The Kinetic Energy of Rolling — Question 10

Question 10

A measured energy fraction gives an inertia ratio directly

In pure rolling, an unknown wheel stores $30\%$ of its total kinetic energy in rotation. Its mass is $5\,\mathrm{kg}$, radius $0.20\,\mathrm m$ and total kinetic energy $35\,\mathrm J$.

1. Infer $I/(MR^2)$ and center speed.

2. Find its angular speed and verify both energy contributions.

Question 10 - Solution

A measured energy fraction gives an inertia ratio directly

(a) Derivation and calculation

The rotational fraction obeys $$f=\frac{\beta}{1+\beta}=.30\Rightarrow \boxed{\beta=\frac{.30}{.70}=\frac37}.$$ The translational energy is $.70(35)=24.5$ J, so $$\boxed{V=\sqrt{\frac{2(24.5)}5}=\sqrt{9.8}=3.130\ \mathrm{m/s}}.$$

(b) Result and interpretation

The angular speed and inertia are $$\boxed{|\omega|=V/R=\sqrt{9.8}/.20=15.65\ \mathrm{rad/s}},$$ $$\boxed{I=\frac37(5)(.20)^2=\frac3{35}\ \mathrm{kg\,m^2}}.$$ Substitution verifies $$K_{\mathrm{rot}}=\tfrac12(3/35)(245)=10.5\ \mathrm J,\quad K_{\mathrm{trans}}=24.5\ \mathrm J,\quad K=35.0\ \mathrm J.$$

Physical check

The inferred $\beta=3/7$ lies within the bound for mass inside the outer radius.