Worked exercise — optimal height for the cue strike

Problem

A billiard ball (solid sphere, radius $r$, mass $m$) is struck by the cue with a horizontal impulse $\vec{I}$. At what height $b$ from the table must it be struck so that the ball rolls without slipping from the very start?

Solution

Translational impulse: the impulse $I_x$ gives the centre of mass a momentum $m\,v_\text{CM} = I_x$ Angular impulse about the contact point P: for the ball to roll, it must have an angular momentum about P that is consistent. Decomposing the angular momentum about P as spin plus the moment of the CM: $\tfrac{2}{5}mr^2\,\omega + m\,v_\text{CM}\,r = I_x\cdot b$ Rolling condition $v_\text{CM} = \omega r$ and $m v_\text{CM} = I_x$; substituting: $\tfrac{2}{5}mr^2\cdot\frac{v_\text{CM}}{r} + m\,v_\text{CM}\,r = m\,v_\text{CM}\cdot b$ $\tfrac{2}{5}r + r = b \quad\Rightarrow\quad \ev{b = \tfrac{7}{5}\,r}$ The ball must be struck at $7/5$ of the radius from the table, i.e. slightly above the centre (which is at height $r$). Striking exactly there, the translational and angular impulses match and the ball sets off in pure rolling, without slipping.

Links

Topics: Rotational dynamics Concepts: Rolling · Angular momentum · Impulse Objects: Rolling sphere