PHYS2250 Chap.7 Rotational Kinematics and Dynamics
Rotational Kinematics and Dynamics
Name the rotation axis before calculating
Angular kinematics relates angle, angular speed and angular acceleration, while torque depends on a perpendicular lever arm and moment of inertia depends on mass distribution.
This chapter uses axis choice to remove unknown reactions, applies the parallel-axis theorem only to parallel axes, and requires both force and torque balance for rigid-body equilibrium.
Reasoning route and exam method
Rotational kinematics mirrors linear kinematics only when the axis is fixed and angular acceleration is treated consistently.
Angular displacement, angular velocity and angular acceleration correspond to displacement, velocity and acceleration; a point at radius r has tangential speed rω and tangential acceleration rα; this relationship belongs in the diagnostic checklist for the chapter. Radial acceleration rω² remains present even when α = 0; this is a useful checkpoint before any numerical substitution.
Dynamics adds the distribution-sensitive quantity moment of inertia I = ∫r²dm; the same result should survive the relevant limiting-case test.
The equation Στ = Iα is about a specified axis, so both the torque and the inertia must refer to the same axis; that distinction determines which quantities may be combined.
A force through the chosen origin has zero moment about that origin; this observation fixes the physical interpretation of the algebra.
In statics, choosing a support point can remove an unknown reaction from the torque equation; the statement supplies a direct check on the final result. In dynamics, choosing a fixed hinge often produces the cleanest angular equation, provided the appropriate hinge-axis inertia is used; this condition should remain visible beside the governing equation.
The parallel-axis theorem I = ICM + Md² shifts an inertia from a parallel axis through the centre of mass; it is not valid between nonparallel axes.
A uniform 3.0 kg rod of length 1.2 m is horizontal and hinged at one end. Gravity acts at its centre, so the initial torque magnitude is Mg(L/2) = 3.0(9.8)(0.60) = 17.64 N m. The hinge-axis inertia is (1/3)ML² = (1/3)(3.0)(1.2²) = 1.44 kg m².
Hence α = τ/I = 12.25 rad s⁻² downward. The hinge reaction is absent from the torque equation because its line of action passes through the chosen origin.
Prompt. A body has ICM = 0.40 kg m² and mass 2.0 kg. Find I about a parallel axis 0.30 m away.
TWO BALANCES Rigid-body equilibrium.
Require both ΣF = 0 and Στ = 0. Either condition alone permits acceleration of a different kind.
Torque magnitude equals force times the perpendicular distance from the pivot to the force line of action. Extending the red force line makes the correct lever arm visible without trigonometric guessing.
Torque depends on a force line, an origin and a sign convention.
Omit any one and the number has no mechanical meaning.
What this chapter covers
- 01
angular variables
- 02
torque and lever arm
- 03
moment of inertia
- 04
parallel-axis theorem
- 05
rigid-body equilibrium
Worked example · free
Release a hinged uniform rod
- 2Find gravitational torque about the hinge.
- 2Use the hinge-axis moment of inertia.
- 3Solve the angular equation and give direction.
Key terms
- Torque
- The moment of a force about a stated point or axis.
- Moment Of Inertia
- The integral of squared perpendicular distance over mass.
- Lever Arm
- The perpendicular distance from a torque origin to the force line of action.
Rotational Kinematics and Dynamics FAQ
Why can a hinge reaction disappear from torque?
Its line of action passes through the hinge, giving zero lever arm about that origin. It still appears in the force balance when translational acceleration or support loading is required.
When is the parallel-axis theorem valid?
When shifting between parallel axes, one through the centre of mass and one a distance d away; it does not rotate the axis into a new spatial direction or change the body's mass distribution.
Is zero net torque enough for equilibrium?
No. Rigid-body equilibrium also requires zero net force; otherwise the body can translate without angular acceleration even though its orientation remains momentarily unchanged in the selected inertial frame.
Exam move
Annotate every inertia formula with its axis. Practise choosing torque origins that eliminate unknown reactions, then use force balance to recover those reactions if asked. Build an axis catalogue for a rod, disk, ring and sphere, drawing the axis through the object rather than listing formulas alone.
Derive one entry directly from the mass integral, then shift it with the parallel-axis theorem and confirm the added term has correct dimensions. In equilibrium questions, take moments about two different points and verify that both routes give the same support forces. In dynamic questions, separate tangential acceleration from radial acceleration of the centre of mass.
Check the sign of angular acceleration against the torque sketch and state whether a negative result reverses the assumed sense or indicates an inconsistent geometry. Close by deriving one angular result from linear motion at radius r, keeping the translation-rotation analogy tied to geometry instead of an isolated formula table.
Working through Rotational Kinematics and Dynamics in PHYS2250? Sia is AskSia’s AI Physics tutor — ask any PHYS2250 Rotational Kinematics and Dynamics question and get a clear, step-by-step explanation grounded in how PHYS2250 is taught and assessed. Read this chapter free, then take your hardest questions to Sia.
