Question 11.10: A billiard ball of massM and radius R is pushed by a cue so ...

Since the momentum direction passes through the center of gravity, the angular momentum with respect to the center of gravity at the time t = 0 equals zero. The friction force f points opposite to the direction of motion (see Fig. 11.18) and causes a torque about the center of gravity

D_{s} = f · R = μMgR.                                    (11.33)

The result is an angular acceleration of the ball, so that

\dot{ω} = \frac{μMgR}{\Theta_{s}}= \frac{μMgR}{(2/5)MR^{2}}= \frac{5}{2} \frac{μg}{R}.                             (11.34)

Moreover, the friction force causes a deceleration of the center of gravity, i.e.,

Ma_{s} =−f     or      as =− \frac{f}{M} =−\frac{μgM}{M}.                                 (11.35)

a_{s} is the acceleration of the center of gravity.

For the rotation velocity of the ball, one gets from (11.34)—after performing the integration—

ω =\int\limits_{0}^{t}{\dot{ω} dt} = \frac{5}{2} \frac{μg}{R}t.                                     (11.36)

The linear velocity of the center of gravity follows from (11.35)—again after integration—as

v_{s} =\int{a_{s}  dt} = v_{0} −μgt .                                 (11.37)

The billiard ball begins rolling when v_{s} = ωR, or

\frac{5}{2}μ \frac{g}{R} t R = v_{0} −gμt                                    (11.38)

or when

v_{0} = \frac{7}{2} μgt        and              t = \frac{2}{7} \frac{v_{0}}{μg}.                                     (11.39)

The distance passed before rolling starts is obtained—by integrating (11.37)—as

s =\int\limits_{0}^{t}{v_{s}  dt} = v_{0}t − \frac{μgt^{2}}{2}                                              (11.40)

and with t from (11.39) finally as

s = \frac{2}{7} \frac{v^{2}_{0}}{μg} − \frac{v^{2}_{0}}{2μg} \left(\frac{2}{7}\right)^{2}= \frac{12}{49} \frac{v^{2}_{0}}{μg}.                                     (11.41)

If the ball is kicked at a distance h above the center of gravity, besides the linear motion there appears a rotational motion with the angular velocity

ω = \frac{Mv_{0}h}{\Theta} = \frac{5}{2} \frac{v_{0}h}{R^{2}} .                                                       (11.42)

If h = (2/5)R, the rolling motion of the ball starts immediately. For h < (2/5)R, one has ω<v_{0}/R, and for h > (2/5)R correspondingly ω>v_{0}/R; in the second case the friction force points forward.

Figure 11.19 shows the change of v_{s}   and   ωR as a function of time for h = 0. If v_{s} = ωR, the rolling motion begins, the friction vanishes, and then v_{s}   and  ω remain constant.