Kinetic Energy of Rotation — Question 1

Question 1

Summing particle energies produces rotational inertia

Two small masses, $2\,\mathrm{kg}$ at radius $0.30\,\mathrm m$ and $1\,\mathrm{kg}$ at radius $0.60\,\mathrm m$, are rigidly joined to a light shaft assembly rotating at $5\,\mathrm{rad/s}$.

1. Find total kinetic energy by summing particle energies and infer the assembly inertia.

2. What fraction of the energy belongs to the lighter mass?

Question 1 - Solution

Summing particle energies produces rotational inertia

(a) Derivation and calculation

Each particle has speed $v_i=r_i\omega$: $$v_1=0.30(5)=1.50,\qquad v_2=0.60(5)=3.00\quad\mathrm{m/s}.$$ $$K_1=\tfrac12(2)(1.50)^2=2.25\ \mathrm J,\quad K_2=\tfrac12(1)(3.00)^2=4.50\ \mathrm J,\quad\boxed{K=6.75\ \mathrm J}.$$ $$\boxed{I=\sum m_ir_i^2=2(0.30)^2+1(0.60)^2=0.54\ \mathrm{kg\,m^2}}.$$

(b) Result and interpretation

The lighter particle’s energy fraction is $$\boxed{\frac{K_2}{K}=\frac{4.50}{6.75}=\frac23}.$$ Its contribution relative to the heavier particle is $K_2/K_1=(1/2)(0.60/0.30)^2=2$; squared radius outweighs the mass reduction.

Physical check

The aggregate formula gives $\tfrac12(0.54)(5)^2=6.75$ J, equal to the particle sum.