Worked exercise — sphere rolling down a slope

Problem

A uniform sphere ($m = 5\;\mathrm{kg}$, $R = 2\;\mathrm{m}$, $I = \tfrac{2}{5}mR^2$) starts from rest at the top of an incline of height $h = 3\;\mathrm{m}$ and rolls without slipping. Find the speed at the bottom.

Solution

Without slipping, friction is static and dissipates no energy: mechanical energy is conserved. The kinetic energy at the bottom, however, has two contributions, translational and rotational. $mgh = \tfrac{1}{2}mv^2 + \tfrac{1}{2}I\omega^2$ Rolling constraint $\omega = v/R$; substituting $I = \tfrac{2}{5}mR^2$: $\tfrac{1}{2}I\omega^2 = \tfrac{1}{2}\cdot\tfrac{2}{5}mR^2\cdot\frac{v^2}{R^2} = \tfrac{1}{5}mv^2$ The balance becomes: $\tfrac{1}{2}mv^2 + \tfrac{1}{5}mv^2 = mgh \quad\Rightarrow\quad \tfrac{7}{10}v^2 = gh$ $v = \sqrt{\frac{10\,gh}{7}} = \sqrt{\frac{10\cdot 9{,}81\cdot 3}{7}} \approx \ev{6{,}5\;\mathrm{m/s}}$ This is slower than a block sliding without friction from the same height ($v = \sqrt{2gh} = 7{,}7\;\mathrm{m/s}$): part of the potential energy has ended up in rotation instead of translation.

Connections

Topics: Dinamica rotazionale Concepts: Rotolamento · Energia cinetica rotazionale · Conservazione dell’energia meccanica Objects: Sfera che rotola · Piano inclinato