The Kinetic Energy of Rolling — Question 3

Question 3

A rolling climb stores both kinds of kinetic energy

A solid sphere rolls up a fixed rough ramp with initial center speed $5\,\mathrm{m/s}$. It maintains pure rolling; there is no dissipation. Use $I=2MR^2/5$ and $g=9.8\,\mathrm{m/s^2}$.

1. Find its maximum vertical rise.

2. Compare with the rise after entering a frictionless ramp smoothly with the same initial translation and spin.

Question 3 - Solution

A rolling climb stores both kinds of kinetic energy

(a) Derivation and calculation

For a rolling solid sphere, $$K_i=\tfrac12MV^2+\tfrac12\left(\frac25MR^2\right)\frac{V^2}{R^2}=\frac7{10}MV^2.$$ Both translation and spin vanish at the rolling turning point: $$Mgh=K_i\Rightarrow\boxed{h=\frac{7(5)^2}{10(9.8)}=\frac{25}{14}=1.786\ \mathrm m}.$$

(b) Result and interpretation

On the smooth straight ramp, $\tau_C=0$, hence $\omega$ and $K_{\mathrm{rot}}$ remain constant. Only translation changes into potential energy: $$Mgh_{\mathrm{smooth}}=\tfrac12MV^2,$$ $$\boxed{h_{\mathrm{smooth}}=\frac{25}{19.6}=\frac{125}{98}=1.276\ \mathrm m}.$$ Energy remains conserved; the unreleased spin energy accounts for the smaller height.

Physical check

The height ratio is $h/h_{\mathrm{smooth}}=7/5$, equal to the initial total-to-translational energy ratio.