Question

A billiard ball (of radius R), initially at rest is given a harp impulse by a cue. The cue is held horizontally a distance h above the central line. The ball leaves the cue with a speed $V_0$, eventually acquires a final speed of $\frac{9}{7}{V}_0$. Find h.

• A. $\frac{7R}{13}$
• B. $\frac{2R}{13}$
• C. $\frac{4R}{13}$
• D. None of these

Answer 1

The correct option is C

$\frac{4R}{13}$

Let $v_0$ be the linear velocity and $w_0$ be the angular velocity imparted.

$\Rightarrow$ linear impulse = $intFdt=m{v}_0$

Angular impulse=$\int Fhdt=I{\omega }_0\Rightarrow h=\frac{\frac{2}{5}m{R}^2{\omega }_0}{m{v}_0}\Rightarrow v_0=\frac{2R^2{\omega }_0}{5h}$ .....(1)

As linear velocity increases to $\frac{9}{7}{v}_0$, friction must be in forward direction & hence opposes angular motion.

For linear motion: friction $\left(\mu mgR\right)$ increases velocity. $(a=\mu g)$ (t = time taken to start rolling)

For rotation: frictional torque $\left(\mu mgR\right)$ opposes rotation & hence decreases

$\omega .\alpha =\frac{-\mu mgR}{I}\Rightarrow \omega ={\omega }_0-\frac{\mu mgR}{\left(\frac{2}{5}m{R}^2\right)}t$

After time t $V=\frac{9}{7}{v}_0=R\omega$

combining all the equation $\frac{9}{7}{v}_0-v_0=-(\frac{9v_0}{7R}-{\omega }_0)\frac{2R}{5}$ (eliminating $\mu gt$)

substituting for $v_0$ from $I.\frac{2}{7}\left(\frac{2}{5}\frac{R^2{\omega }_0}{h}\right)=({\omega }_0-\frac{9}{7}\frac{2R{\omega }_0}{5h})\frac{2R}{5}\Rightarrow h=\frac{4R}{5}$