System of Particles and Rotational Motion · Lesson 12 of 15
Kinematics of Rotation about a Fixed Axis
“Derive and apply angular equations for constant angular acceleration with consistent units and signs.”
• Specify angular position and positive rotation sense. • Derive all three constant-α equations. • Convert rpm to rad/s. • Check the interval over which the acceleration model applies.
One coordinate describes the fixed-axis rotation
With an axis fixed in space and body, angular position θ specifies orientation relative to a chosen reference direction. A full turn adds 2π radians. θ can accumulate through multiple revolutions even though the body eventually repeats its orientation. Choose a positive rotational sense, usually anticlockwise as viewed from the positive axis end. A negative angular velocity means the opposite sense, not a negative angular speed.
Deriving the constant-acceleration equations
Assume α is constant over the entire interval and let ω₀ and θ₀ be the values at t = 0. Integrate dω = α dt from the initial to the final state: ω − ω₀ = αt. This gives the first equation. Integrate dθ = (ω₀ + αt) dt to obtain θ − θ₀ = ω₀t + αt²/2. The initial angle must be retained when the requested quantity is final angular position.
The ω–t graph is a straight line, so its signed trapezium area gives angular displacement Δθ = (ω₀ + ω)t/2. Substitute t = (ω − ω₀)/α, for nonzero α, and use (ω + ω₀)(ω − ω₀) = ω² − ω₀². Rearrangement produces the third relation. The calculus route is α = ω dω/dθ, followed by integrating ω dω = α dθ. Both methods give the same result.
Problem
A motor wheel increases from 1200 rpm to 3120 rpm in 16 s under constant angular acceleration. Find α and revolutions.
- 1.Convert: ω₀ = 1200(2π/60) = 40π rad/s and ω = 3120(2π/60) = 104π rad/s.
- 2.α = (104π − 40π)/16 = 4π rad/s².
- 3.Δθ = (40π + 104π)(16)/2 = 1152π rad. Divide by 2π to obtain 576 revolutions.
- 4.Check that the mean rotation frequency is (20 + 52)/2 = 36 rev/s. In 16 s this also gives 576 revolutions.
Signs, reversals and choosing an equation
Use ω = ω₀ + αt if angle is absent; use Δθ = ω₀t + αt²/2 when final velocity is absent; use the squared equation if time is absent. A wheel slowing in the positive direction has negative α. If that acceleration continues after it stops, it reverses, so angular distance and signed angular displacement differ. Split at the stopping instant when calculating total turns travelled. A constant braking model normally ends when the wheel comes to rest.
Problem
A wheel initially has ω₀ = 12 rad/s and α = −3 rad/s² until it stops.
- 1.Set final ω = 0, giving t = 4 s.
- 2.Δθ = 12(4) + (−3)(4²)/2 = 24 rad, about 3.82 revolutions.
- 3.If the initial angle were 5 rad, its final accumulated angle would be 29 rad. Position and displacement are different quantities.
Quiz
One revolution equals how many radians?
Radians are arc length divided by radius; a circumference gives 2π.
1200 rpm equals which angular speed?
1200/60 = 20 revolutions per second, each worth 2π radians.
The standard three angular equations require what?
They are kinematic consequences of constant α.
Area under ω–t represents what?
Integrating angular velocity accumulates angular position change.
A positive-turning wheel has negative α. Initially it does what?
Its positive angular velocity decreases.
Key Takeaways
• Fixed-axis rotation can be described by one signed angular coordinate. • Derive angular equations by integrating constant α. • Convert rpm before using SI dynamical formulas. • Keep initial angle, direction and reversal boundaries explicit.