Why does pulling mass inward increase angular speed?
Why does pulling mass inward increase angular speed?
- Record angular speed at r=2 m.
- Halve radius without changing angular momentum.
- Compare the kinetic-energy ratio.
What to measure
I in kg m²; L in kg m²/s; ω in rad/s.
The idea underneath
With no external torque, angular momentum remains constant. A point mass on a circular track has I=mr², so decreasing radius requires a larger angular speed. Rotational kinetic energy increases during inward motion because the mechanism pulling the mass does work. Conservation of angular momentum does not require kinetic energy to stay constant while the radius is changed. The laboratory compares equilibrium circular states rather than modelling the radial actuator.
I=mr² · L=Iω · K=L²/(2I)I in kg m²; L in kg m²/s; ω in rad/s.
Build the argument
- Draw the forces and choose coordinates before writing a differential equation.
- Check the result against energy conservation and dimensional consistency.
- Change one parameter at a time and compare a ratio rather than relying on visual impressions.
Worked example 1
L=1.5, m=1 kg, r=1.5 m gives ω=0.667 rad/s.
Worked example 2
Halving radius multiplies both ω and K by four.
Applications
Oscillators appear in clocks, vibration isolation, sensors and mechanical structures. Conserved quantities make complicated motion easier to predict.
Common misconception
A force changes velocity, not just speed. A massless rod or ideal spring is a model assumption, and small-angle pendulum formulas have a limited range.
A further investigation
Explain where the extra kinetic energy comes from.
Check your understanding
At fixed L, halving radius multiplies ω by…
- 4
- 2
- 1/2
Explanation
I scales as r², so ω=L/I increases fourfold. The actuator supplies work.
Model limits
A point mass on a prescribed circular track; radius changes select new ideal states.