times F$ where $r$ is position vector and $F$ is force vector
• Counterclockwise torques are typically considered positive
• Clockwise torques are typically considered negative
• The perpendicular distance from the rotation axis to the force line is the lever arm

Rotational Analog of Newton's Laws

Newton's laws can be rewritten to describe rotational motion with remarkable symmetry to their linear forms.

• First law: An object maintains constant angular velocity unless acted upon by a net torque. This is the rotational analog of Newton's first law—a system will have a constant angular velocity only if the net torque exerted on the system is zero.
• If the net torque on a rigid system is not zero, the angular velocity must change. In other words, unbalanced torque causes angular acceleration, so the object speeds up, slows down, or changes its rotational motion. This means $\sum \tau = I \alpha$, where a nonzero net torque produces a nonzero angular acceleration.
• For this topic, focus on the rotational form of Newton's first law: constant angular velocity requires zero net torque.

Applying these principles to rotating systems requires defining a clear axis of rotation.

• The axis of rotation serves as the reference point for all torque calculations
• In AP Physics 1, problems focus on fixed-axis rotation
• The moment of inertia depends on the mass distribution relative to the axis

Practice Problem 1: Rotational Equilibrium

To solve this problem, we need to find the position where the net torque about the pivot is zero.

Let's establish a coordinate system with the pivot at the origin, positive direction to the right.

Step 1: Calculate the torque due to the 25 kg child. Position of child from pivot = 1.5 m - 2.5 m = -1.0 m (left of pivot)

Torque = mass × gravity × lever arm = 25 kg × 9.8 m/s² × (-1.0 m) = -245 N·m

Step 2: The seesaw's center of mass is at the pivot, so it creates no torque.

Step 3: Find where the 40 kg child should sit. Let x = distance from pivot (positive to the right) Torque due to second child = 40 kg × 9.8 m/s² × x = 392x N·m

Step 4: For equilibrium, sum of torques equals zero. -245 N·m + 392x N·m = 0 392x = 245 x = 0.625 m

The 40 kg child should sit 0.625 m to the right of the pivot.

Practice Problem 2: Zero Net Torque

Step 1: Define the pivot point and coordinate system.

• Pivot at 50 cm mark (x = 0)
• Positions left of pivot are negative, right are positive
• Position of 200 g mass: x = -30 cm = -0.3 m

Step 2: Calculate the torque due to the 200 g mass.

• Torque = mass × gravity × lever arm
• Torque = 0.2 kg × 9.8 m/s² × (-0.3 m) = -0.588 N·m

Step 3: For zero net torque, the 150 g mass must create an equal and opposite torque.

• Let x = position of 150 g mass
• Torque = 0.15 kg × 9.8 m/s² × x = 1.47x N·m
• For equilibrium: -0.588 N·m + 1.47x N·m = 0
• Solving for x: x = 0.4 m = 40 cm from the pivot

The 150 g mass should be placed at the 90 cm mark of the meter stick.

Practice Problem 3: Applying Rotational Equilibrium

Step 1: Draw a free-body diagram. The forces on the beam are:

• The weight of the sign (80 N downward) acting at the midpoint of the beam (1.0 m from the hinge)
• The tension $T$ in the cable acting at the end of the beam (2.0 m from the hinge)
• The hinge force at the wall (which we can ignore by choosing the hinge as our pivot)

Step 2: Take torques about the hinge. Since the hinge force acts at the pivot, it produces zero torque.

• The sign's weight produces a clockwise torque: $\tau_{\text{sign}} = 80 \text{ N} \times 1.0 \text{ m} = 80 \text{ N·m}$
• The cable tension produces a counterclockwise torque. Only the vertical component of the tension ($T \sin 30°$) creates torque about the hinge, with a lever arm of 2.0 m: $\tau_{\text{cable}} = T \sin(30°) \times 2.0 \text{ m} = T(0.5)(2.0) = 1.0T \text{ N·m}$

Step 3: For rotational equilibrium, set $\sum \tau = 0$: $1.0T - 80 = 0$

$T = 80 \text{ N}$

The tension in the cable is 80 N. The sign remains at rest because the net torque about the hinge is zero—this is rotational equilibrium in action. Even though the hinge also exerts a force on the beam, choosing the hinge as the pivot point simplifies the problem since that force contributes no torque. This is a powerful strategy whenever you can place the pivot at the location of an unknown force.