The Kinetic Energy of Rolling — Question 2

Question 2

A measured downhill speed identifies mass distribution

A body with $I=\beta MR^2$ rolls from rest through vertical drop $h=0.80\,\mathrm m$ without slipping or dissipating energy. Its final speed is $3.20\,\mathrm{m/s}$. Use $g=9.8\,\mathrm{m/s^2}$.

1. Infer $\beta$.

2. Find the fraction of final energy in rotation and test consistency with mass confined within radius $R$.

Question 2 - Solution

A measured downhill speed identifies mass distribution

(a) Derivation and calculation

Conservation of energy gives $$Mgh=\tfrac12M(1+\beta)V^2.$$ Solving for the dimensionless inertia, $$\boxed{\beta=\frac{2gh}{V^2}-1=\frac{2(9.8)(.80)}{3.20^2}-1=.53125=\frac{17}{32}}.$$

(b) Result and interpretation

The rotational fraction is $$\boxed{\frac{K_{\mathrm{rot}}}K=\frac{\beta}{1+\beta}=\frac{17}{49}=.347}.$$ For all mass at radii $r\le R$, $$0\le I=\int r^2\,dm\le MR^2\Rightarrow0\le\beta\le1.$$ The inferred value is physically admissible under the stated mass-distribution constraint.

Physical check

The remaining fraction $32/49$ is translational energy; the two fractions sum to one.