Rigid Body Mechanics

Introduction

Up to this point in our mechanics journey, we have treated everyday objects as simple, localized point masses. We assumed that if you push an object, it moves in a straight line. But consider an actual, physical object in your classroom—like a long wooden meter ruler or a heavy door. If you apply a force to the very center of a door, it pushes flat. If you apply that same force far away from the hinge, the door swings open in an arc. If you push right at the hinge, nothing happens at all.

This introduces us to the mechanics of extended, Rigid Bodies. In this realm, where a force is applied is just as critical as how much force is applied. We transition our understanding from translational motion (moving through space) to rotational motion (spinning around a fixed pivot).

Key Concepts

1. Torque and Rotational Equilibrium

A force applied to a rigid body can cause it to rotate about a pivot axis. The turning effectiveness of this force is defined as Torque (τ).

τ = F * r * sin(θ)

Where: τ = torque (Newton-metres, N m), F = applied force (N), r = distance from the pivot to the point of force application (m), and θ = the angle between the force vector and the position vector r.

• Maximum Torque (θ = 90°): Occurs when the force is applied perfectly perpendicular to the lever arm (e.g., pushing a wrench handle squarely).
• Zero Torque (θ = 0° or 180°): Occurs when the force line passes directly through the pivot point. You cannot open a door by pulling or pushing along its edge toward the hinge.

For an extended rigid body to be in absolute Mechanical Equilibrium, it must satisfy two independent structural criteria:

1. Translational Equilibrium: The vector sum of all external forces must be zero (ΣF = 0).
2. Rotational Equilibrium: The vector sum of all external torques must be zero (Στ = 0). This means the total clockwise torques must perfectly balance the total counter-clockwise torques.

2. Moment of Inertia (I)

In linear dynamics, mass ($m$) represents an object’s structural resistance to acceleration. In rotational dynamics, an object’s resistance to angular acceleration depends not only on its total mass, but also on how that mass is distributed relative to the axis of rotation. This property is the Moment of Inertia (I).

For a collection of point masses: I = Σ(m * r²) (Units: kg m²)

• A hoop of mass $M$ and radius $R$ has all its mass concentrated far from the center, giving it a high moment of inertia ($I = M * R²$).
• A solid disk of the same mass and radius has its mass distributed closer to the center, cutting its rotational resistance in half ($I = 0.5 * M * R²$).

3. Newton’s Second Law for Rotation

Just as a net linear force accelerates a mass ($F = ma$), a net external torque causes a rigid body to undergo an angular acceleration ($\alpha$, measured in rad s⁻²).

τ_net = I * α

4. Rotational Kinematics

If a rigid body spins with a completely constant angular acceleration, its motion can be predicted using relationships that mirror our linear suvat equations:

• ω = ω0 + α * t
• θ = ((ω0 + ω) / 2) * t
• θ = ω0 * t + 0.5 * α * t²
• ω² = ω0² + 2 * α * θ

Where: θ = angular displacement (radians, rad), ω0 = initial angular velocity (rad s⁻¹), ω = final angular velocity (rad s⁻¹), α = angular acceleration (rad s⁻²), and t = time elapsed (s).

5. Angular Momentum (L)

• Angular Momentum (L): The rotational analogue of linear momentum. For a rigid body rotating about a fixed axis, it is the product of its moment of inertia and its angular velocity.

L = I * ω (Units: kg m² s⁻¹)

• The Conservation of Angular Momentum: If the net external torque acting on a system is exactly zero, the total angular momentum of the system remains perfectly constant.

I_initial * ω_initial = I_final * ω_final

• The Figure Skater Effect: If a spinning ice skater pulls her arms inward close to her body, she redistributes her mass closer to the rotation axis. This drops her moment of inertia ($I$). To conserve angular momentum ($L = I * ω$), her angular velocity ($ω$) must instantly shoot upward, causing her to spin drastically faster.

6. Rotational Kinetic Energy

A rolling or spinning object possesses kinetic energy locked within its rotational motion.

Ek_rotational = 0.5 * I * ω²

If an object is rolling without slipping down a hill, it transitions its gravitational potential energy into both forms of kinetic energy simultaneously: Total Ek = Ek_translational + Ek_rotational Total Ek = 0.5 * m * v² + 0.5 * I * ω²

Common Mistakes

• Mixing up Newton-metres (N m) with Joules (J): Although torque ($F * r$) has the same fundamental base units as mechanical work ($F * s$), torque is not energy. Torque is a vector tracking a turning effort; work is a scalar tracking energy transfer. Always write torque units as N m, never as Joules.
• Forgetting that r is Measured from the Pivot: When calculating torque, students often use the total length of the object instead of the precise distance from the designated pivot point to where the force arrow lands. Always locate your pivot point first.
• Assuming v = r * ω Applies in Every Scenario: This conversion factor holds true only for points resting exactly on the outer boundary of a rolling object, or when an object rolls completely without slipping. If a wheel spins in place on slick ice, $v$ does not match $r * ω$.

Exam Tips

• Choosing a Clever Pivot Point: In statics problems (like a beam balanced on supports), you can choose any point on the body as your reference pivot to write your $\Sigma\tau = 0$ equation. A great tip is to place your imaginary pivot directly on top of an unknown force line. Since $r = 0$ at that spot, that unknown force creates zero torque and drops out of the algebra completely!
• The Rolling Race Trap: If a solid cylinder and a hollow hoop of the same mass are released from the top of an incline, which wins? Students often guess a tie due to mass. However, the hoop has a larger moment of inertia, meaning it forces more of its available energy pool into spinning rather than moving forward. The solid cylinder splits less energy into rotation, leaving more for forward velocity—meaning the solid cylinder always wins the race.
• Radians Only: Never input degree values into rotational kinematics or power equations. Ensure your calculator is handling angular layouts cleanly and always map angular displacement ($\theta$) in pure radians.

Practice Questions

Question 1 (Multiple Choice)

A uniform horizontal beam of weight W is pinned to a vertical wall at a pivot point P and held stable by a cable attached to its opposite end. The cable exerts a tension force T at an angle of 30° relative to the beam line. Which expression represents the correct clockwise torque balance around pivot P due to the beam’s own weight if the total length of the beam is L?

A. W * L B. W * L * sin(30°) C. W * (L / 2) D. T * L * cos(30°)

Question 2 (Structured Paper 2 Style)

A solid uniform flywheel of mass 8.0 kg and radius 0.25 metres is mounted on a frictionless horizontal axle. The moment of inertia of a solid cylinder is given by the expression I = 0.5 * M * R². A cord wrapped tightly around the rim of the flywheel is pulled with a constant horizontal tension force of 20.0 Newtons.

(a) Calculate the moment of inertia of the flywheel. [2 marks]

(b) Determine the net torque applied to the flywheel and its resulting angular acceleration. [3 marks]

(c) If the flywheel starts completely from rest, calculate its rotational kinetic energy store after a timeline of 4.0 seconds has elapsed. [3 marks]