PHYS2250 Chap.8 Angular Momentum and Rolling Motion
Angular Momentum and Rolling Motion
Use the origin and no-slip condition deliberately
Angular momentum is measured about a chosen origin, allowing an impact calculation to ignore an impulse whose line passes through that origin. Pure rolling combines centre translation with rotation and enforces a no-slip relation.
The chapter compares point speeds on a wheel and partitions kinetic energy into translational and rotational parts without treating slower descent as energy loss.
Reasoning route and exam method
Angular momentum about an origin is L = r × p for a particle.
For a rigid body rotating about a fixed principal axis it reduces to L = Iω along that axis; this relationship belongs in the diagnostic checklist for the chapter. The general conservation statement follows from external torque: dL/dt = τext; this is a useful checkpoint before any numerical substitution.
Conservation therefore depends on the chosen origin and on external torque about it; the same result should survive the relevant limiting-case test.
An impulsive force through a pivot can change linear momentum while delivering zero angular impulse about that pivot, which is why angular momentum about the hinge is the useful conserved quantity in many impact problems; that distinction determines which quantities may be combined.
For pure rolling vCM = Rω and aCM = Rα along the surface; this observation fixes the physical interpretation of the algebra.
Static friction may be required to enforce that relation, but it does no work at an instantaneous contact point that is at rest on a fixed surface; the statement supplies a direct check on the final result. Total kinetic energy is ½MvCM² + ½ICMω².
Which object reaches the bottom of an incline first depends on I/(MR²): more energy locked in rotation leaves less translational speed for the same height loss; this condition should remain visible beside the governing equation.
A solid cylinder rolls without slipping down through height 1.8 m. With ICM = ½MR² and ω = v/R, energy gives Mgh = ½Mv² + ¼Mv² = ¾Mv². Therefore v = √(4gh/3) = √[4(9.8)(1.8)/3] = 4.85 m s⁻¹.
A sliding particle from the same height would have √(2gh) = 5.94 m s⁻¹. The difference is not energy loss; part of the same conserved energy is rotational.
Prompt. A wheel rolls right with centre speed 3 m s⁻¹. Find the ground-frame speeds of its top and contact points.
Answer. Its lever arm about the hinge is zero, so its angular impulse about that point is zero.
ABOUT WHICH POINT? Origin audit.
State the origin in every angular-momentum equation. A quantity conserved about one point need not be conserved about another.
Pure rolling combines translation of the centre with rotation about the centre.
The top point moves at twice the centre speed while the instantaneous contact point is at rest relative to the ground.
Conserve angular momentum about the point where unknown impulsive forces have zero moment.
What this chapter covers
- 01
particle angular momentum
- 02
angular impulse
- 03
conservation about an origin
- 04
pure rolling
- 05
translation and rotation energy
Worked example · free
Find the speed of a rolling cylinder
- 2Write both kinetic-energy terms.
- 2Apply no slip and conserve energy.
- 2Calculate and compare with sliding.
Key terms
- Angular Impulse
- The time integral of external torque, equal to angular-momentum change.
- Pure Rolling
- Motion satisfying the no-slip relation between centre speed and angular speed.
- No-Slip Condition
- The kinematic relation equating centre speed with radius times angular speed during pure rolling.
Angular Momentum and Rolling Motion FAQ
Why state the origin for angular momentum?
External torque and angular momentum depend on that origin, so conservation may hold about one point but not another during the same physical interaction and time interval.
What is the speed of the contact point in pure rolling?
It is instantaneously zero relative to the fixed ground, while the top point moves at twice the centre speed because translational and rotational velocities add there.
Does static friction dissipate rolling energy?
Not in ideal pure rolling on a fixed surface because the instantaneous contact point has no displacement relative to the surface, despite static friction possibly being nonzero.
Exam move
Decompose point velocity into translation plus rotation and mark the chosen angular-momentum origin. Use energy and torque methods on the same rolling case to see what each reveals. Draw velocities at the top, centre, side and contact point of a rolling wheel and reconstruct each from centre translation plus rotation.
For impulsive problems, list the large unknown forces and choose an origin through their common line of action before testing conservation. Compare a hoop, cylinder and sphere descending the same height, expressing their speeds through the dimensionless ratio I divided by MR squared. Then solve one rolling incline with force and torque equations to recover friction and acceleration.
Energy alone gives speed efficiently, but the second method reveals the contact force and verifies the no-slip condition. Calculate every external angular impulse about the selected origin once, so the omitted hinge impulse is a demonstrated zero rather than an unexplained disappearance.
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